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sales-analysis-iii | EASY PANDAS SOLLUTION, BEATS 99% | easy-pandas-sollution-beats-99-by-grzego-pvmg | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem seems to involve analyzing sales data and filtering it based on specific cr | grzegorzgatkowski | NORMAL | 2023-08-24T05:55:19.024726+00:00 | 2023-08-24T05:55:19.024755+00:00 | 920 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem seems to involve analyzing sales data and filtering it based on specific criteria. We are given two DataFrames: one containing product information and another containing sales data. The goal appears to be to filter the sales data to include only records that fall within a certain date range (January 1, 2019, to March 31, 2019) and then merge this filtered data with the product information to get a final DataFrame.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. First, we need to organize the sales data so we can work with it more easily. To do this, we group the sales data by each product (using \'product_id\') and find the earliest (\'min\') and latest (\'max\') sale dates for each product.\n2. Next, we need to filter the sales data to include only the records where the sale date falls between January 1, 2019, and March 31, 2019. This step ensures we\'re only looking at sales within our target date range.\n3. Now, we have a list of products that were sold during the specific time frame. We want to know the names of these products too, not just their IDs. To do this, we merge (combine) the filtered sales data with the product information. We match them using the \'product_id\', and we keep only the \'product_id\' and \'product_name\' for our final result.\n4. Return the resulting DataFrame as the answer.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nOverall time complexity is O(n).\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nimport pandas as pd\n\ndef sales_analysis(product: pd.DataFrame, sales: pd.DataFrame) -> pd.DataFrame:\n # Group the \'sales\' DataFrame by \'product_id\' and calculate the minimum and maximum sale dates for each product\n sales = sales.groupby(\'product_id\')[\'sale_date\'].agg([\'min\', \'max\']).reset_index()\n \n # Filter the sales data to include only records with sale dates between January 1, 2019, and March 31, 2019\n sales = sales[(sales[\'min\'] >= \'2019-01-01\') & (sales[\'max\'] <= \'2019-03-31\')]\n \n # Merge the filtered sales data with the \'product\' DataFrame based on \'product_id\', keeping only \'product_id\' and \'product_name\' columns\n result = pd.merge(sales, product, on=\'product_id\', how=\'inner\')[[\'product_id\', \'product_name\']]\n \n # Return the resulting DataFrame\n return result\n\n``` | 8 | 0 | ['Pandas'] | 4 |
sales-analysis-iii | MySQL, MS SQL and Oracle solution with explanations | mysql-ms-sql-and-oracle-solution-with-ex-u8m6 | We may have 4 cases here:\n1) product has no any sales\n2) product sold in the spring of 2019 only\n3) product sold in other periods, but not in the spring of 2 | vazhev | NORMAL | 2022-04-23T22:48:01.734681+00:00 | 2022-04-23T22:48:01.734712+00:00 | 2,172 | false | We may have 4 cases here:\n1) product has no any sales\n2) product sold in the spring of 2019 only\n3) product sold in other periods, but not in the spring of 2019\n4) product sold in the spring of 2019 and in other periods\n\nThe 2nd case only is correct for the task condition (**only sold in the spring of 2019, between 2019-01-01 and 2019-03-31 inclusive**)\n\nAt the first we have to extract all lines from sales and join them with products, group them by product id and name (the 1st case goes away).\n\nWe may calculate count of sale lines that out of the spring 2019 period:\n```SUM(CASE WHEN sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\' THEN 0 ELSE 1 END)```\nIf this sum is 0, it means that all sales of this product belong to the spring 2019 period (3rd and 4th cases went away)\n\nFinally we have the solution:\n\n\n```\nSELECT\n p.product_id,\n p.product_name\nFROM Sales s\nINNER JOIN Product p ON s.product_id = p.product_id\nGROUP BY\n p.product_id,\n p.product_name\nHAVING\n SUM(CASE WHEN sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\' THEN 0 ELSE 1 END) = 0\n``` | 8 | 0 | ['MySQL', 'Oracle', 'MS SQL Server'] | 4 |
sales-analysis-iii | Using Having, no sub query needed | using-having-no-sub-query-needed-by-jani-6aho | SELECT s.product_id, p.product_name\nFROM Sales s JOIN Product p\nON p.product_id = s.product_id\nGROUP BY s.product_id\nHAVING MIN(s.sale_date) >= \'2019-01-01 | JaniceJ | NORMAL | 2021-06-17T04:41:50.188563+00:00 | 2021-06-17T04:41:50.188610+00:00 | 772 | false | SELECT s.product_id, p.product_name\nFROM Sales s JOIN Product p\nON p.product_id = s.product_id\nGROUP BY s.product_id\nHAVING MIN(s.sale_date) >= \'2019-01-01\' AND MAX(s.sale_date) <=\'2019-03-31 \'\n | 8 | 0 | [] | 0 |
sales-analysis-iii | SQL Beginner: Simple Logic | sql-beginner-simple-logic-by-harshada202-0hu7 | \tselect\n\t\ts.product_id, p.product_name\n\tfrom \n\t\tSales s join Product p\n\ton\n\t\ts.product_id = p.product_id\n\tgroup by\n\t\ts.product_id \n\thaving\ | harshada2022 | NORMAL | 2022-09-19T17:30:37.318911+00:00 | 2022-09-19T17:30:37.318954+00:00 | 2,695 | false | \tselect\n\t\ts.product_id, p.product_name\n\tfrom \n\t\tSales s join Product p\n\ton\n\t\ts.product_id = p.product_id\n\tgroup by\n\t\ts.product_id \n\thaving\n\t\tmin(s.sale_date) >= "2019-01-01" \n\t\tAND\n\t\tmax(s.sale_date) <= "2019-03-31" | 7 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | Easiest Mysql Solution | easiest-mysql-solution-by-sofiaisasavage-0h1r | select \nproduct_id, product_name\nfrom product\nwhere product_id not in\n(select product_id\nfrom sales\nwhere sale_date <\'2019-01-01\'or sale_date>\'2019-03- | sofiaisasavage | NORMAL | 2020-05-31T17:33:58.710033+00:00 | 2020-05-31T17:33:58.710071+00:00 | 860 | false | select \nproduct_id, product_name\nfrom product\nwhere product_id not in\n(select product_id\nfrom sales\nwhere sale_date <\'2019-01-01\'or sale_date>\'2019-03-31\') | 7 | 1 | [] | 2 |
sales-analysis-iii | Beginner-Friendly Solution Using SQL Server || Clear | beginner-friendly-solution-using-sql-ser-vxxq | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | truongtamthanh2004 | NORMAL | 2024-06-01T08:57:52.449491+00:00 | 2024-06-01T08:57:52.449525+00:00 | 798 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n/* Write your T-SQL query statement below */\nselect p.product_id, product_name\nfrom Product p\ninner join Sales sa on sa.product_id = p.product_id\ngroup by p.product_id, product_name\nhaving min(sale_date) >= \'2019-01-01\' and max(sale_date) <= \'2019-03-31\'\n``` | 6 | 0 | ['MS SQL Server'] | 1 |
sales-analysis-iii | SIMPLE MSOLUTION WITH NOT EXISTS ( SQL SERVER + MYSQL ) | simple-msolution-with-not-exists-sql-ser-j50d | Intuition\r\n Describe your first thoughts on how to solve this problem. \r\n\r\n# Approach\r\n Describe your approach to solving the problem. \r\n\r\n# Complex | cooking_Guy_9ice | NORMAL | 2023-03-04T04:29:01.709762+00:00 | 2023-11-21T02:14:35.721661+00:00 | 695 | false | # Intuition\r\n<!-- Describe your first thoughts on how to solve this problem. -->\r\n\r\n# Approach\r\n<!-- Describe your approach to solving the problem. -->\r\n\r\n# Complexity\r\n- Time complexity:\r\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\r\n\r\n- Space complexity:\r\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\r\n\r\n# Code\r\n```\r\n/* Write your T-SQL query statement below */\r\n\r\nSELECT\r\nDISTINCT\r\n P.product_id,\r\n P.product_name\r\nFROM\r\n Product P\r\nINNER JOIN\r\n Sales S ON P.product_id = S.product_id\r\nAND\r\n NOT EXISTS\r\n (\r\n SELECT\r\n 1\r\n FROM\r\n Sales SS\r\n WHERE\r\n SS.product_id = S.product_id\r\n AND\r\n (SS.sale_date < \'2019-01-01\' OR SS.sale_date > \'2019-03-31\') \r\n )\r\n\r\n\r\n\r\n``` | 6 | 0 | ['MySQL', 'MS SQL Server'] | 1 |
sales-analysis-iii | Simplest solution using HAVING | simplest-solution-using-having-by-sukuma-vevo | \nSELECT Product.product_id, Product.product_name\nFROM Product \nINNER JOIN Sales \nON Product.product_id = Sales.product_id \nGROUP BY Sales.product_id \nHAVI | sukumar-satapathy | NORMAL | 2022-09-21T19:35:41.938212+00:00 | 2023-01-30T10:30:15.669300+00:00 | 1,470 | false | ```\nSELECT Product.product_id, Product.product_name\nFROM Product \nINNER JOIN Sales \nON Product.product_id = Sales.product_id \nGROUP BY Sales.product_id \nHAVING MIN(Sales.sale_date) >= "2019-01-01" AND MAX(Sales.sale_date) <= "2019-03-31"\n```\nIf it helped or you learned a new way, kindly upvote. Thanks :) | 6 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | ✅[Accepted] Solution for MySQL | Clean & Simple Code | accepted-solution-for-mysql-clean-simple-7fur | \nSELECT p.product_id, p.product_name\nFROM Product p JOIN Sales s ON p.product_id=s.product_id\nGROUP BY p.product_id\nHAVING MIN(s.sale_date) >= \'2019-01-01\ | axitchandora | NORMAL | 2022-05-24T17:57:28.029248+00:00 | 2022-05-24T17:57:28.029296+00:00 | 802 | false | ```\nSELECT p.product_id, p.product_name\nFROM Product p JOIN Sales s ON p.product_id=s.product_id\nGROUP BY p.product_id\nHAVING MIN(s.sale_date) >= \'2019-01-01\' \nAND MAX(s.sale_date) <= \'2019-03-31\';\n``` | 6 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | simple solution | simple-solution-by-cindy1516-rmhx | \nselect a.product_id, product_name\nfrom Sales a\ninner join Product b\non a.product_id = b.product_id\ngroup by a.product_id\nhaving min(sale_date)>=\'2019-01 | cindy1516 | NORMAL | 2019-06-17T07:20:52.599219+00:00 | 2019-06-17T07:20:52.599290+00:00 | 1,183 | false | ```\nselect a.product_id, product_name\nfrom Sales a\ninner join Product b\non a.product_id = b.product_id\ngroup by a.product_id\nhaving min(sale_date)>=\'2019-01-01\' and max(sale_date)<=\'2019-03-31\'\n;\n``` | 6 | 0 | [] | 0 |
sales-analysis-iii | My solution | my-solution-by-vnsemkin-re3z | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | vnsemkin | NORMAL | 2024-10-05T12:18:04.976956+00:00 | 2024-10-05T12:18:04.976979+00:00 | 741 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```postgresql []\nSELECT p.product_id, p.product_name\nFROM product p JOIN sales s USING(product_id)\nGROUP BY p.product_id, p.product_name\nHAVING MIN(s.sale_date) >= \'2019-01-01\' AND MAX(s.sale_date) <= \'2019-03-31\';\n\n``` | 5 | 0 | ['PostgreSQL'] | 1 |
sales-analysis-iii | Easy to understand || SQL | easy-to-understand-sql-by-yashwardhan24_-h6fv | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | yashwardhan24_sharma | NORMAL | 2023-03-09T06:02:43.227577+00:00 | 2023-03-09T06:02:43.227626+00:00 | 3,455 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect s.product_id, p.product_name\nfrom sales s, product p\nwhere s.product_id = p.product_id\ngroup by s.product_id, p.product_name\nhaving min(s.sale_date) >= \'2019-01-01\' AND max(s.sale_date) <= \'2019-03-31\';\n``` | 5 | 0 | ['Database', 'MySQL'] | 1 |
sales-analysis-iii | EASY MySQL Solution || LEFT JOIN || INNER JOINS | easy-mysql-solution-left-join-inner-join-l322 | \n# Write your MySQL query statement below\nWITH rejected_ids AS (\n SELECT s.product_id\n FROM sales s\n WHERE s.sale_date NOT BETWEEN DATE(\'2019-01- | raghavdabra | NORMAL | 2022-10-18T12:47:30.229295+00:00 | 2022-10-18T12:47:30.229337+00:00 | 4,216 | false | ```\n# Write your MySQL query statement below\nWITH rejected_ids AS (\n SELECT s.product_id\n FROM sales s\n WHERE s.sale_date NOT BETWEEN DATE(\'2019-01-01\') AND DATE(\'2019-03-31\')\n)\nSELECT DISTINCT p.product_id, p.product_name\nFROM product p\nINNER JOIN sales s ON s.product_id = p.product_id\nLEFT JOIN rejected_ids r ON r.product_id = p.product_id\nWHERE s.sale_date BETWEEN DATE(\'2019-01-01\') AND DATE(\'2019-03-31\') AND r.product_id IS NULL\n``` | 5 | 0 | ['MySQL'] | 2 |
sales-analysis-iii | My straight forward solution using [MySQL]MIN | my-straight-forward-solution-using-mysql-l9ru | \nSELECT P.PRODUCT_ID, P.PRODUCT_NAME\n FROM PRODUCT P\nJOIN SALES S\n USING (PRODUCT_ID)\nGROUP BY PRODUCT_ID\n HAVING MIN(S.SALE_DATE) >= \'2019-01-0 | shivamleet07 | NORMAL | 2022-07-30T17:00:00.164905+00:00 | 2022-07-30T17:01:55.309300+00:00 | 1,305 | false | ```\nSELECT P.PRODUCT_ID, P.PRODUCT_NAME\n FROM PRODUCT P\nJOIN SALES S\n USING (PRODUCT_ID)\nGROUP BY PRODUCT_ID\n HAVING MIN(S.SALE_DATE) >= \'2019-01-01\' \n AND \n MAX(S.SALE_DATE) <= \'2019-03-31\'\n``` | 5 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | Don't know the answer ! | dont-know-the-answer-by-gogole-tsye | \n\n\n\nselect distinct p.product_id, product_name \nfrom product p left join \nsales s on p.product_id=s.product_id\ngroup by p.product_id\nhaving min(sale_dat | GOGOLE | NORMAL | 2022-06-21T04:00:45.137968+00:00 | 2022-06-21T04:00:45.138005+00:00 | 386 | false | \n\n\n```\nselect distinct p.product_id, product_name \nfrom product p left join \nsales s on p.product_id=s.product_id\ngroup by p.product_id\nhaving min(sale_date) >= \'2019-01-01\' \n and max(sale_date) <= \'2019-03-31\';\n``` | 5 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | MySQL 2 simple solutions with comment. (Runtime: 846ms, Faster than 99%) | mysql-2-simple-solutions-with-comment-ru-r1kv | Where and Select\nsql\nSELECT\n product_id,\n product_name\nFROM Product\n# get product where only sold in first quarter 2019\nWHERE product_id NOT IN (\n | NovaldiS | NORMAL | 2022-05-20T16:58:28.066475+00:00 | 2022-05-20T16:58:28.066523+00:00 | 493 | false | Where and Select\n```sql\nSELECT\n product_id,\n product_name\nFROM Product\n# get product where only sold in first quarter 2019\nWHERE product_id NOT IN (\n # get all product_id in table Sales Where not sold in first quarter 2019\n SELECT\n product_id\n FROM Sales\n WHERE sale_date < "2019-01-01" OR sale_date > \'2019-03-31\'\n);\n```\nHaving and Quarter\n```sql\nSELECT\n p.product_id,\n p.product_name\nFROM Product AS p\n# join product and sales table\nJOIN Sales AS s ON p.product_id = s.product_id\nGROUP BY p.product_id\n# where in quarter only 1\nHAVING MAX(QUARTER(sale_date)) = 1;\n```\n# Upvote for your support | 5 | 0 | ['MySQL'] | 2 |
sales-analysis-iii | Simple MySQL using HAVING SUM()=0 | simple-mysql-using-having-sum0-by-joyele-dxvi | \nSELECT s.product_id,\nproduct_name\nFROM sales s\nJOIN product p\nON s.product_id=p.product_id\nGROUP BY s.product_id\nHAVING SUM(IF(sale_date BETWEEN \'2019- | joyelee2466 | NORMAL | 2019-12-19T02:55:09.381228+00:00 | 2020-04-28T08:55:18.369609+00:00 | 931 | false | ```\nSELECT s.product_id,\nproduct_name\nFROM sales s\nJOIN product p\nON s.product_id=p.product_id\nGROUP BY s.product_id\nHAVING SUM(IF(sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\',0,1))=0\n``` | 5 | 0 | [] | 5 |
sales-analysis-iii | Simple Solution using CTE | simple-solution-using-cte-by-shalinipm_4-20kw | IntuitionIntuition
The goal is to identify products that were exclusively sold in the first quarter of 2019. The first step is to filter sales occurring within | Shalinipm_416 | NORMAL | 2025-01-03T06:53:57.387808+00:00 | 2025-01-03T06:53:57.387808+00:00 | 1,027 | false | # Intuition
Intuition
The goal is to identify products that were exclusively sold in the first quarter of 2019. The first step is to filter sales occurring within this timeframe, and then exclude products with sales outside this range. The key insight is leveraging SQL constructs like WITH clauses (for Common Table Expressions) and set operations (NOT IN) to achieve this.
# Approach
Step 1: Filter Sales in Q1 2019
Create a temporary list of products sold between 2019-01-01 and 2019-03-31 using a WITH clause. This captures all relevant products during the desired timeframe.
Step 2: Identify Sales Outside Q1 2019
Create another temporary list of products sold before or after Q1 2019. This step ensures we can exclude products that are not exclusive to the Q1 period.
Step 3: Exclude Non-Exclusive Products
Use the two temporary lists to filter products sold only in Q1 2019. This is achieved by joining the first list with the Product table and excluding products present in the second list using NOT IN.
Final Query:
Return the product_id and product_name for products that meet the above criteria.
# Code
```mysql []
WITH FirstQuarterSales AS (
SELECT DISTINCT product_id
FROM Sales
WHERE sale_date BETWEEN '2019-01-01' AND '2019-03-31'
),
OtherSales AS (
SELECT DISTINCT product_id
FROM Sales
WHERE sale_date < '2019-01-01' OR sale_date > '2019-03-31'
)
SELECT p.product_id, p.product_name
FROM Product AS p
JOIN FirstQuarterSales AS fqs
ON p.product_id = fqs.product_id
WHERE p.product_id NOT IN (SELECT product_id FROM OtherSales);
``` | 4 | 0 | ['MySQL'] | 2 |
sales-analysis-iii | Sales Analysis III | Pandas | Easy | sales-analysis-iii-pandas-easy-by-khosiy-ffbs | see the successfully Accepted Submission\n\nimport pandas as pd\n\ndef sales_analysis(product: pd.DataFrame, sales: pd.DataFrame) -> pd.DataFrame:\n # First, | Khosiyat | NORMAL | 2023-09-19T18:31:19.949300+00:00 | 2023-10-01T17:38:52.098932+00:00 | 332 | false | [see the successfully Accepted Submission](https://leetcode.com/submissions/detail/1053850673/)\n```\nimport pandas as pd\n\ndef sales_analysis(product: pd.DataFrame, sales: pd.DataFrame) -> pd.DataFrame:\n # First, we convert \'sale_date\' column to datetime if it\'s not already in datetime format\n sales[\'sale_date\'] = pd.to_datetime(sales[\'sale_date\'])\n\n # Secondly, the \'product\' with \'sales\' based on \'product_id\' are merged\n merged_data = pd.merge(product, sales, on=\'product_id\', how=\'inner\')\n\n # Then, we apply group by for \'product_id\' and apply aggregation functions\n result = merged_data.groupby(\'product_id\').agg(\n min_sale_date=(\'sale_date\', \'min\'),\n max_sale_date=(\'sale_date\', \'max\'),\n product_name=(\'product_name\', \'first\') # Include \'product_name\' in the aggregation\n )\n \n # In the next step, the dates after January (including January) are filtered\n after_january = (result[\'min_sale_date\'] >= \'2019-01-01\')\n \n # At the same time, the dates before March (including March) are also filtered\n before_march = (result[\'max_sale_date\'] <= \'2019-03-31\')\n\n # After that, the results based on date criteria are filtered. The criteria is reporting the products that were only sold in the first quarter of 2019. That is, between 2019-01-01 and 2019-03-31 inclusive.\n filtered_df = result[after_january & before_march]\n\n # Next, the index to make \'product_id\' a regular column is reset.\n new_df = filtered_df.reset_index(inplace=False)\n\n # Finally, we select the desired columns \'product_id\' and \'product_name\'\n quarter_products = new_df[[\'product_id\', \'product_name\']]\n\n return quarter_products\n \n```\n\n**SQL**\n\n[see the successfully Accepted Submission](https://leetcode.com/submissions/detail/1061692619/)\n\n```\nSELECT _sales.product_id, product_name\nFROM Sales _sales \n\nLEFT JOIN Product _product\nON _sales.product_id = _product.product_id\n\nGROUP BY _sales.product_id\nHAVING MIN(sale_date) >= CAST(\'2019-01-01\' AS DATE) \nAND\nMAX(sale_date) <= CAST(\'2019-03-31\' AS DATE);\n```\n\n```\n-- Select the product_id and product_name columns\nSELECT _sales.product_id, product_name\n-- From the Sales table, alias it as "_sales"\nFROM Sales _sales \n\n-- Perform a LEFT JOIN with the Product table, alias it as "_product"\nLEFT JOIN Product _product\n-- Match records where the product_id in Sales matches the product_id in Product\nON _sales.product_id = _product.product_id\n\n-- Group the result set by the product_id from Sales\nGROUP BY _sales.product_id\n\n-- Filter the grouped results based on certain conditions\n-- Keep groups where the minimum sale_date is greater than or equal to \'2019-01-01\'\n-- AND where the maximum sale_date is less than or equal to \'2019-03-31\'\nHAVING MIN(sale_date) >= CAST(\'2019-01-01\' AS DATE) \nAND\nMAX(sale_date) <= CAST(\'2019-03-31\' AS DATE);\n```\n\n\n\n# Pandas & SQL | SOLVED & EXPLAINED LIST\n\n**EASY**\n\n- [Combine Two Tables](https://leetcode.com/problems/combine-two-tables/solutions/4051076/pandas-sql-easy-combine-two-tables/)\n\n- [Employees Earning More Than Their Managers](https://leetcode.com/problems/employees-earning-more-than-their-managers/solutions/4051991/pandas-sql-easy-employees-earning-more-than-their-managers/)\n\n- [Duplicate Emails](https://leetcode.com/problems/duplicate-emails/solutions/4055225/pandas-easy-duplicate-emails/)\n\n- [Customers Who Never Order](https://leetcode.com/problems/customers-who-never-order/solutions/4055429/pandas-sql-easy-customers-who-never-order-easy-explained/)\n\n- [Delete Duplicate Emails](https://leetcode.com/problems/delete-duplicate-emails/solutions/4055572/pandas-sql-easy-delete-duplicate-emails-easy/)\n- [Rising Temperature](https://leetcode.com/problems/rising-temperature/solutions/4056328/pandas-sql-easy-rising-temperature-easy/)\n\n- [ Game Play Analysis I](https://leetcode.com/problems/game-play-analysis-i/solutions/4056422/pandas-sql-easy-game-play-analysis-i-easy/)\n\n- [Find Customer Referee](https://leetcode.com/problems/find-customer-referee/solutions/4056516/pandas-sql-easy-find-customer-referee-easy/)\n\n- [Classes More Than 5 Students](https://leetcode.com/problems/classes-more-than-5-students/solutions/4058381/pandas-sql-easy-classes-more-than-5-students-easy/)\n- [Employee Bonus](https://leetcode.com/problems/employee-bonus/solutions/4058430/pandas-sql-easy-employee-bonus/)\n\n- [Sales Person](https://leetcode.com/problems/sales-person/solutions/4058824/pandas-sql-easy-sales-person/)\n\n- [Biggest Single Number](https://leetcode.com/problems/biggest-single-number/solutions/4063950/pandas-sql-easy-biggest-single-number/)\n\n- [Not Boring Movies](https://leetcode.com/problems/not-boring-movies/solutions/4065350/pandas-sql-easy-not-boring-movies/)\n- [Swap Salary](https://leetcode.com/problems/swap-salary/solutions/4065423/pandas-sql-easy-swap-salary/)\n\n- [Actors & Directors Cooperated min Three Times](https://leetcode.com/problems/actors-and-directors-who-cooperated-at-least-three-times/solutions/4065511/pandas-sql-easy-explained-step-by-step-actors-directors-cooperated-min-three-times/)\n\n- [Product Sales Analysis I](https://leetcode.com/problems/product-sales-analysis-i/solutions/4065545/pandas-sql-easy-product-sales-analysis-i/)\n\n- [Project Employees I](https://leetcode.com/problems/project-employees-i/solutions/4065635/pandas-sql-easy-project-employees-i/)\n- [Sales Analysis III](https://leetcode.com/problems/sales-analysis-iii/solutions/4065755/sales-analysis-iii-pandas-easy/)\n\n- [Reformat Department Table](https://leetcode.com/problems/reformat-department-table/solutions/4066153/pandas-sql-easy-explained-step-by-step-reformat-department-table/)\n\n- [Top Travellers](https://leetcode.com/problems/top-travellers/solutions/4066252/top-travellers-pandas-easy-eaxplained-step-by-step/)\n\n- [Replace Employee ID With The Unique Identifier](https://leetcode.com/problems/replace-employee-id-with-the-unique-identifier/solutions/4066321/pandas-sql-easy-replace-employee-id-with-the-unique-identifier/)\n- [Group Sold Products By The Date](https://leetcode.com/problems/group-sold-products-by-the-date/solutions/4066344/pandas-sql-easy-explained-step-by-step-group-sold-products-by-the-date/)\n\n- [Customer Placing the Largest Number of Orders](https://leetcode.com/problems/customer-placing-the-largest-number-of-orders/solutions/4068822/pandas-sql-easy-explained-step-by-step-customer-placing-the-largest-number-of-orders/)\n\n- [Article Views I](https://leetcode.com/problems/article-views-i/solutions/4069777/pandas-sql-easy-article-views-i/)\n\n- [User Activity for the Past 30 Days I](https://leetcode.com/problems/user-activity-for-the-past-30-days-i/solutions/4069797/pandas-sql-easy-user-activity-for-the-past-30-days-i/)\n\n- [Find Users With Valid E-Mails](https://leetcode.com/problems/find-users-with-valid-e-mails/solutions/4069810/pandas-sql-easy-find-users-with-valid-e-mails/)\n\n- [Patients With a Condition](https://leetcode.com/problems/patients-with-a-condition/solutions/4069817/pandas-sql-easy-patients-with-a-condition/)\n\n- [Customer Who Visited but Did Not Make Any Transactions](https://leetcode.com/problems/customer-who-visited-but-did-not-make-any-transactions/solutions/4072542/pandas-sql-easy-customer-who-visited-but-did-not-make-any-transactions/)\n\n\n- [Bank Account Summary II](https://leetcode.com/problems/bank-account-summary-ii/solutions/4072569/pandas-sql-easy-bank-account-summary-ii/)\n\n- [Invalid Tweets](https://leetcode.com/problems/invalid-tweets/solutions/4072599/pandas-sql-easy-invalid-tweets/)\n\n- [Daily Leads and Partners](https://leetcode.com/problems/daily-leads-and-partners/solutions/4072981/pandas-sql-easy-daily-leads-and-partners/)\n\n- [Recyclable and Low Fat Products](https://leetcode.com/problems/recyclable-and-low-fat-products/solutions/4073003/pandas-sql-easy-recyclable-and-low-fat-products/)\n\n- [Rearrange Products Table](https://leetcode.com/problems/rearrange-products-table/solutions/4073028/pandas-sql-easy-rearrange-products-table/)\n- [Calculate Special Bonus](https://leetcode.com/problems/calculate-special-bonus/solutions/4073595/pandas-sql-easy-calculate-special-bonus/)\n\n- [Count Unique Subjects](https://leetcode.com/problems/number-of-unique-subjects-taught-by-each-teacher/solutions/4073666/pandas-sql-easy-count-unique-subjects/)\n\n- [Count Unique Subjects](https://leetcode.com/problems/number-of-unique-subjects-taught-by-each-teacher/solutions/4073666/pandas-sql-easy-count-unique-subjects/)\n\n- [Fix Names in a Table](https://leetcode.com/problems/fix-names-in-a-table/solutions/4073695/pandas-sql-easy-fix-names-in-a-table/)\n- [Primary Department for Each Employee](https://leetcode.com/problems/primary-department-for-each-employee/solutions/4076183/pandas-sql-easy-primary-department-for-each-employee/)\n\n- [The Latest Login in 2020](https://leetcode.com/problems/the-latest-login-in-2020/solutions/4076240/pandas-sql-easy-the-latest-login-in-2020/)\n\n- [Find Total Time Spent by Each Employee](https://leetcode.com/problems/find-total-time-spent-by-each-employee/solutions/4076313/pandas-sql-easy-find-total-time-spent-by-each-employee/)\n\n- [Find Followers Count](https://leetcode.com/problems/find-followers-count/solutions/4076342/pandas-sql-easy-find-followers-count/)\n- [Percentage of Users Attended a Contest](https://leetcode.com/problems/percentage-of-users-attended-a-contest/solutions/4077301/pandas-sql-easy-percentage-of-users-attended-a-contest/)\n\n- [Employees With Missing Information](https://leetcode.com/problems/employees-with-missing-information/solutions/4077308/pandas-sql-easy-employees-with-missing-information/)\n\n- [Average Time of Process per Machine](https://leetcode.com/problems/average-time-of-process-per-machine/solutions/4077402/pandas-sql-easy-average-time-of-process-per-machine/)\n | 4 | 0 | [] | 2 |
sales-analysis-iii | Fast and efficient solution, step by step explanation. | fast-and-efficient-solution-step-by-step-zym6 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires finding the unique product IDs and names from the first quarter sa | andreypomortsev | NORMAL | 2023-09-09T10:58:26.775813+00:00 | 2023-11-14T02:36:54.058857+00:00 | 588 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires finding the unique product IDs and names from the first quarter sales that are not present in the rest of the sales data.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nTo solve the problem of finding unique product IDs and names from the first quarter sales that are not present in the rest of the sales data, the following approach is taken:\n- Filter the `sales` DataFrame to obtain the sales data for the first quarter, defined as the period from `"2019-01-01"` to `"2019-03-31"`. This is done by creating a boolean mask using the `>=` and `<=` operators on the `"sale_date"` column.\n- Drop any duplicate rows based on the `product_id` in the first quarter sales using the `drop_duplicates` method. This ensures that only unique product IDs are considered.\n- Separate the rest of the sales data by filtering the `sales` DataFrame based on the condition that the `"sale_date"` is either before the start of the first quarter or after the end of the first quarter. This is done using the `|` (`OR`) operator.\n- Use the `isin` method to create **a boolean mask** that identifies the product IDs from the first quarter sales that are also present in the rest of the sales data.\n- Invert the boolean mask using the `~` operator to select the product IDs from the first quarter sales that are not present in the rest of the sales data.\n- Merge the filtered first quarter sales with the `product` DataFrame based on the `product_id` using the `merge` method. This ensures that the resulting DataFrame contains the corresponding product names.\n- Return the resulting DataFrame with only the `product_id` and `product_name` columns.\n\nThis approach efficiently filters and merges the data to obtain the desired result. By dropping duplicate rows and using the `isin` method, we ensure that only unique product IDs and names are returned. The use of boolean masks allows for easy identification of product IDs that are present or not present in the rest of the sales data.\n# Complexity\n#### Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(n + m)$$ The time complexity of this approach is dependent on the size of the `sales` DataFrame. The filtering and merging operations have a time complexity of $$O(n)$$, where $$n$$ is the number of rows in the DataFrame. The use of `isin` has a time complexity of $$O(m)$$, where $$m$$ is the number of unique product IDs in the first quarter sales (`first_q_sales`). \n#### Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nIn the given approach, the space complexity can be analyzed as follows:\n- The `sales` DataFrame for the first quarter is filtered and stored in memory. The size of this DataFrame depends on the number of rows and columns in the original data, but for the purpose of space complexity analysis, we can assume it takes $$O(n)$$ space, where $$n$$ is the number of records in the first quarter sales data.\n- The duplicate rows are dropped from the first quarter `sales` DataFrame, resulting in a new DataFrame with unique product IDs. The space required for this operation is proportional to the number of unique product IDs, which we can assume to be $$O(m)$$, where $$m$$ is the number of unique product IDs in the first quarter sales data.\n- Another DataFrame is created to store the rest of the sales data that is not in the first quarter. The space required for this DataFrame is again proportional to the number of records in the rest of the sales data, which we can assume to be $$O(k)$$, where $$k$$ is the number of records in the rest of the sales data.\n- A boolean mask is created to identify the product IDs from the first quarter sales that are also present in the rest of the sales data. This boolean mask requires additional space to store the boolean values for each product ID. Since the size of the mask is equal to the number of unique product IDs in the first quarter sales data, we can assume its space complexity to be $$O(m)$$.\n- The inverted boolean mask is used to select the product IDs from the first quarter sales that are not present in the rest of the sales data. This selection operation does not require any additional memory as it operates on the existing boolean mask.\n- Finally, the resulting DataFrame with the selected product IDs and names is returned. The space required for this DataFrame is again proportional to the number of unique product IDs in the first quarter sales data, which we can assume to be $$O(m)$$.\n\n**TL;DR** \nOverall, the space complexity of the approach can be approximated as $$O(n + m + k)$$, where $$n$$ is the number of records in the first quarter sales data, $$m$$ is the number of unique product IDs in the first quarter sales data, and $$k$$ is the number of records in the rest of the sales data.\n# Running & Memory\n\n\n# Code\n```\nimport pandas as pd\n\n\ndef sales_analysis(product_df: pd.DataFrame, sales_df: pd.DataFrame) -> pd.DataFrame:\n """\n Perform sales analysis for a specific quarter.\n\n Parameters:\n - product_df (pd.DataFrame): DataFrame containing product information.\n - sales_df (pd.DataFrame): DataFrame containing sales data.\n\n Returns:\n - pd.DataFrame: DataFrame with product_id and product_name for products\n that had sales in the specified quarter.\n """\n # Define the start and finish dates for the analysis quarter\n start_date, finish_date = "2019-01-01", "2019-03-31"\n\n # Select products with sales within the specified quarter\n first_q_sales = sales_df[\n (sales_df["sale_date"] >= start_date) & (sales_df["sale_date"] <= finish_date)\n ][["product_id"]].drop_duplicates()\n\n # Select products with sales outside the specified quarter\n other_sales = sales_df[\n (sales_df["sale_date"] < start_date) | (sales_df["sale_date"] > finish_date)\n ][["product_id"]]\n\n # Create a boolean mask to identify products that had sales outside the first quarter\n mask = first_q_sales["product_id"].isin(other_sales["product_id"])\n\n # Identify products with sales only in the specified quarter\n selected_products = first_q_sales[~mask].merge(product_df, on="product_id")[\n ["product_id", "product_name"]\n ]\n\n return selected_products\n\n``` | 4 | 0 | ['Database', 'Python', 'Python3', 'Pandas'] | 0 |
sales-analysis-iii | Solution with except | solution-with-except-by-alideveloper-6pcl | \n\n\nselect product_id, product_name from product where product_id in (\n select product_id from sales \n where sale_date between \'2019-01-01\' and \'20 | alideveloper | NORMAL | 2023-02-20T06:13:25.440272+00:00 | 2023-02-20T06:13:25.440315+00:00 | 588 | false | \n\n\n**select product_id, product_name from product where product_id in (\n select product_id from sales \n where sale_date between \'2019-01-01\' and \'2019-03-31\'\n except \n select product_id from sales \n where sale_date not between \'2019-01-01\' and \'2019-03-31\'\n)**\n\n\n\n\n``` | 4 | 0 | ['MS SQL Server'] | 0 |
sales-analysis-iii | MySQL Solution | mysql-solution-by-abstractconnoisseurs-9bbd | Code\n\nSELECT s.product_id, product_name\nFROM Sales s\nLEFT JOIN Product p\nON s.product_id = p.product_id\nGROUP BY s.product_id\nHAVING MIN(sale_date) >= CA | abstractConnoisseurs | NORMAL | 2023-01-27T09:10:02.444177+00:00 | 2023-01-27T09:10:02.444239+00:00 | 3,861 | false | # Code\n```\nSELECT s.product_id, product_name\nFROM Sales s\nLEFT JOIN Product p\nON s.product_id = p.product_id\nGROUP BY s.product_id\nHAVING MIN(sale_date) >= CAST(\'2019-01-01\' AS DATE) AND\n MAX(sale_date) <= CAST(\'2019-03-31\' AS DATE)\n``` | 4 | 0 | ['Database', 'MySQL'] | 1 |
sales-analysis-iii | Faster than 98.73%, left join is faster than in | faster-than-9873-left-join-is-faster-tha-1qlq | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | allen43 | NORMAL | 2023-01-10T02:46:27.269052+00:00 | 2023-01-10T02:46:27.269082+00:00 | 3,357 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n/* Write your T-SQL query statement below */\n\nSELECT DISTINCT A.product_id, B.product_name\nFROM Sales A LEFT JOIN Product B ON A.product_id = B.product_id LEFT JOIN (SELECT DISTINCT product_id FROM Sales WHERE sale_date not between \'2019-01-01\' and \'2019-03-31\') C ON A.product_id = C.product_id\nWHERE C.product_id is null\n``` | 4 | 0 | ['MS SQL Server'] | 0 |
sales-analysis-iii | Using INNER JOIN and GROUP BY | using-inner-join-and-group-by-by-shikhar-jd3i | SELECT \n p.product_id,\n p.product_name\n FROM Product p INNER JOIN Sales s\n ON p.product_id = s.product_id\n GROUP BY product_id\n HAVING M | shikharsaxena34 | NORMAL | 2022-07-18T17:46:35.605822+00:00 | 2022-07-18T17:46:35.605860+00:00 | 280 | false | SELECT \n p.product_id,\n p.product_name\n FROM Product p INNER JOIN Sales s\n ON p.product_id = s.product_id\n GROUP BY product_id\n HAVING MIN(sale_date) >= \'2019-01-01\' AND MAX(sale_date) <= \'2019-03-31\'; | 4 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | Simple sol with one Subquery | simple-sol-with-one-subquery-by-zakirh-yhou | \nSELECT DISTINCT p.product_id, product_name \nFROM Product p \nJOIN Sales s ON s.product_id = p.product_id \nWHERE s.sale_date BETWEEN \'2019-01-01\' AND \'201 | zakirh | NORMAL | 2022-07-08T17:24:19.211391+00:00 | 2022-07-11T12:24:43.599932+00:00 | 346 | false | ```\nSELECT DISTINCT p.product_id, product_name \nFROM Product p \nJOIN Sales s ON s.product_id = p.product_id \nWHERE s.sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\' \n AND p.product_id NOT IN (\n SELECT product_id \n FROM Sales \n WHERE sale_date NOT BETWEEN \'2019-01-01\' AND \'2019-03-31\'\n );\n``` | 4 | 0 | [] | 3 |
sales-analysis-iii | Using INNER JOIN and GROUP BY | using-inner-join-and-group-by-by-suprati-2txi | \nSELECT s.product_id,p.product_name FROM Sales as s INNER JOIN Product as p ON s.product_id=p.product_id GROUP BY s.product_id HAVING MIN(sale_date)>=\'2019-01 | supratim_code | NORMAL | 2022-05-29T19:07:16.718617+00:00 | 2022-05-29T19:07:16.718656+00:00 | 310 | false | ```\nSELECT s.product_id,p.product_name FROM Sales as s INNER JOIN Product as p ON s.product_id=p.product_id GROUP BY s.product_id HAVING MIN(sale_date)>=\'2019-01-01\' AND MAX(sale_date)<=\'2019-03-31\';\n``` | 4 | 0 | [] | 0 |
sales-analysis-iii | Simple MSSQL Solution | simple-mssql-solution-by-codingkida-pfli | \nselect product_id, product_name\nfrom product \nwhere product_id not in(select product_id from sales where sale_date<\'2019-01-01\' or sale_date>\'2019-03-31\ | codingkida | NORMAL | 2022-04-06T00:38:58.775725+00:00 | 2022-04-06T00:38:58.775756+00:00 | 606 | false | ```\nselect product_id, product_name\nfrom product \nwhere product_id not in(select product_id from sales where sale_date<\'2019-01-01\' or sale_date>\'2019-03-31\')\n``` | 4 | 0 | ['MS SQL Server'] | 1 |
sales-analysis-iii | easy to understand my sql solution without join or cte | easy-to-understand-my-sql-solution-witho-96lk | easy my sql solution \n\nselect product_id, product_name from product where product_id not in ( \nselect product_id from sales\nwhere sale_date not between \'20 | vasudha25 | NORMAL | 2022-02-22T19:21:26.646113+00:00 | 2022-02-22T19:21:26.646161+00:00 | 324 | false | easy my sql solution \n```\nselect product_id, product_name from product where product_id not in ( \nselect product_id from sales\nwhere sale_date not between \'2019-01-01\' and \'2019-03-31\')\n``` | 4 | 0 | [] | 0 |
sales-analysis-iii | MySQL Min and Max Approach | mysql-min-and-max-approach-by-imranmahmo-bxwa | select p.product_id, p.product_name\nfrom product p\ninner join sales s on s.product_id = p.product_id\ngroup by p.product_id\nhaving min(sale_date) >= \'2019-0 | imranmahmood-mba | NORMAL | 2021-11-19T06:18:57.422666+00:00 | 2021-11-19T06:18:57.422700+00:00 | 283 | false | select p.product_id, p.product_name\nfrom product p\ninner join sales s on s.product_id = p.product_id\ngroup by p.product_id\nhaving min(sale_date) >= \'2019-01-01\' and max(sale_date) <= \'2019-03-31\' | 4 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | Elegant Select Solution | elegant-select-solution-by-vishkk-bw98 | \nSELECT \n\tproduct_id,\n\tproduct_name \nFROM\n\tProduct\nWHERE product_id NOT IN\n\t(SELECT\n\t\tproduct_id\n\tFROM \n\t\tSales\n\tWHERE\n\t\tsale_date \n | vishkk | NORMAL | 2021-11-12T00:26:29.639511+00:00 | 2021-11-12T00:26:29.639556+00:00 | 317 | false | ```\nSELECT \n\tproduct_id,\n\tproduct_name \nFROM\n\tProduct\nWHERE product_id NOT IN\n\t(SELECT\n\t\tproduct_id\n\tFROM \n\t\tSales\n\tWHERE\n\t\tsale_date \n NOT BETWEEN \n \'2019-01-01\' AND \'2019-03-31\')\n``` | 4 | 0 | [] | 2 |
sales-analysis-iii | Faster than 100% of MS SQL Server online submissions | faster-than-100-of-ms-sql-server-online-8rmlu | \nSELECT p.product_id,\n p.product_name\nFROM product p\n JOIN sales s\n ON p.product_id = s.product_id\nGROUP BY p.product_id,\n | neudatageek | NORMAL | 2021-03-08T22:12:29.519158+00:00 | 2021-03-08T22:12:29.519196+00:00 | 518 | false | ```\nSELECT p.product_id,\n p.product_name\nFROM product p\n JOIN sales s\n ON p.product_id = s.product_id\nGROUP BY p.product_id,\n p.product_name\nHAVING Min(sale_date) >= \'2019-01-01\'\n AND Max(sale_date) <= \'2019-03-31\' \n``` | 4 | 0 | [] | 0 |
sales-analysis-iii | ✔️✔️✔️easy solution - MySQL || PostgreSQL || MS SQL Server⚡⚡⚡ | easy-solution-mysql-postgresql-ms-sql-se-ekcp | Code - MySQL\n\nSELECT p.product_id, p.product_name\nFROM Product p\nLEFT JOIN Sales s\nON p.product_id = s.product_id\nGROUP BY s.product_id\nHAVING MIN(s.sale | anish_sule | NORMAL | 2024-03-12T13:32:38.930189+00:00 | 2024-03-12T13:32:38.930223+00:00 | 361 | false | # Code - MySQL\n```\nSELECT p.product_id, p.product_name\nFROM Product p\nLEFT JOIN Sales s\nON p.product_id = s.product_id\nGROUP BY s.product_id\nHAVING MIN(s.sale_date) > \'2018-12-31\' AND MAX(s.sale_date) < \'2019-04-01\'\n```\n\n# Code - MS SQL Server | PostgreSQL\n```\nSELECT p.product_id, p.product_name\nFROM Product p\nLEFT JOIN Sales s\nON p.product_id = s.product_id\nGROUP BY p.product_id, p.product_name\nHAVING MIN(s.sale_date) > \'2018-12-31\' AND MAX(s.sale_date) < \'2019-04-01\'\n``` | 3 | 0 | ['MySQL', 'MS SQL Server'] | 2 |
sales-analysis-iii | MySQL Solution for Sales Analysis III Problem | mysql-solution-for-sales-analysis-iii-pr-7jq8 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the given solution is to select the product_id and product_name of | Aman_Raj_Sinha | NORMAL | 2023-05-18T03:41:32.732464+00:00 | 2023-05-18T03:41:32.732508+00:00 | 1,718 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the given solution is to select the product_id and product_name of products that had sales between a specific date range and had sales overall. The approach involves performing an inner join between the Sales table and the Product table using the product_id column. The results are then grouped by product_id. The HAVING clause is used to compare the sum of sales within the specific date range with the overall sum of sales.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Perform an inner join between the Sales table and the Product table using the product_id column.\n1. Group the results by product_id.\n1. Use the HAVING clause to filter the grouped results based on the comparison of the sum of sales within the date range and the overall sum of sales.\n1. Return the product_id and product_name columns of the selected products.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of this solution depends on the size of the Sales table, the indexing of the relevant columns, and the efficiency of the join and grouping operations. If the tables are properly indexed on the product_id and sale_date columns, and the date range filter can be efficiently applied, the time complexity can be estimated as O(n * log n), where n is the number of rows in the Sales table.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of this solution depends on the size of the resulting dataset, which includes the product_id and product_name columns.\n\n# Code\n```\n# Write your MySQL query statement below\n select product_id, product_name from Sales\n inner join product using(product_id)\n group by product_id\n having sum(if(sale_date between \'2019-01-01\' and \'2019-03-31\', 1, 0)) = sum(if(sale_date, 1, 0))\n``` | 3 | 0 | ['MySQL'] | 2 |
sales-analysis-iii | MySQL Solution | mysql-solution-by-pranto1209-vhdn | Code\n\n# Write your MySQL query statement below\nselect product_id, product_name from Product natural join Sales\ngroup by product_id \nhaving min(sale_date) > | pranto1209 | NORMAL | 2023-03-09T12:57:20.574851+00:00 | 2023-03-13T15:59:14.114032+00:00 | 2,324 | false | # Code\n```\n# Write your MySQL query statement below\nselect product_id, product_name from Product natural join Sales\ngroup by product_id \nhaving min(sale_date) >= \'2019-01-01\' and max(sale_date) <= \'2019-03-31\'; \n``` | 3 | 0 | ['MySQL'] | 1 |
sales-analysis-iii | MYSQL Easy Solution ✅✅ | mysql-easy-solution-by-shubhamjain287-jj5s | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Shubhamjain287 | NORMAL | 2023-03-08T05:48:28.778930+00:00 | 2023-03-08T05:48:28.778988+00:00 | 4,538 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect s.product_id, p.product_name\nfrom sales s, product p\nwhere s.product_id = p.product_id\ngroup by s.product_id, p.product_name\nhaving min(s.sale_date) >= \'2019-01-01\' AND max(s.sale_date) <= \'2019-03-31\'\n``` | 3 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | Simple Query || Sales Analysis III | simple-query-sales-analysis-iii-by-himsh-j3e6 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | himshrivas | NORMAL | 2023-01-26T08:51:27.947706+00:00 | 2023-02-07T13:52:01.492919+00:00 | 3,213 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT DISTINCT A.product_id, B.product_name FROM Sales A LEFT JOIN Product B \nON A.product_id = B.product_id LEFT JOIN \n(SELECT DISTINCT product_id FROM Sales WHERE sale_date not between \'2019-01-01\' and \'2019-03-31\') \nC ON A.product_id = C.product_id\nWHERE C.product_id is null;\n\n```\nPlease upvote!! | 3 | 0 | ['MySQL'] | 2 |
sales-analysis-iii | Simple Solution | simple-solution-by-mediocre-coder-kruk | \nSELECT\n DISTINCT p.product_id, p.product_name\nFROM\n Product p INNER JOIN Sales s ON p.product_id = s.product_id\nWHERE\n sale_date BETWEEN \'2019- | mediocre-coder | NORMAL | 2022-09-19T19:14:36.166973+00:00 | 2022-09-19T19:14:36.167013+00:00 | 771 | false | ```\nSELECT\n DISTINCT p.product_id, p.product_name\nFROM\n Product p INNER JOIN Sales s ON p.product_id = s.product_id\nWHERE\n sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\' AND p.product_id NOT IN (\n SELECT\n p.product_id\n FROM\n Product p INNER JOIN Sales s ON p.product_id = s.product_id \n WHERE\n sale_date < \'2019-01-01\' or sale_date > \'2019-03-31\'\n )\n``` | 3 | 0 | ['Oracle'] | 1 |
sales-analysis-iii | Simple solution without joins | simple-solution-without-joins-by-ajinkya-d6gg | select product_id ,product_name\n from product \n where product_id not in (select product_id from sales where sale_date | ajinkyanhatolkar | NORMAL | 2022-09-13T14:21:24.666191+00:00 | 2022-09-13T14:21:24.666241+00:00 | 662 | false | select product_id ,product_name\n from product \n where product_id not in (select product_id from sales where sale_date not between \'2019-01-01\' AND \'2019-03-31\' ) and product_id in (select product_id from sales)\n | 3 | 0 | [] | 2 |
sales-analysis-iii | Simple Solution | simple-solution-by-krex0r-krcz | sql\nSELECT Sales.product_id, Product.product_name\nFROM Sales\nLEFT JOIN Product\nUSING(product_id)\nGROUP BY product_id\nHAVING MIN(Sales.sale_date) >= \'2019 | krex0r | NORMAL | 2022-09-07T19:57:08.156512+00:00 | 2022-09-07T19:57:08.156549+00:00 | 499 | false | ```sql\nSELECT Sales.product_id, Product.product_name\nFROM Sales\nLEFT JOIN Product\nUSING(product_id)\nGROUP BY product_id\nHAVING MIN(Sales.sale_date) >= \'2019-01-01\'\nAND MAX(Sales.sale_date) <= \'2019-03-31\'\n``` | 3 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | Faster than 93.61% | MySQL Easy Solution | faster-than-9361-mysql-easy-solution-by-iemof | ```\nSELECT product_id, product_name\nFROM Product\nWHERE product_id IN (\n SELECT product_id FROM Sales\n GROUP BY product_id\n HAVING MIN(sale_date) | arghapaul002 | NORMAL | 2022-08-21T01:09:36.661173+00:00 | 2022-08-21T01:09:36.661210+00:00 | 742 | false | ```\nSELECT product_id, product_name\nFROM Product\nWHERE product_id IN (\n SELECT product_id FROM Sales\n GROUP BY product_id\n HAVING MIN(sale_date) >= \'2019-01-01\' AND MAX(sale_date) <= \'2019-03-31\'); | 3 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | Easy to understand mySQL solution using a subquery | easy-to-understand-mysql-solution-using-yyzl2 | \nSELECT product_id, product_name FROM product\n\nWHERE product_id IN \n\n\t(SELECT DISTINCT product_id FROM sales\n\n GROUP BY product_id \n\n HAVING M | atalayfa | NORMAL | 2022-08-10T11:30:44.876000+00:00 | 2022-08-10T11:30:44.876050+00:00 | 623 | false | ```\nSELECT product_id, product_name FROM product\n\nWHERE product_id IN \n\n\t(SELECT DISTINCT product_id FROM sales\n\n GROUP BY product_id \n\n HAVING MIN(sale_date)>= \'2019-01-01\' \n AND MAX(sale_date) <= \'2019-03-31\')\n``` | 3 | 0 | ['MySQL'] | 1 |
sales-analysis-iii | ✅ MySql, 2 solutions | mysql-2-solutions-by-ranbeer_singh-9bi5 | \nselect p.product_id, p.product_name\nfrom Product p\nleft join Sales s\non p.product_id = s.product_id\ngroup by p.product_id\nhaving sum(s.sale_date between | Ranbeer_Singh | NORMAL | 2022-07-16T02:39:41.762023+00:00 | 2022-07-16T02:39:41.762063+00:00 | 209 | false | ```\nselect p.product_id, p.product_name\nfrom Product p\nleft join Sales s\non p.product_id = s.product_id\ngroup by p.product_id\nhaving sum(s.sale_date between \'2019-01-01\' and \'2019-03-31\') = count(s.sale_date) \n\nOR\n\nselect p.product_id, p.product_name\nfrom Product p\ninner join Sales s on p.product_id = s.product_id\ngroup by s.product_id\nhaving min(sale_date) >= \'2019-01-01\'\nand max(sale_date) <= \'2019-03-31\';\n``` | 3 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | Faster than 95.27% of Oracle online submissions for Sales Analysis III | faster-than-9527-of-oracle-online-submis-9gqz | \nselect p.product_id\n ,p.product_name\nfrom product p\njoin (select s.product_id from sales s\n group by s.product_id\n having min(s.sale_dat | hoangdang | NORMAL | 2022-07-13T13:45:13.359919+00:00 | 2022-07-13T13:45:13.359948+00:00 | 371 | false | ```\nselect p.product_id\n ,p.product_name\nfrom product p\njoin (select s.product_id from sales s\n group by s.product_id\n having min(s.sale_date) >= \'2019-01-01\'\n and max(s.sale_date) <= \'2019-03-31\') s1\non\np.product_id = s1.product_id\n``` | 3 | 0 | ['Oracle'] | 0 |
sales-analysis-iii | simple having solution | simple-having-solution-by-hardawaylong-9h4s | select product_id, product_name\nfrom Product\nwhere product_id in \n(select product_id\nfrom Sales\ngroup by product_id\nhaving max(sale_date) <= \'2019-03-31\ | HardawayLong | NORMAL | 2022-06-10T01:37:53.981901+00:00 | 2022-06-10T01:37:53.981925+00:00 | 241 | false | select product_id, product_name\nfrom Product\nwhere product_id in \n(select product_id\nfrom Sales\ngroup by product_id\nhaving max(sale_date) <= \'2019-03-31\' and min(sale_date) >= \'2019-01-01\') | 3 | 0 | [] | 0 |
sales-analysis-iii | Multiple ways of solving in MySQL | multiple-ways-of-solving-in-mysql-by-iam-k227 | Approach 1(With CTE)\n\n\nWITH sell_history AS (\n SELECT\n s.product_id,\n SUM(CASE WHEN sale_date NOT BETWEEN \'2019-01-01\' AND \'2019-03-31 | iamrafiul | NORMAL | 2022-05-05T21:06:57.058085+00:00 | 2022-05-05T21:06:57.058126+00:00 | 331 | false | **Approach 1(With CTE)**\n\n```\nWITH sell_history AS (\n SELECT\n s.product_id,\n SUM(CASE WHEN sale_date NOT BETWEEN \'2019-01-01\' AND \'2019-03-31\' THEN 1 ELSE 0 END) sold_outside,\n SUM(CASE WHEN sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\' THEN 1 ELSE 0 END) sold_inside\n FROM sales s \n GROUP BY s.product_id\n)\n\nSELECT\n s.product_id,\n p.product_name\nFROM sell_history s \nLEFT JOIN product p ON p.product_id = s.product_id\nWHERE s.sold_outside = 0 \nAND s.sold_inside > 0;\n```\n\n**Approach 2(Using subquery with JOIN & GROUP BY)**\n\n```\nSELECT\n s.product_id,\n p.product_name\nFROM sales s \nLEFT JOIN product p ON p.product_id = s.product_id\nGROUP BY s.product_id\nHAVING s.product_id NOT IN (\n SELECT product_id FROM sales WHERE sale_date NOT BETWEEN \'2019-01-01\' AND \'2019-03-31\'\n);\n```\n\n**Approach 3(Using subquery with no JOIN)**\n\n```\nSELECT\n product_id,\n product_name\nFROM product\nWHERE product_id NOT IN (\n SELECT product_id FROM sales WHERE sale_date NOT BETWEEN \'2019-01-01\' AND \'2019-03-31\'\n)\n``` | 3 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | MYSQL very simple solution | mysql-very-simple-solution-by-sonicak-2d6h | \n# Write your MySQL query statement below\nselect p.product_id, p.product_name\nfrom product p\njoin sales s\non p.product_id=s.product_id\ngroup by s.product_ | sonicak | NORMAL | 2022-01-25T07:26:06.639993+00:00 | 2022-01-25T07:26:06.640024+00:00 | 330 | false | ```\n# Write your MySQL query statement below\nselect p.product_id, p.product_name\nfrom product p\njoin sales s\non p.product_id=s.product_id\ngroup by s.product_id\nhaving min(s.sale_date)>=\'2019-01-01\' and max(s.sale_date)<=\'2019-03-31\'\n``` | 3 | 0 | [] | 1 |
sales-analysis-iii | MYSQL | mysql-by-zizhengwang-ah40 | ```\nSELECT Product.product_id,product_name\nFROM Product\nINNER JOIN Sales USING(product_id)\nGROUP BY Product.product_id\nHAVING SUM(CASE WHEN\n sal | ZizhengWang | NORMAL | 2021-02-01T04:01:46.868558+00:00 | 2021-02-01T04:01:46.868645+00:00 | 284 | false | ```\nSELECT Product.product_id,product_name\nFROM Product\nINNER JOIN Sales USING(product_id)\nGROUP BY Product.product_id\nHAVING SUM(CASE WHEN\n sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\' THEN 0\n ELSE 1\n END)=0; | 3 | 0 | [] | 0 |
sales-analysis-iii | EASY MySQL solution | easy-mysql-solution-by-baixiaofanya123-ql42 | \nSELECT sales.product_id, product_name\nFROM product\n JOIN sales ON product.product_id = sales.product_id\nGROUP BY product_id\nHAVING SUM(sale_date BETWEE | baixiaofanya123 | NORMAL | 2020-12-04T04:55:28.732339+00:00 | 2020-12-04T04:55:28.732382+00:00 | 318 | false | ```\nSELECT sales.product_id, product_name\nFROM product\n JOIN sales ON product.product_id = sales.product_id\nGROUP BY product_id\nHAVING SUM(sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\') = COUNT(sale_date)\n```\n | 3 | 0 | [] | 0 |
sales-analysis-iii | Can someone please tell me what's wrong with this code? | can-someone-please-tell-me-whats-wrong-w-g77i | \nselect p.product_id,p.product_name from Product p join Sales s on p.product_id=s.product_id \nwhere s.sale_date between \'2019-01-01\' and \'2019-03-31\';\n\n | anshuman_05 | NORMAL | 2020-06-24T12:23:57.569343+00:00 | 2020-06-24T12:23:57.569375+00:00 | 227 | false | ```\nselect p.product_id,p.product_name from Product p join Sales s on p.product_id=s.product_id \nwhere s.sale_date between \'2019-01-01\' and \'2019-03-31\';\n```\n\n\n | 3 | 0 | [] | 2 |
sales-analysis-iii | beats 80% | beats-80-by-shinymer-p9jy | \nselect temp.product_id, Product.product_name\nfrom (select product_id, \n sum(case when sale_date between \'2019-01-01\' and \'2019-03-31\' then 0 else | shinymer | NORMAL | 2019-08-31T06:44:33.002999+00:00 | 2019-08-31T07:18:29.036564+00:00 | 415 | false | ```\nselect temp.product_id, Product.product_name\nfrom (select product_id, \n sum(case when sale_date between \'2019-01-01\' and \'2019-03-31\' then 0 else 1 end) as val\nfrom Sales\ngroup by product_id\nhaving val = 0) temp\njoin Product\non Product.product_id = temp.product_id\n``` | 3 | 0 | [] | 1 |
sales-analysis-iii | simple not in solution | simple-not-in-solution-by-akumammx-rsw0 | select product_id,product_name from Product where product_id not in(select product_id from Sales where sale_date < \'2019-01-01\' or sale_date>\'2019-03-31\') | akumammx | NORMAL | 2019-08-14T15:00:41.725359+00:00 | 2019-08-14T15:00:41.725390+00:00 | 360 | false | select product_id,product_name from Product where product_id not in(select product_id from Sales where sale_date < \'2019-01-01\' or sale_date>\'2019-03-31\') | 3 | 0 | [] | 1 |
sales-analysis-iii | Using First_value( ), MAX( ) and CTE | using-first_value-max-and-cte-by-amxnkz-en7s | IntuitionWhen the problem specified that the product_id should be sold only in the first quarter of 2019, my initial thought was to use the FIRST_VALUE() functi | amxnkz | NORMAL | 2025-02-09T22:03:52.116300+00:00 | 2025-02-09T22:03:52.116300+00:00 | 457 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
When the problem specified that the product_id should be sold only in the first quarter of `2019`, my initial thought was to use the `FIRST_VALUE()` function along with `MAX()`. Although you could use `MIN()` instead, I chose to utilize `FIRST_VALUE()`. This solution is not the most elegant, but it reflects my first approach to writing the query.
# Approach
<!-- Describe your approach to solving the problem. -->
After creating the CTE, all we had to do was specify our conditions in the `WHERE` clause using `BETWEEN` and `AND` for the first_sale_date and `<=` for the last_sale_date.
# Code
```mysql []
# Write your MySQL query statement below
WITH CTE AS (
SELECT s.product_id, p.product_name,
FIRST_VALUE(s.sale_date) OVER (PARTITION BY s.product_id ORDER BY s.sale_date
) AS first_sale_date,
MAX(s.sale_date) OVER (PARTITION BY s.product_id) AS last_sale_date
FROM Sales s
LEFT JOIN Product p
ON s.product_id = p.product_id
)
SELECT DISTINCT product_id, product_name
FROM CTE
WHERE first_sale_date BETWEEN '2019-01-01' AND '2019-03-31'
AND last_sale_date <= '2019-03-31'
;
``` | 2 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | 👉🏻FAST AND EASY TO UNDERSTAND SOLUTION || MySQL | fast-and-easy-to-understand-solution-mys-7xpz | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | djain7700 | NORMAL | 2025-01-08T18:11:02.028024+00:00 | 2025-01-08T18:11:02.028024+00:00 | 345 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```mysql []
# Write your MySQL query statement below
SELECT product_id, product_name
FROM Product
WHERE product_id IN (SELECT DISTINCT product_id
FROM Sales
GROUP BY product_id
HAVING MIN(sale_date) >= '2019-01-01' AND MAX(sale_date) <= '2019-03-31')
``` | 2 | 0 | ['Database', 'MySQL'] | 0 |
sales-analysis-iii | 100% Beats✅✅Easiest MySQL Solution✅🔥💯Without Joins | 100-beatseasiest-mysql-solutionwithout-j-hp4r | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | chinnibhargavi17 | NORMAL | 2024-11-12T02:34:23.434937+00:00 | 2024-11-12T02:34:23.434970+00:00 | 349 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```mysql []\nSELECT product_id, product_name\nFROM Product\nWHERE product_id IN (\n SELECT product_id\n FROM Sales\n GROUP BY product_id\n HAVING MIN(sale_date) >= "2019-01-01" AND MAX(sale_date) <="2019-03-31"\n);\n``` | 2 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | PostgreSQL 92% solution | postgresql-92-solution-by-phiduyanh-ylcb | \n\n# Code\npostgresql []\n-- Write your PostgreSQL query statement below\nWITH cte AS\n( \n SELECT\n product_id,\n COUNT(sale_date) FILTER (WH | phiduyanh | NORMAL | 2024-10-24T02:48:59.228098+00:00 | 2024-10-24T02:48:59.228128+00:00 | 161 | false | \n\n# Code\n```postgresql []\n-- Write your PostgreSQL query statement below\nWITH cte AS\n( \n SELECT\n product_id,\n COUNT(sale_date) FILTER (WHERE sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\') AS first_quarter_sales,\n COUNT(sale_date) AS all_sales\n FROM Sales\n GROUP BY product_id\n)\nSELECT\n cte.product_id,\n Product.product_name\nFROM cte \nJOIN Product ON cte.product_id = Product.product_id\nWHERE first_quarter_sales = all_sales;\n``` | 2 | 0 | ['PostgreSQL'] | 0 |
sales-analysis-iii | Easiest Solution ~ | easiest-solution-by-sheetaltondwalkar-krvh | Intuition\nWe need to report Product IDs exclusively sold in first quarter 2019. So if a product is sold in first quarter and also in later or prior then it sho | sheetaltondwalkar | NORMAL | 2024-06-10T17:09:28.367299+00:00 | 2024-06-10T17:09:28.367340+00:00 | 623 | false | # Intuition\nWe need to report Product IDs exclusively sold in first quarter 2019. So if a product is sold in first quarter and also in later or prior then it should not be reported.\n\n# Approach\nSince the output has product name, product id and sale date we join Product and Sales table. \nNext we check for 2 conditions \n a) Product has been sold between dates \'2019-01-01\' and \'2019-03-31\'\n b) Product has not been sold outside above range\n\n# Code\n```\n# Write your MySQL query statement below\n\nselect s.product_id product_id,p.product_name product_name from Sales s join Product p on s.product_id = p.product_id\n where s.Sale_date between \'2019-01-01\' and \'2019-03-31\'\n and s.product_id not in (select product_id from Sales where Sale_date > \'2019-03-31\' or Sale_date < \'2019-01-01\' )\n group by 1,2\n\n``` | 2 | 0 | ['MySQL'] | 2 |
sales-analysis-iii | Easy Solution Using subquery | easy-solution-using-subquery-by-i_manish-zfnj | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | I_ManishSharma | NORMAL | 2024-02-05T10:12:45.294635+00:00 | 2024-02-05T10:12:45.294664+00:00 | 1,037 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT product_id, product_name\nFROM Product p\nWHERE EXISTS (\n SELECT product_id\n FROM Sales\n WHERE product_id = p.product_id AND sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\'\n)\nAND NOT EXISTS (\n SELECT product_id\n FROM Sales\n WHERE product_id = p.product_id AND (sale_date < \'2019-01-01\' OR sale_date > \'2019-03-31\')\n);\n``` | 2 | 0 | ['MySQL'] | 2 |
sales-analysis-iii | Beginner Friendly Solution | MySQL | No Joins | beginner-friendly-solution-mysql-no-join-yxki | \n\n# Code\n\nSELECT product_id, product_name\nFROM Product\nWHERE product_id IN (\n SELECT product_id\n FROM Sales\n GROUP BY product_id\n HAVING | Newbie_Nerd | NORMAL | 2023-11-28T09:44:24.426675+00:00 | 2023-11-28T09:44:24.426702+00:00 | 977 | false | \n\n# Code\n```\nSELECT product_id, product_name\nFROM Product\nWHERE product_id IN (\n SELECT product_id\n FROM Sales\n GROUP BY product_id\n HAVING MIN(sale_date) >= "2019-01-01" AND MAX(sale_date) <="2019-03-31"\n);\n\n``` | 2 | 0 | ['MySQL'] | 1 |
sales-analysis-iii | JOIN + LEFT JOIN | join-left-join-by-sergeyzhirkov-ju1m | \nSQL []\nSELECT p.product_id, p.product_name\nFROM product p\nJOIN sales s \n ON p.product_id = s.product_id AND s.sale_date between \'2019-01-01\' and \'2019 | SergeyZhirkov | NORMAL | 2023-10-24T02:51:25.850128+00:00 | 2024-09-22T19:14:41.754623+00:00 | 388 | false | \n```SQL []\nSELECT p.product_id, p.product_name\nFROM product p\nJOIN sales s \n ON p.product_id = s.product_id AND s.sale_date between \'2019-01-01\' and \'2019-03-31\'\nLEFT JOIN sales s2\n ON p.product_id = s2.product_id AND s2.sale_date not between \'2019-01-01\' and \'2019-03-31\'\nWHERE s2.product_id is null\nGROUP BY p.product_id, p.product_name\n```\n\n\n\n | 2 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | EASY TO UNDERSTAND, BEGINNER FRIENDLY 🔥 | easy-to-understand-beginner-friendly-by-6dhop | EASY TO UNDERSTAND, BEGINNER FRIENDLY \uD83D\uDD25\n x your first thoughts on how to solve this problem. \n easy to understand \n\n# Code\n\n# Write your MySQL | Jaloldinov | NORMAL | 2023-08-25T15:05:31.472300+00:00 | 2023-08-25T15:05:31.472323+00:00 | 755 | false | # EASY TO UNDERSTAND, BEGINNER FRIENDLY \uD83D\uDD25\n<!-- x your first thoughts on how to solve this problem. -->\n easy to understand \n\n# Code\n```\n# Write your MySQL query statement below\nselect p.product_id, p.product_name \nfrom Product p\nWHERE p.product_id IN (SELECT s.product_id FROM Sales s \n WHERE sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\') \nAND p.product_id NOT IN (SELECT s.product_id FROM Sales s \n WHERE sale_date NOT BETWEEN \'2019-01-01\' AND \'2019-03-31\')\n\n``` | 2 | 0 | ['MySQL'] | 2 |
sales-analysis-iii | where not between easy solution | where-not-between-easy-solution-by-jdu9a-n041 | Intuition\nFast and clearly to understanf for beginners\n\n\n# Code\n\nSELECT product_id, product_name\nFROM product\nWHERE product_id IN (\n SELECT product_ | jdu9a0009 | NORMAL | 2023-08-25T15:04:10.244072+00:00 | 2023-08-25T15:05:41.244798+00:00 | 293 | false | # Intuition\nFast and clearly to understanf for beginners\n\n\n# Code\n```\nSELECT product_id, product_name\nFROM product\nWHERE product_id IN (\n SELECT product_id\n FROM sales\n WHERE sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\'\n) AND product_id NOT IN (\n SELECT product_id\n FROM sales\n WHERE sale_date NOT BETWEEN \'2019-01-01\' AND \'2019-03-31\'\n);\n``` | 2 | 0 | ['MySQL'] | 2 |
sales-analysis-iii | Pandas solution | pandas-solution-by-teemacia-stnb | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Teemacia | NORMAL | 2023-08-19T10:29:12.412509+00:00 | 2023-08-19T10:29:12.412543+00:00 | 362 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nimport pandas as pd\n\ndef sales_analysis(product: pd.DataFrame, sales: pd.DataFrame) -> pd.DataFrame:\n data = product.merge(sales, how =\'left\', on = \'product_id\').query("sale_date < \'2019-01-01\' or sale_date>\'2019-03-31\' or sale_date.isnull()")\n data_1 = product[[\'product_id\',\'product_name\']].query("~product_id.isin(@data[\'product_id\'])")\n return data_1\n``` | 2 | 0 | ['Pandas'] | 1 |
sales-analysis-iii | Fast and simple query by MySQL!!! | fast-and-simple-query-by-mysql-by-nourim-r5ow | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | nourimanov | NORMAL | 2023-06-02T11:57:36.655684+00:00 | 2023-06-02T11:57:36.655726+00:00 | 540 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nselect s.product_id, p.product_name from Product p\njoin Sales s on p.product_id = s.product_id\ngroup by product_id\nhaving max(s.sale_date) <= \'2019-03-31\' and min(s.sale_date) >= \'2019-01-01\'\n\n\n\n\n``` | 2 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | 👀🥷MySQL Solution | mysql-solution-by-dipesh_12-7nvm | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | dipesh_12 | NORMAL | 2023-04-14T15:40:58.143468+00:00 | 2023-04-14T15:40:58.143502+00:00 | 1,643 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect p.product_id,p.product_name from Product p,Sales s where p.product_id=s.product_id group by s.product_id having min(s.sale_date)>=\'2019-01-01\' and max(s.sale_date)<=\'2019-03-31\';\n``` | 2 | 0 | ['MySQL'] | 1 |
sales-analysis-iii | MYSQL || MIN-MAX || HAVING | mysql-min-max-having-by-utkarshsakarwar-7kyh | \n\n# Code\n\n# Write your MySQL query statement below\nSELECT p.product_id,p.product_name\nFROM Product p\nLEFT JOIN Sales s\nON p.product_id=s.product_id\nGRO | utkarshsakarwar | NORMAL | 2023-01-22T09:03:12.500780+00:00 | 2023-01-22T09:03:12.500814+00:00 | 648 | false | \n\n# Code\n```\n# Write your MySQL query statement below\nSELECT p.product_id,p.product_name\nFROM Product p\nLEFT JOIN Sales s\nON p.product_id=s.product_id\nGROUP BY s.product_id\nHAVING MIN(s.sale_date)>\'2018-12-31\' AND MAX(s.sale_date)<\'2019-04-01\'\n``` | 2 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | Solution for 1084. Sales Analysis III | solution-for-1084-sales-analysis-iii-by-e847l | Intuition\r\nObtain all the product id that were sold apart from requested dates.\r\noutput unique pid, pnames that are not in above result\r\n\r\n# Approach\r\ | vigneshrajgopalnayak | NORMAL | 2023-01-01T06:00:45.454717+00:00 | 2023-01-01T06:00:45.454761+00:00 | 932 | false | # Intuition\r\nObtain all the product id that were sold apart from requested dates.\r\noutput unique pid, pnames that are not in above result\r\n\r\n# Approach\r\nUse temp table to join both sales and pproduct table\r\nInner join is needed.\r\n\r\n# Complexity\r\n- Time complexity:\r\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\r\n\r\n- Space complexity:\r\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\r\n\r\n# Code\r\n```\r\n# Write your MySQL query statement below\r\n/* creating temp table*/\r\nWITH new_tbl AS\r\n\r\n(SELECT a.product_id AS pid,a.product_name AS pname, b.sale_date AS sd\r\nFROM Product AS a\r\nINNER JOIN Sales AS b\r\nON a.product_id = b.product_id)\r\n\r\n/*main query*/\r\nSELECT pid AS product_id,pname AS product_name\r\nFROM new_tbl\r\nWHERE pid NOT IN \r\n/* sub query to identify all the product id that were sold apart from requested dates.*/\r\n(\r\nSELECT pid FROM new_tbl \r\nWHERE sd NOT BETWEEN \'2019-01-01\' AND \'2019-03-31\'\r\n)\r\n\r\n/* use group by to obtain unique values */\r\nGROUP BY pid\r\n;\r\n``` | 2 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | MySQL || beats 100% || only with one select || inner join || easy-to-understand || best-solution | mysql-beats-100-only-with-one-select-inn-ncur | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Muzaffar_khuja | NORMAL | 2022-12-13T08:43:39.129279+00:00 | 2022-12-13T08:43:39.129316+00:00 | 931 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n100%\n\n- Space complexity:\n100%\n\n# Code\n```\nselect p.product_id, p.product_name from Product as p inner join Sales as s on \np.product_id=s.product_id group by p.product_id \nhaving min(sale_date)>=\'2019-01-01\' and max(sale_date)<=\'2019-03-31\'\n``` | 2 | 0 | ['MySQL'] | 1 |
sales-analysis-iii | MySQL | Somple solution w/ DISTINCT | mysql-somple-solution-w-distinct-by-lulz-18tq | That\'s all:\n\nSELECT DISTINCT s.product_id, p.product_name\nFROM Product p\nLEFT JOIN Sales s ON p.product_id = s.product_id\nWHERE\n sale_date BETWEEN \'2 | lulzseq | NORMAL | 2022-09-11T17:33:24.141166+00:00 | 2022-09-11T17:33:24.141205+00:00 | 542 | false | That\'s all:\n```\nSELECT DISTINCT s.product_id, p.product_name\nFROM Product p\nLEFT JOIN Sales s ON p.product_id = s.product_id\nWHERE\n sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\'\n AND s.product_id NOT IN (\n SELECT product_id\n FROM sales\n WHERE sale_date NOT BETWEEN \'2019-01-01\' AND \'2019-03-31\'\n );\n``` | 2 | 0 | ['MySQL'] | 1 |
sales-analysis-iii | easy solution | sql | easy-solution-sql-by-rohan_fasale-tfv3 | ~~~\nselect product.product_id,product_name from product\njoin sales on product.product_id=sales.product_id\ngroup by product.product_id\nhaving max(sale_date) | Rohan_Fasale | NORMAL | 2022-09-03T06:44:18.726107+00:00 | 2022-09-03T06:44:18.726135+00:00 | 226 | false | ~~~\nselect product.product_id,product_name from product\njoin sales on product.product_id=sales.product_id\ngroup by product.product_id\nhaving max(sale_date) <= "2019-03-31" and min(sale_date)>= "2019-01-01"\n~~~\n | 2 | 0 | [] | 0 |
sales-analysis-iii | 👇 MySQL 4 solutions | very easy 🔥 if | having | join | mysql-4-solutions-very-easy-if-having-jo-xa8b | \uD83D\uDE4B\uD83C\uDFFB\u200D\u2640\uFE0F Hello, here are my solutions to the problem.\n### Please upvote to motivate me post future solutions. HAPPY CODING \u | rusgurik | NORMAL | 2022-08-28T04:05:52.768898+00:00 | 2022-08-28T04:05:52.768925+00:00 | 527 | false | ### \uD83D\uDE4B\uD83C\uDFFB\u200D\u2640\uFE0F Hello, here are my solutions to the problem.\n### Please upvote to motivate me post future solutions. HAPPY CODING \u2764\uFE0F\n##### Any suggestions and improvements are always welcome.\n##### Solution 1: If, join, long \uD83E\uDD26\uD83C\uDFFB\u200D\u2640\uFE0F\n##### \u2705 Runtime: 1642 ms, faster than 32.63% of MySQL .\n\n```\nwith q as (\nselect \n product_id,\n sum(if(sale_date between \'2019-01-01\' and \'2019-03-31\', 1, 0)) sold_in_period,\n sum(if(sale_date not between \'2019-01-01\' and \'2019-03-31\', 1, 0)) sold_not_in_period\nfrom sales\ngroup by 1\nhaving sold_in_period > 0 and sold_not_in_period = 0\n)\nselect\n p.product_id,\n product_name\nfrom product p join q\non p.product_id = q.product_id\n```\n##### Solution 2: Same, but withoun join with from twice and work faster \uD83E\uDD37\uD83C\uDFFB\u200D\u2640\uFE0F\n##### \u2705 Runtime: 1016 ms, faster than 88.22% of MySQL.\n\n```\n# Write your MySQL query statement below\nwith q as (\nselect \n product_id,\n sum(if(sale_date between \'2019-01-01\' and \'2019-03-31\', 1, 0)) sold_in_period,\n sum(if(sale_date not between \'2019-01-01\' and \'2019-03-31\', 1, 0)) sold_not_in_period\nfrom sales\ngroup by 1\nhaving sold_in_period > 0 and sold_not_in_period = 0\n)\nselect\n p.product_id,\n product_name\nfrom product p, q\nwhere p.product_id = q.product_id\n```\n##### Solution 3: Same, with in close \uD83E\uDD37\uD83C\uDFFB\u200D\u2640\uFE0F\n##### \u2705 Runtime: 1157 ms, faster than 67.46% of MySQL.\n\n```\nwith q as (\nselect \n product_id,\n sum(if(sale_date between \'2019-01-01\' and \'2019-03-31\', 1, 0)) sold_in_period,\n sum(if(sale_date not between \'2019-01-01\' and \'2019-03-31\', 1, 0)) sold_not_in_period\nfrom sales\ngroup by 1\nhaving sold_in_period > 0 and sold_not_in_period = 0\n)\nselect\n product_id,\n product_name\nfrom product\nwhere product_id in (select product_id from q)\n```\n##### Solution 4: MIn and max, with join, work fast, but you can do it with where close or in close (work longer) \uD83E\uDD37\uD83C\uDFFB\u200D\u2640\uFE0F\n##### I really don\'t like joins, but work faster\n##### \u2705 Runtime: 996 ms, faster than 91.25% of MySQL.\n```\nwith q as (\nselect product_id\nfrom sales\ngroup by 1\nhaving min(sale_date) >= \'2019-01-01\' and max(sale_date) <= \'2019-03-31\'\n)\nselect\n p.product_id,\n product_name\nfrom product p join q\non p.product_id = q.product_id\n```\n\n##### If you like the solutions, please upvote \uD83D\uDD3C\n##### For any questions, or discussions, comment below. \uD83D\uDC47\uFE0F | 2 | 0 | ['MySQL'] | 2 |
sales-analysis-iii | Simple & Easy-to-Understand Solution | simple-easy-to-understand-solution-by-fr-wuc4 | \'\'\'\n\n\n\tselect s.product_id, p.product_name\n\tfrom sales s, product p\n\twhere s.product_id = p.product_id\n\tgroup by s.product_id, p.product_name\n\tha | freezing_stone | NORMAL | 2022-08-06T08:29:11.000356+00:00 | 2022-08-06T08:29:11.000382+00:00 | 208 | false | \'\'\'\n\n\n\tselect s.product_id, p.product_name\n\tfrom sales s, product p\n\twhere s.product_id = p.product_id\n\tgroup by s.product_id, p.product_name\n\thaving min(s.sale_date) >= \'2019-01-01\' \n\t\tand max(s.sale_date) <= \'2019-03-31\'\n\'\'\' | 2 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | MySQL Solution | mysql-solution-by-raito_16_01_2001-cyz3 | select distinct p1.product_id, p1.product_name\nfrom Product p1, Sales s1\nwhere p1.product_id not in(\n select product_id \n from Sales\n where sale_d | Raito_16_01_2001 | NORMAL | 2022-08-03T08:30:15.997981+00:00 | 2022-08-03T08:30:15.998015+00:00 | 235 | false | select distinct p1.product_id, p1.product_name\nfrom Product p1, Sales s1\nwhere p1.product_id not in(\n select product_id \n from Sales\n where sale_date not between \'2019-01-01\' and \'2019-03-31\')\n and p1.product_id = s1.product_id | 2 | 0 | [] | 0 |
sales-analysis-iii | ✅✅ EXPLAINED MySQL || JOINS && HAVING && GROUP BY | explained-mysql-joins-having-group-by-by-fr6i | \n\n# So this command will give wrong answer because as there are duplicate\n#in the sales table therefore if you see in example acc. to this query , G4 will\n | bhomik23 | NORMAL | 2022-07-27T09:48:22.171181+00:00 | 2022-07-27T09:49:49.149886+00:00 | 216 | false | ```\n\n# So this command will give wrong answer because as there are duplicate\n#in the sales table therefore if you see in example acc. to this query , G4 will\n#also be printed because it has been bought in the spring but if you see it is \n#also been bought after the spring therefore we have to use group by so that \n#we get onlt those products that have only been bought in the spring\n\n# SELECT s.product_id, p.product_name \n#FROM Product AS p INNER JOIN Sales AS s\n#ON s.product_id=p.product_id \n#WHERE s.sale_date >= \'2019-01-01\' AND s.sale_date <=\'2019-03-31\';\n\n\n#*************************************************************************************\n\n\n# therefore we will group on the basis of product ID and we will then check\n#what are the minimum and maximum sale date for each product and \n#thereofore if that product sale is in spring\n\nSELECT p.product_id, p.product_name \nFROM Product AS p INNER JOIN Sales AS s \nON s.product_id=p.product_id \nGROUP BY s.product_id \nHAVING MIN(s.sale_date)>=\'2019-01-01\' AND MAX(s.sale_date) <=\'2019-03-31\';\n``` | 2 | 0 | [] | 0 |
sales-analysis-iii | SQL for dummies! :) | sql-for-dummies-by-dennisdeneve-eoml | \tselect p.product_id, product_name \n\tfrom product p, sales s\n\twhere p.product_id = s.product_id\n\tgroup by p.product_id\n\thaving min(s.sale_date)>= \'201 | dennisdeneve | NORMAL | 2022-07-19T23:30:38.475826+00:00 | 2022-07-19T23:30:38.475858+00:00 | 165 | false | \tselect p.product_id, product_name \n\tfrom product p, sales s\n\twhere p.product_id = s.product_id\n\tgroup by p.product_id\n\thaving min(s.sale_date)>= \'2019-01-01\' and max(s.sale_date) <= \'2019-03-31\'; | 2 | 0 | [] | 0 |
sales-analysis-iii | [MySQL] Easy solution with explanation in comment | mysql-easy-solution-with-explanation-in-1cy3j | \nSELECT\n product_id, product_name\nFROM\n Product\nWHERE\n\t# First, find out the product that were sold in 2019 Q1\n product_id IN (\n SELECT | kevin-shu | NORMAL | 2022-07-07T07:02:00.068716+00:00 | 2022-07-10T16:52:20.147131+00:00 | 246 | false | ```\nSELECT\n product_id, product_name\nFROM\n Product\nWHERE\n\t# First, find out the product that were sold in 2019 Q1\n product_id IN (\n SELECT product_id\n FROM Sales\n WHERE\n sale_date BETWEEN \'2019-01-01\' AND \'2019-03-31\'\n )\n AND \n\t# Then, we have to eliminate the products which were sold besides 2019 Q1 \n\t# That is, we only pick those products which had never been sold "except 2019 Q1" (not between 2019/1/1 and 2019/3/31)\n product_id NOT IN (\n SELECT product_id\n FROM Sales\n WHERE\n sale_date NOT BETWEEN \'2019-01-01\' AND \'2019-03-31\'\n )\n``` | 2 | 0 | [] | 1 |
sales-analysis-iii | MySQL solution, using MIN only | mysql-solution-using-min-only-by-evanfan-5yx6 | \nselect s.product_id, p.product_name\nfrom Sales s\njoin Product p using (product_id)\ngroup by s.product_id having min(s.sale_date between \'2019-01-01\' and | evanfang | NORMAL | 2022-07-06T10:58:53.266624+00:00 | 2022-07-06T10:58:53.266663+00:00 | 170 | false | ```\nselect s.product_id, p.product_name\nfrom Sales s\njoin Product p using (product_id)\ngroup by s.product_id having min(s.sale_date between \'2019-01-01\' and \'2019-03-31\') = 1\n``` | 2 | 0 | [] | 2 |
sales-analysis-iii | MySQL | Simple Approach | mysql-simple-approach-by-divine_a-2mlg | ```\nSELECT s.product_id, product_name\nFROM Sales s LEFT JOIN Product p\nON s.product_id = p.product_id\nGROUP BY s.product_id\nHAVING MIN(sale_date) >= CAST(\ | Divine_A | NORMAL | 2022-07-02T10:25:57.105120+00:00 | 2022-07-02T10:25:57.105149+00:00 | 293 | false | ```\nSELECT s.product_id, product_name\nFROM Sales s LEFT JOIN Product p\nON s.product_id = p.product_id\nGROUP BY s.product_id\nHAVING MIN(sale_date) >= CAST(\'2019-01-01\' AS DATE) AND\nMAX(sale_date) <= CAST(\'2019-03-31\' AS DATE); | 2 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | My Simple SQL Solution | my-simple-sql-solution-by-user6774u-0d5j | \n\n\nSELECT\n p.product_id, p.product_name \nFROM \n Product p\nJOIN \n Sales s on p.product_id = s.product_id\nGROUP BY \n s.product_id\nHAVING \n | user6774u | NORMAL | 2022-06-27T07:02:04.533307+00:00 | 2022-06-27T07:02:04.533340+00:00 | 289 | false | \n```\n\nSELECT\n p.product_id, p.product_name \nFROM \n Product p\nJOIN \n Sales s on p.product_id = s.product_id\nGROUP BY \n s.product_id\nHAVING \n min(s.sale_date) >= \'2019-01-01\' \n and \n max(s.sale_date) <= \'2019-03-31\' \n; \n```\n\nPlease Upvote if you like the solution.\n | 2 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | MySQL solution faster than 97.93% | mysql-solution-faster-than-9793-by-darko-bpnm | \nSELECT \nDISTINCT # return just once if more than 1 is sold in requested time\nProduct.product_id, Product.product_name\nFROM\nProduct\nINNER JOIN\nSales\nON | darkoracic | NORMAL | 2022-06-24T19:12:51.472559+00:00 | 2022-06-24T19:12:51.472604+00:00 | 253 | false | ```\nSELECT \nDISTINCT # return just once if more than 1 is sold in requested time\nProduct.product_id, Product.product_name\nFROM\nProduct\nINNER JOIN\nSales\nON Product.product_id = Sales.product_id\nWHERE\nSales.sale_date BETWEEN CAST(\'2019-01-01\' AS DATE) AND CAST(\'2019-03-31\' AS DATE) # sale has to be in requested period\nAND\nProduct.product_id NOT IN # sale MUST NOT BE before OR after requested dates\n(SELECT Product.product_id\nFROM\nProduct\nINNER JOIN\nSales\nON Product.product_id = Sales.product_id\nWHERE\nSales.sale_date < CAST(\'2019-01-01\' AS DATE) \nOR \nSales.sale_date > CAST(\'2019-03-31\' AS DATE))\n``` | 2 | 0 | ['MySQL'] | 0 |
sales-analysis-iii | Mysql | mysql-by-ayush-sinha-gcp6 | SELECT product_id,product_name FROM Product\nLEFT JOIN Sales USING(product_id)\nGROUP BY product_id\nHAVING MIN(sale_date) >= \'2019-01-01\'AND MAX(sale_date) < | Ayush-sinha | NORMAL | 2022-06-16T07:03:40.470328+00:00 | 2022-06-16T07:03:40.470358+00:00 | 182 | false | SELECT product_id,product_name FROM Product\nLEFT JOIN Sales USING(product_id)\nGROUP BY product_id\nHAVING MIN(sale_date) >= \'2019-01-01\'AND MAX(sale_date) < \'2019-04-01\'\n \n | 2 | 0 | [] | 0 |
sales-analysis-iii | Two solution inspired by other solutions and as beginner | two-solution-inspired-by-other-solutions-z618 | \n\n\n# 1St Method\nSELECT product_id, product_name \nFROM Product\nWHERE product_id NOT IN\n(\nSELECT product_id FROM Sales WHERE sale_date < "2019-01-01" OR s | pk80103 | NORMAL | 2022-06-07T03:39:55.804320+00:00 | 2022-06-07T03:39:55.804362+00:00 | 176 | false | \n\n```\n# 1St Method\nSELECT product_id, product_name \nFROM Product\nWHERE product_id NOT IN\n(\nSELECT product_id FROM Sales WHERE sale_date < "2019-01-01" OR sale_date > "2019-03-31"\n);\n\n\n\n# 2nd Method using Group by\nSELECT Sales.product_id, Product.product_name \nFROM Sales \nJOIN Product \nON Product.product_id = Sales.product_id\nGROUP BY product_id \nHAVING MIN(sale_date) >= "2019-01-01" AND MAX(sale_date) <= "2019-03-31";\n``` | 2 | 0 | [] | 1 |
sales-analysis-iii | Sub Query and between | sub-query-and-between-by-aviskumar95-quim | select product_id,product_name from product\nwhere product_id not in\n(select product_id from sales\nwhere sale_date not between \'2019-01-01\' and \'2019-03-31 | aviskumar95 | NORMAL | 2022-05-03T17:38:57.638489+00:00 | 2022-05-03T17:38:57.638517+00:00 | 233 | false | select product_id,product_name from product\nwhere product_id not in\n(select product_id from sales\nwhere sale_date not between \'2019-01-01\' and \'2019-03-31\'); | 2 | 0 | ['MySQL'] | 0 |
exam-room | [C++/Java/Python] Straight Forward | cjavapython-straight-forward-by-lee215-l6wp | Very straight forward idea.\nUse a list L to record the index of seats where people sit.\n\nseat():\n 1. find the biggest distance at the start, at the end a | lee215 | NORMAL | 2018-06-17T03:12:49.176338+00:00 | 2018-10-24T00:13:02.450829+00:00 | 46,035 | false | Very straight forward idea.\nUse a list `L` to record the index of seats where people sit.\n\n`seat()`:\n 1. find the biggest distance at the start, at the end and in the middle.\n 2. insert index of seat\n 3. return index\n\n`leave(p)`: pop out `p`\n\n\n**Time Complexity**:\nO(N) for `seat()` and `leave()`\n\n\n**C++:**\n```\n int N;\n vector<int> L;\n ExamRoom(int n) {\n N = n;\n }\n\n int seat() {\n if (L.size() == 0) {\n L.push_back(0);\n return 0;\n }\n int d = max(L[0], N - 1 - L[L.size() - 1]);\n for (int i = 0; i < L.size() - 1; ++i) d = max(d, (L[i + 1] - L[i]) / 2);\n if (L[0] == d) {\n L.insert(L.begin(), 0);\n return 0;\n }\n for (int i = 0; i < L.size() - 1; ++i)\n if ((L[i + 1] - L[i]) / 2 == d) {\n L.insert(L.begin() + i + 1, (L[i + 1] + L[i]) / 2);\n return L[i + 1];\n }\n L.push_back(N - 1);\n return N - 1;\n }\n\n void leave(int p) {\n for (int i = 0; i < L.size(); ++i) if (L[i] == p) L.erase(L.begin() + i);\n }\n```\n\n**Java:**\n```\n int N;\n ArrayList<Integer> L = new ArrayList<>();\n public ExamRoom(int n) {\n N = n;\n }\n\n public int seat() {\n if (L.size() == 0) {\n L.add(0);\n return 0;\n }\n int d = Math.max(L.get(0), N - 1 - L.get(L.size() - 1));\n for (int i = 0; i < L.size() - 1; ++i) d = Math.max(d, (L.get(i + 1) - L.get(i)) / 2);\n if (L.get(0) == d) {\n L.add(0, 0);\n return 0;\n }\n for (int i = 0; i < L.size() - 1; ++i)\n if ((L.get(i + 1) - L.get(i)) / 2 == d) {\n L.add(i + 1, (L.get(i + 1) + L.get(i)) / 2);\n return L.get(i + 1);\n }\n L.add(N - 1);\n return N - 1;\n }\n\n public void leave(int p) {\n for (int i = 0; i < L.size(); ++i) if (L.get(i) == p) L.remove(i);\n }\n```\n**Python:**\n```\n def __init__(self, N):\n self.N, self.L = N, []\n\n def seat(self):\n N, L = self.N, self.L\n if not L: res = 0\n else:\n d, res = L[0], 0\n for a, b in zip(L, L[1:]):\n if (b - a) / 2 > d:\n d, res = (b - a) / 2, (b + a) / 2\n if N - 1 - L[-1] > d: res = N - 1\n bisect.insort(L, res)\n return res\n\n def leave(self, p):\n self.L.remove(p)\n``` | 202 | 31 | [] | 51 |
exam-room | [Python] O(log n) time for both seat() and leave() with heapq and dicts - Detailed explanation | python-olog-n-time-for-both-seat-and-lea-ytdo | We will name a range of unoccupied seats from i to j an Available segment with first index i and last index j. For each available segment we can say how far the | spatar | NORMAL | 2018-06-17T06:24:42.995019+00:00 | 2018-10-23T12:35:31.767334+00:00 | 23,176 | false | We will name a range of unoccupied seats from `i` to `j` an *Available segment* with first index `i` and last index `j`. For each available segment we can say how far the "best" seat in this segment is from the closest occupied seat. The number of empty seats in between is **priority** of the segment. The higher the priority the better seat you can get from a segment. For the edge cases when segment starts with index `0` or ends with index `N - 1` priority equals to `segment_size - 1`. For segments in the middle of the row priority can be calculated as `(segment_size - 1) // 2` or `(last - first) // 2`. Please note that two segments of different size may have equal priority. For example, segments with 3 seats and with 4 seats have the same priority "1".\n\n\nWe will use priority queue `self.heap` to store all currently available segments. Python implements `heapq` as **min heap**, so we will use negated priority to keep the best availabe segment on top of the queue. If two segments have equal priority, then the one with lower `first` index is better. Taken this into account, we will store availabale segment in `self.heap` as 4-items list: `[-segment_priority, first_index_of_segment, last_index_of_segment, is_valid]`. The first two items `-segment_priority` and `first_index_of_segment` guarantee correct priority queue order.\n\n\nA helper function `put_segment()` takes `first` and `last` index of the available segment, calculates its priority and pushes a list object into `self.heap`. In addition, it puts this list object into two dicts: `self.avail_first[first] = segment` and `self.avail_last[last] = segment`. These dicts will be used later in `leave()`.\n\n\nWe start with only one available segment [0, N - 1]. When `seat()` is called, we pop best available segment from `self.heap`. If segment\'s `is_valid` flag is `False` then we pop another one, until we get a valid available segment. There are two edge cases when popped segment starts at 0 or ends at N - 1. For these cases we return the edge seat number (0 or N - 1 respectively) and push new segment into `self.heap`. Otherwize, when the popped segment is in the middle of the row, we return its middle seat and create up to two new available segments of smaller size, and push them into `self.heap`.\n\n\nNow, `leave()` implementation is quite interesting. When seat `p` is vacated, we need to check if there are adjacent available segment(s) in the heap, and merge them together with `p`. We use dicts `self.avail_first` and `self.avail_last` to check for adjacent available segments. If these segment(s) are found, they need to be excluded from `self.heap`. Deleting items in `self.heap` will break heap invariant and requires subsequent `heapify()` call that executes in O(n log n) time. Instead we can just mark segments as invalid by setting `is_valid` flag: `segment[3] = False`. Invalid segments will be skipped upon `heappop()` in `seat()`.\n\n\n```python\nfrom heapq import heappop, heappush\n\n\nclass ExamRoom(object):\n\n def __init__(self, N):\n """\n :type N: int\n """\n self.N = N\n self.heap = []\n self.avail_first = {}\n self.avail_last = {}\n self.put_segment(0, self.N - 1)\n\n def put_segment(self, first, last):\n\n if first == 0 or last == self.N - 1:\n priority = last - first\n else:\n priority = (last - first) // 2\n\n segment = [-priority, first, last, True]\n\n self.avail_first[first] = segment\n self.avail_last[last] = segment\n\n heappush(self.heap, segment)\n\n def seat(self):\n """\n :rtype: int\n """\n while True:\n _, first, last, is_valid = heappop(self.heap)\n\n if is_valid:\n del self.avail_first[first]\n del self.avail_last[last]\n break\n\n if first == 0:\n ret = 0\n if first != last:\n self.put_segment(first + 1, last)\n\n elif last == self.N - 1:\n ret = last\n if first != last:\n self.put_segment(first, last - 1)\n\n else:\n ret = first + (last - first) // 2\n\n if ret > first:\n self.put_segment(first, ret - 1)\n\n if ret < last:\n self.put_segment(ret + 1, last)\n\n return ret\n\n def leave(self, p):\n """\n :type p: int\n :rtype: void\n """\n first = p\n last = p\n\n left = p - 1\n right = p + 1\n\n if left >= 0 and left in self.avail_last:\n segment_left = self.avail_last.pop(left)\n segment_left[3] = False\n first = segment_left[1]\n\n if right < self.N and right in self.avail_first:\n segment_right = self.avail_first.pop(right)\n segment_right[3] = False\n last = segment_right[2]\n\n self.put_segment(first, last)\n``` | 175 | 1 | [] | 23 |
exam-room | Python heapq clean solution with explanation, seat O(log n) leave O(n) | python-heapq-clean-solution-with-explana-qng8 | seat: O(log n)\nleave: O(n)\n\nThis solution is largely inspired by:\nhttps://leetcode.com/problems/exam-room/discuss/148595/Java-PriorityQueue-with-customized- | sfdye | NORMAL | 2018-10-25T12:11:15.651001+00:00 | 2020-03-01T01:43:20.366959+00:00 | 9,899 | false | seat: `O(log n)`\nleave: `O(n)`\n\nThis solution is largely inspired by:\nhttps://leetcode.com/problems/exam-room/discuss/148595/Java-PriorityQueue-with-customized-object.-seat:-O(logn)-leave-O(n)-with-explanation\n\nwith a few modification to work in Python. A few things I\'ve learned during the process:\n\n1. `heapq` implements a `minheap`, to use a `maxheap`, simpy negate the proirity (x -> -x)\n2. If you want to store the priority together with the data, you can use tuple as the element data structure, e.g. `heapq.heapify([(1, "data")])`\n3. In case there is a tie, you can specify a second priority, if you are using tuple that would be the second item, e.g. `(1, 1, 3) < (1, 2, 2)`\n4. If you remove an item from the underlying list (using either `del` or `.remove`), you have do `heapify` again on the list, otherwise it will lose the heap property.\n\n```python\nclass ExamRoom:\n \n def dist(self, x, y): # length of the interval (x, y)\n if x == -1: # note here we negate the value to make it maxheap\n return -y\n elif y == self.N:\n return -(self.N - 1 - x)\n else:\n return -(abs(x-y)//2) \n \n def __init__(self, N):\n self.N = N\n self.pq = [(self.dist(-1, N), -1, N)] # initialize heap\n \n def seat(self):\n _, x, y = heapq.heappop(self.pq) # current max interval \n if x == -1:\n seat = 0\n elif y == self.N:\n seat = self.N - 1\n else:\n seat = (x+y) // 2\n heapq.heappush(self.pq, (self.dist(x, seat), x, seat)) # push two intervals by breaking at seat\n heapq.heappush(self.pq, (self.dist(seat, y), seat, y))\n return seat\n \n def leave(self, p):\n head = tail = None\n for interval in self.pq: # interval is in the form of (d, x, y)\n if interval[1] == p: \n tail = interval\n if interval[2] == p: \n head = interval\n if head and tail:\n break\n self.pq.remove(head)\n self.pq.remove(tail)\n\t\theapq.heapify(self.pq) # important! re-heapify after deletion\n heapq.heappush(self.pq, (self.dist(head[1], tail[2]), head[1], tail[2]))\n``` | 142 | 0 | [] | 14 |
exam-room | C++ O(logn) seat() and O(logn) leave() with STL set and map | c-ologn-seat-and-ologn-leave-with-stl-se-5cm3 | Much similar to the solution of @spatar, but my solution uses set rather than priority_queue to achieve exactly O(logn) seat() instead of potentially calling he | szp14 | NORMAL | 2018-06-23T11:45:15.579669+00:00 | 2018-10-25T10:33:41.589962+00:00 | 10,836 | false | Much similar to the solution of [@spatar](https://leetcode.com/problems/exam-room/discuss/139941/Python-O(log-n)-time-for-both-seat()-and-leave()-with-heapq-and-dicts-Detailed-explanation), but my solution uses **set** rather than **priority_queue** to achieve exactly O(logn) seat() instead of potentially calling heappop() many times when calling seat() in spatar\'s solution.\n\nLike *Available segment* in spatar\'s solution, I choose *Interval* to indicate which seats are vacant, and implement it in **struct IntV**. My IntV has a static member called **N**, meaning there are N seats in total. I number these seats from 0 to N - 1, consistent with the result of test cases. IntV has two members named **l** and **r**, denoting that all seats in the interval \\[l, r\\] are empty, and either **l == 0** or the seat **l - 1** is taken, as well as either **r == N - 1** or the seat **r + 1** is taken. According to the definition of **interval** here, we can see no two intervals will intersect during the whole process of seating or leaving of students. Struct IntV needs three useful functions as described bellow:\n\n* **getD()**: Say we have an interval \\[l, r\\], this function returns the maximal distance to the closest person a student can achieve when he sits in one of the seats in this interval.\n* **getP()**: Like getD(), but this function returns the seat\'s number that student chooses to sit which can achieve the maximal distance to the closest person.\n* **operator <**: IntV does operator overloading to determine which interval we consider first: those intervals with greater return value of getD() should go first, and when several intervals share the same return value of getD(), the interval with the smallest left end (**l**) should go first.\n\nThese three functions are easy to implement, according to their descriptions.\n\nThen we turn to face **seat()** and **leave()**. I use a **set** (later on I will give the reason why I don\'t use **priority_queue**) named **tab** to record all intervals we have, and at the very beginning we only have \\[0, N - 1\\], a single interval. When we call **seat()**, we can get the first interval of **tab** by visiting iterator **tab.begin()**, which gives us maximal distance to the closest person currently. Then this interval splits into two smaller intervals (it\'s possible one or two of these smaller intervals have a length of 0, meaning no seats at all), so we need to delete the original interval from **tab** and insert two smaller intervals into **tab**. In case there are intervals with zero length, we can see that those intervals must have **l** > **r**, which will give it the lowest priority compared with other intervals (since nobody can sit there), so the implement of **getD()** will take this case into consideration.\n\nWhen we call **leave(int p)**, this question guarantees to have a student currently sitting in seat number p, so we now need to combine two intervals together to make a larger one. The question is: how could we know what these two intervals are? More precisely, among all intervals now in **tab**, we know there is a interval whose **r** is **p - 1**, what\'s its **l**? And there is a interval whose **l** is **p + 1**, what\'s its **r**? This requires us to have two **map**s called **l2r** and **r2l** to record the intervals currently in **tab**, so once we have a left end of a intervals, we can get the interval\'s right end by visiting **l2r**, and the same with **r2l**. So we know what these two interval are, one to the left of **p** and another to the right of **p**, and combine them together to have a larger interval. Then we delete these two smaller intervals from **tab** and insert the larger one. Since here we delete two certain elements from a **set**, it\'s much faster and more convenient than deleting them in **priority_queue**. The **l2r** and **r2l** will change according to the implement of **seat()** and **leave()**; \n\n```\nstruct IntV\n{\n static int N;\n int l, r;\n IntV(int l, int r) : l(l), r(r) {}\n int getD() const\n {\n if(l == 0) return r;\n if(r == N - 1) return N - 1 - l;\n if(r < l) return -1;\n else return (r - l) / 2;\n }\n int getP() const\n {\n if(l == 0) return 0;\n if(r == N - 1) return N - 1;\n else return l + (r - l) / 2;\n }\n bool operator < (const IntV& a) const\n {\n int d1 = getD(), d2 = a.getD();\n if(d1 != d2) return d1 > d2;\n return l < a.l;\n }\n};\n\nint IntV :: N = 0;\n\nclass ExamRoom {\npublic:\n set<IntV> tab;\n map<int, int> l2r, r2l;\n \n ExamRoom(int N)\n {\n IntV :: N = N;\n tab.clear(); l2r.clear(); r2l.clear();\n tab.insert(IntV(0, N - 1));\n l2r[0] = N - 1;\n r2l[N - 1] = 0;\n }\n \n int seat()\n {\n IntV cur = *(tab.begin());\n tab.erase(tab.begin());\n \n int pos = cur.getP();\n tab.insert(IntV(cur.l, pos - 1));\n tab.insert(IntV(pos + 1, cur.r));\n l2r[cur.l] = pos - 1; r2l[pos - 1] = cur.l;\n l2r[pos + 1] = cur.r; r2l[cur.r] = pos + 1;\n return pos;\n }\n \n void leave(int p)\n {\n int l = r2l[p - 1], r = l2r[p + 1];\n tab.erase(IntV(l, p - 1));\n tab.erase(IntV(p + 1, r));\n tab.insert(IntV(l, r));\n l2r[l] = r; r2l[r] = l;\n r2l.erase(p - 1); l2r.erase(p + 1);\n }\n};\n``` | 101 | 0 | [] | 9 |
exam-room | [Java] Two TreeSets Solution. Both Seat and Leave O(log n) Time Complexity | java-two-treesets-solution-both-seat-and-b000 | I was inspired by this solution. Please upvote his solution as well! Credit to @wangdeve.\n\nIn leave(seat) operation, we could take advantage of higher(seat) a | masonzhou | NORMAL | 2019-10-14T03:27:00.337398+00:00 | 2019-10-14T05:23:21.601305+00:00 | 6,695 | false | I was inspired by this [solution](https://leetcode.com/problems/exam-room/discuss/148595/Java-PriorityQueue-with-customized-object.-seat:-O(logn)-leave-O(n)-with-explanation). Please upvote his solution as well! Credit to [@wangdeve](https://leetcode.com/wangdeve "User").\n\nIn leave(seat) operation, we could take advantage of `higher(seat)` and `lower(seat)` methods in `TreeSet` to get the start and end position of the new merged interval.\n\n```\nclass ExamRoom {\n // O(logn) seat, O(logn) leave, n is the size of treesets.\n private TreeSet<Integer> takenSeats;\n private TreeSet<Interval> descIntervals;\n private int N;\n \n class Interval {\n int x, y, dis;\n String id; // for O(1) lookup\n\n public Interval(int x, int y) {\n this.x = x;\n this.y = y;\n if (x == -1) {\n this.dis = y;\n } else if (y == N) {\n this.dis = N - 1 - x;\n } else {\n this.dis = Math.abs(x - y) / 2; // works for both odd and even distance. The dis is to the closest person, no matter on the left or right.\n }\n this.id = x + "_" + y;\n }\n\n // this is for TreeSet remove (Object o) method.\n public boolean equals(Interval i) {\n return this.id.equals(i.id);\n }\n }\n\n public ExamRoom(int N) {\n this.takenSeats = new TreeSet<>();\n this.descIntervals = new TreeSet<>((a, b) -> a.dis != b.dis ? b.dis - a.dis : a.x - b.x);\n this.N = N;\n this.descIntervals.add(new Interval(-1, N));\n }\n \n\t// O(log n) time complexity with TreeSet add methods\n public int seat() {\n int pos = 0;\n Interval interval = this.descIntervals.pollFirst();\n if (interval.x == -1) {\n descIntervals.add(new Interval(0, interval.y));\n } else if (interval.y == this.N) {\n pos = N - 1;\n descIntervals.add(new Interval(interval.x, N - 1));\n } else {\n pos = interval.x + (interval.y - interval.x) / 2;\n descIntervals.add(new Interval(interval.x, pos));\n descIntervals.add(new Interval(pos, interval.y));\n }\n this.takenSeats.add(pos);\n return pos;\n }\n \n\t// O(log n) time complexity. TreeSet lower, higher, remove and add methods all have O(log n) time complexity.\n public void leave(int p) {\n int startIndex = this.takenSeats.lower(p) == null ? -1 : this.takenSeats.lower(p);\n int endIndex = this.takenSeats.higher(p) == null ? N : this.takenSeats.higher(p);\n mergeIntervals(new Interval(startIndex, p), new Interval(p, endIndex), this.descIntervals);\n this.takenSeats.remove(p);\n }\n \n // Time complexity of add and remove operations in TreeSet is O(log n)\n private void mergeIntervals(Interval i1, Interval i2, TreeSet<Interval> descIntervals) {\n descIntervals.remove(i1);\n descIntervals.remove(i2);\n descIntervals.add(new Interval(i1.x, i2.y));\n }\n}\n``` | 55 | 1 | [] | 9 |
exam-room | Java PriorityQueue Solution O(logN) for seat() O(N) for leave() | java-priorityqueue-solution-ologn-for-se-9n0f | seat(): Get the longest segment in pq, get the middle, update the pq\nleave(): Iterate the pq, get the two segments beside the p, update the pq\n\nSpent too muc | wangzi6147 | NORMAL | 2018-06-17T03:54:43.068599+00:00 | 2018-09-25T15:46:48.038678+00:00 | 11,971 | false | `seat()`: Get the longest segment in `pq`, get the middle, update the `pq`\n`leave()`: Iterate the `pq`, get the two segments beside the `p`, update the `pq`\n\nSpent too much time dealing with the the edge cases. So sad.\n\n```\nclass ExamRoom {\n \n private PriorityQueue<int[]> pq;\n private int N;\n\n public ExamRoom(int N) {\n this.N = N;\n pq = new PriorityQueue<>((a, b) -> {\n if ((b[1] - b[0]) / 2 == (a[1] - a[0]) / 2) {\n return a[0] - b[0];\n }\n return (b[1] - b[0]) / 2 - (a[1] - a[0]) / 2;\n });\n }\n \n public int seat() {\n if (pq.size() == 0) {\n pq.offer(new int[]{0, 2 * (N - 1)});\n return 0;\n } else {\n int[] longest = pq.poll();\n int result = longest[0] + (longest[1] - longest[0]) / 2;\n if (result != 0) { // result = 0, we don\'t need to add the left side\n pq.offer(new int[]{longest[0], result});\n }\n if (result != N - 1) { // result = N - 1, we don\'t need to add the right side\n pq.offer(new int[]{result, longest[1]});\n }\n return result;\n }\n }\n \n public void leave(int p) {\n if (pq.size() == 1 && (pq.peek()[1] >= N || pq.peek()[0] < 0)) { // Edge cases: Only [0, 2N] or [-N , N] in pq\n pq.clear();\n return;\n }\n int[] p1 = null, p2 = null; // p1: left side, p2: right side\n for (int[] pair : pq) {\n if (pair[1] == p) {\n p1 = pair;\n }\n if (pair[0] == p) {\n p2 = pair;\n }\n }\n if (p1 != null) {\n pq.remove(p1);\n }\n if (p2 != null) {\n pq.remove(p2);\n }\n if (p1 == null || p1[0] < 0) { // No left side found or p is the left most position in current seats.\n p1 = new int[]{-p2[1], p};\n }\n if (p2 == null || p2[1] >= N) { // No right side found or p is the right most position in current seats.\n p2 = new int[]{p, p1[0] + 2 * (N - p1[0] - 1)};\n }\n pq.offer(new int[]{p1[0], p2[1]});\n }\n}\n``` | 42 | 0 | [] | 10 |
exam-room | Java Solution based on treeset | java-solution-based-on-treeset-by-jainch-vct8 | Idea is to Use a TreeSet to store the current seat index in use. Whenever we want to add a seat, check the max distance adjacent indices in the treeset, whichev | jainchethan87 | NORMAL | 2018-06-17T03:28:23.406946+00:00 | 2018-10-25T01:49:00.917487+00:00 | 6,310 | false | Idea is to Use a [TreeSet](https://docs.oracle.com/javase/7/docs/api/java/util/TreeSet.html) to store the current seat index in use. Whenever we want to add a seat, check the max distance adjacent indices in the treeset, whichever is the max that is our answer.\nFor edge cases : \nWhen zero student: return 0;\nWhen there is one student., return 0 if the student is in second half or return N - 1 if the student is in first half.\nOtherwise, apply the idea,\n```\nclass ExamRoom {\n\n TreeSet<Integer> s = new TreeSet<>();\n int N;\n\n public ExamRoom(int N) {\n this.N = N;\n }\n\n public int seat() {\n\t//When no student\n if (s.isEmpty()) {\n s.add(0);\n return 0;\n }\n\t//When One student\n if (s.size() == 1) {\n int num = s.first();\n if (num < (N / 2)) {\n s.add(N - 1);\n return N - 1;\n } else {\n s.add(0);\n return 0;\n }\n }\n\t//When more than one student\n Iterator<Integer> it = s.iterator();\n int dist = -1;\n int position = -1;\n int left = it.next();\n\t//check the distance between 0 and first student\n if (left > 0) {\n position = 0;\n dist = left;\n }\n int right = -1;\n\t//Check the distance between adjacent indices,(already sorted)\n while (it.hasNext()) {\n right = it.next();\n if ((right - left) / 2 > dist) {\n dist = (right - left) / 2;\n position = left + dist;\n }\n left = right;\n }\n\t//check the distance between last student and (N - 1)\n if ((N - 1) - left > dist) {\n position = N - 1;\n }\n s.add(position);\n return position;\n }\n\n public void leave(int p) {\n s.remove(p);\n }\n}\n``` | 40 | 2 | [] | 2 |
exam-room | Easy to Understand Java Solution with Explanation - TreeSet | easy-to-understand-java-solution-with-ex-cbgw | What we really need in discussion section is a readable solution with detailed comments possible to make it look easier to understand, rather than just talking | rahul-raj | NORMAL | 2021-07-11T16:21:52.972708+00:00 | 2021-07-11T16:21:52.972755+00:00 | 3,750 | false | What we really need in discussion section is a readable solution with detailed comments possible to make it look easier to understand, rather than just talking about the same constraints already mentioned in the question! It is important to discuss the optimized solution, but it is also important to come up with a viable solution that can be formed within 20-25 minutes in the actual interview.\n\n```\n\nclass ExamRoom {\n\n int capacity; // this is just to keep track of the max allowed size\n TreeSet<Integer> seats; \n\t/* why treeset, because seats will be sorted automatically \n\t and get first()/last() element in log(n) time. \n\t*/\n \n public ExamRoom(int n) {\n this.capacity=n;\n this.seats = new TreeSet<>();\n }\n \n public int seat() {\n int seatNumber=0;\n\t\t/*\n\t\t Return 0 for first attempt ( as mentioned in question)\n\t\t Otherwise, we need to calculate the max distance by checking the whole treeset : O(n) time. \n\t\t Note that "distance" variable is initialized to first appearing seat,\n\t\t why because the distance calculation is based on current seat and the seat before that. \n\t\t Find the maximum distance and update the seat number accordingly.\n\t\t distance calculation -> (current seat - previous seat )/ 2\n\t\t Update the max distance at each step.\n\t\t New seat number will be -> previous seat number + max distance \n\t\t \n\t\t Now, before returning the end result, check for one more edge case:\n\t\t That is, if the max distance calculated is less than -> capacity-1-seats.last()\n\t\t \n\t\t Why because -> if last seat number in treeset is far from last position, \n\t\t then the largest distance possible is the last position.\n\t\t \n\t\t*/\n if(seats.size() > 0){\n Integer prev=null;\n int distance = seats.first();\n for(Integer seat : seats){\n if(prev != null){\n int d = (seat-prev)/2;\n if(distance < d){\n distance=d;\n seatNumber=prev+distance;\n }\n }\n prev = seat;\n }\n \n if(distance < capacity-1-seats.last()){\n seatNumber = capacity-1;\n }\n }\n seats.add(seatNumber);\n return seatNumber;\n }\n \n public void leave(int p) {\n seats.remove(p); \n\t\t/* simply remove the seat number from treeset\n\t\t and treeset will be automatically adjust its order in log(n) time. \n\t\t*/\n }\n}\n\n\n\n``` | 35 | 0 | ['Java'] | 6 |
exam-room | C++: Using ordered set | c-using-ordered-set-by-vmk1802-hp9v | Logic\n1. Ordered set is used to store occupied seats\n2. Distance between candidate seat and the nearest seat is calculated as below\n\t If end seats 0 or n-1 | vmk1802 | NORMAL | 2021-05-09T09:47:54.929832+00:00 | 2021-07-06T16:41:14.150252+00:00 | 3,372 | false | # Logic\n1. Ordered set is used to store occupied seats\n2. Distance between candidate seat and the nearest seat is calculated as below\n\t* If end seats 0 or n-1 are empty, the distance between the nearest seat is same as difference between occupied seat and the end candidate seat\n\t* For seats in between. the distance is half way between two occupied seats\n\nTime complexity of seat() routine is N seats multiplied by log(N) which is Nlog(N) in worst case.\nleave() routine time complexity is log(N)\n\n```\nclass ExamRoom {\nprivate:\n\tset<int> seats;\n\tint capacity;\n\npublic:\n\n\tExamRoom(int N)\n\t{\n\t\tcapacity = N;\n\t}\n\n\tint seat()\n\t{\n\t\tint dist = 0;\n\t\tint curr = 0;\n\n\t\tif (!seats.empty()) {\n\t\t\tauto itr = seats.begin();\n \n //calculate distance at the begining\n\t\t\tdist = *itr;\n\n\t\t\tif (dist == 0) {\n\t\t\t\titr++;\n\t\t\t}\n \n //calculate distance in between\n\t\t\twhile (itr != seats.end()) {\n int mid_dist = (*itr - *prev(itr)) / 2;\n\t\t\t\tif (dist < mid_dist) {\n\t\t\t\t\tdist = mid_dist;\n\t\t\t\t\tcurr = *prev(itr) + dist;\n\t\t\t\t}\n\t\t\t\titr++;\n\t\t\t}\n \n //calculate distance at the end\n\t\t\tif (dist < ((capacity - 1) - *(seats.rbegin()))) {\n\t\t\t\tcurr = capacity - 1;\n\t\t\t}\n\t\t}\n\n\t\treturn *(seats.insert(curr).first);\n\t}\n\n\tvoid leave(int p)\n\t{\n\t\tseats.erase(p);\n\t}\n};\n``` | 35 | 1 | ['C', 'Ordered Set', 'C++'] | 7 |
exam-room | My Elegant Java Solution Beats 99.84% | my-elegant-java-solution-beats-9984-by-s-qerh | Like other posts, the seat() is O(logn) and leave() is O(n). The pq only save Intervals that are available for later use.\n\nclass ExamRoom {\n class Interva | Shark-Chili | NORMAL | 2018-11-14T01:34:31.715956+00:00 | 2018-11-14T01:34:31.715995+00:00 | 3,308 | false | Like other posts, the `seat()` is O(logn) and `leave()` is O(n). The pq only save Intervals that are available for later use.\n```\nclass ExamRoom {\n class Interval {\n int start;\n int end;\n int length;\n public Interval(int start, int end) {\n this.start = start;\n this.end = end;\n if (start == 0 || end == N - 1) {\n this.length = end - start;\n } else {\n this.length = (end - start) / 2;\n }\n }\n }\n private PriorityQueue<Interval> pq;\n private int N;\n \n public ExamRoom(int N) {\n this.pq = new PriorityQueue<>((a, b) -> a.length != b.length ? b.length - a.length : a.start - b.start);\n this.N = N;\n pq.offer(new Interval(0, N - 1));\n }\n \n public int seat() {\n Interval in = pq.poll();\n int result;\n if (in.start == 0) {\n result = 0;\n } else if (in.end == N - 1) {\n result = N - 1;\n } else {\n result = in.start + in.length;\n }\n \n if (result > in.start) {\n pq.offer(new Interval(in.start, result - 1)); \n }\n if (in.end > result) {\n pq.offer(new Interval(result + 1, in.end)); \n }\n return result;\n }\n \n public void leave(int p) {\n List<Interval> list = new ArrayList(pq);\n Interval prev = null;\n Interval next = null;\n for (Interval in: list) {\n if (in.end + 1 == p) {\n prev = in;\n }\n if (in.start - 1 == p) {\n next = in;\n }\n }\n pq.remove(prev);\n pq.remove(next);\n pq.offer(new Interval(prev == null ? p : prev.start, next == null ? p : next.end));\n \n }\n}\n``` | 32 | 0 | [] | 4 |
exam-room | clean Java TreeSet solution | clean-java-treeset-solution-by-dictatorx-9n3n | \'\'\'\nclass ExamRoom {\n TreeSet set;\n int n;\n public ExamRoom(int N) {\n set = new TreeSet();\n n = N;\n }\n \n public int | dictatorx | NORMAL | 2019-01-24T01:20:39.050095+00:00 | 2019-01-24T01:20:39.050159+00:00 | 1,695 | false | \'\'\'\nclass ExamRoom {\n TreeSet<Integer> set;\n int n;\n public ExamRoom(int N) {\n set = new TreeSet();\n n = N;\n }\n \n public int seat() { \n if(set.size() == 0){\n set.add(0);\n return 0;\n } \n \n int gap = set.first(), sit = 0, pre = set.first();\n for(int cur: set){\n if(gap < (cur - pre) / 2){\n gap = (cur - pre) / 2;\n sit = (cur + pre) / 2;\n }\n pre = cur;\n }\n if(n - 1 - pre > gap)\n sit = n - 1;\n \n set.add(sit);\n return sit;\n }\n \n public void leave(int p) {\n set.remove(p);\n }\n}\n\'\'\'\nthe idea is to iterate the sorted set, find out the biggest gap and then the sit spot would be the middle of the gap.\n\nfor example: \n0, 9 -> the gap is (9 - 0) / 2 = 4, the seat would be (0 + 9) / 2 = 4\n0, 4, 9 -> two gaps are same, pick the front gap\n0, 2, 4, 9 -> gap between 4 and 9 is bigger\n\n0,1,2,3,4,5,6,7,8,9\nthe reason we divde (cur - pre) by 2 is to make the seat always closer to previous occupied seat. Even though 9 - 4 > 4 - 0, the shortest spare seats are same.\n\n | 21 | 0 | [] | 2 |
exam-room | Python - Intervals Based Problem - Max Heap ✅ | python-intervals-based-problem-max-heap-t0bgu | I was not able to come up with an optimal solution myself. I had to look at the discussion section but I hardly found any solution that had a good explanation f | itsarvindhere | NORMAL | 2023-10-22T16:53:50.507478+00:00 | 2023-10-22T17:01:21.564340+00:00 | 629 | false | I was not able to come up with an optimal solution myself. I had to look at the discussion section but I hardly found any solution that had a good explanation for the Max Heap approach.\n\nSo, after spending almost 2 hours understanding the approach, I will try my best to explain what is going on here.\n\nSee, first, we need to understand some things before we move on to Max Heap approach.\n\nLet\'s take an example.\n\n\tSuppose, we have 10 seats.\n\t\n\tNow, if a person comes, he should sit at seat "0" since no seat is occupied.\n\t\n\tAnother person comes. He should sit at seat "9" that is the last seat.\n\t\n\tNow, we have this interval [0,9] where "0" and "9" have person sitting\n\tAnd between these two, all the seats are available\n\t\n\tSo, next time a person comes, we can avoid manually checking each seat to see which is optimal one.\n\t\n\tWhen seats "0" and "9" are occupied, there are two seats that have same maximum closest distance\n\tThose are seats "4" and "5".\n\t\n\tFor seat "4", the closest occupied seat is seat "0" and so distance is "4"\n\tSimilarly For seat "5", the closest occupied seat is seat "9" and so distance is "4"\n\t\n\tBut, as the problem statement says, we have to choose the seat that has a smaller number.\n\tSo here, the person should sit at seat "4".\n\t\n\tNow, you can quickly get this optimal seat for any interval by just doing floor of (left+right) / 2\n\t\n\tHere, floor of (0 + 9)/2 => floor of 4.5 => 4 => MOST OPTIMAL SEAT for interval [0,9]\n\t\nSo, quickly finding the most optimal seat is one thing we need to figure out.\n\n\tNow that a person sits at seat "4" it means, the interval [0,9] is no longer valid\n\tBecause remember, an interval [left,right] means all seats between left and right are empty.\n\t\n\tSo, now that seat "4" is occupied, this means we now have two new intervals.\n\t\n\t[0,4] and [4,9]\n\t\n\tMakes sense, right?\n\t\n\tNow, the next time a person comes, we have to choose the most optimal seat from these two intervals.\n\t\n\tFor interval [0,4], the most optimal seat is floor of (0 + 4) / 2 => seat 2\n\tAnd for interval [4,9] the most optimal seat is floor of (4 + 9) / 2 => seat 6\n\t\n\tNow the thing is, for both the seats, the closest distance is the same.\n\t\n\tFor seat "2" closest occupied seat is "0" so distance => 2\n\tFor seat "6" closest occupied seat is "4" so distance => 2\n\t\n\tBut again, we have to choose the lower seat number.\n\t\n\tNow that seat "2" is occupied as well, it means interval [0,4] is no longer valid\n\tAnd now, two new intervals are there\n\t\n\t[0,2] and [2,4]\n\t\n\tThe issue now is, for large test cases, at any time, there can be a lot of intervals\n\tSo, we cannot just go over every single interval, \n\tget the most optimal seat, get the closest distance and so on\n\t\n\tWe somehow need a way such that at any time, we can pick that one interval \n\tfor which the closest distance is maximum. And even if multiple intervals have same distance\n\tthe interval that appears first is chosen.\n\t\nAnd that\'s where a Max Heap comes into the picture.\n\t\nWe can put our intervals in a Max heap which will order them based on the distance between most optimal seat and the seat closest to that most optimal seat. \n\n\tFor any interval we choose, we can not only find the most optimal seat but we can quickly find the closest distance as well.\n\t\n\tFor example, if we have interval [0,4]\n\t\n\tthe most optimal seat is floor of (0 + 4) / 2 => floor of (2) => 2\n\t\n\tAnd the closest distance is between "2" and "0" => 2\n\t\n\tThis can be calculated as floor of (4 - 0) / 2 => floor of 2 => 2\n\t\nAnd well, that\'s all we need to know before we start writing the code.\n\nNote that there will be some edge cases while calculating the most optimal seat and closest distance.\n\n\tFor example,\n\t\n\tInitially when no seat is occupied, the whole row from 0 to 9 is available\n\t\n\tIt means, the interval is [-1,10]\n\t\n\tSo, here, as per our formula, most optimal seat is floor of (-1 + 10) / 2 => floor of (9/2) => floor of 4.5 => 4\n\t\n\tBut that\'s wrong. Since all sets are empty, the most optimal is seat "0".\n\t\n\tAnd similarly, the formula for closest distance is floor of (10 - (-1)) / 2 -> floor of 11/2 => 5\n\tBut, the closest distance is actually "0" because there is no seat occupied yet apart from seat "0"\n\tSo there is no distance value yet.\n\t\n\tSo, that\'s one edge case.\n\n\tNow, suppose person sits at seat "0".\n\t\n\tNow, there will again be two new intervals [-1,0] and [0,10]\n\t\n\tFor interval [-1,0] the closest distance is "0" because "left" is -1\n\t\n\tIn fact, for any interval that starts with -1, the closest distance will be "right" value\n\tFor example if we have [-1,5] or [-1,4] and so on...\n\t\n\tBecause it makes sense right? If we have [-1,5] then the most optimal seat is "0" (since left is -1)\n\tand it will have closest occupied seat at number "5". So distance will be "5 - 0" => 5\n\t\n\tSimilarly, for interval [0,10] most optimal seat as per our formula is floor of (0 + 10) / 2 => 5\n\t\n\tBut, "10" here is the number of seats available. It means, "10" is not even a valid seat number.\n\tIt just means that right now, we just have a person sitting at seat "left" which here is seat "0".\n\t\n\tSo, in this case, the most optimal seat is at "9". \n\t\n\tAnd that\'s another edge case to take care of.\n\n\tAnd similarly, when we calculate the closest distance in this case, we don\'t do floor of (10 - 0) / 2\n\t\n\tAs that will again give is 5. \n\n\tRather, since the most optimal seat is "9", the closest distance will be "9 - 0" => 9 (Or 10 - 0 - 1)\n\t\n\tSo, for any interval that has "right" value as "n" or the number of seats,\n\tThe most optimal seat is "n - 1"\n\tand the closest distance is "n - 1 - left"\n\nSo far, I have talked about the scenario when a person sits at a seat.\n\nWhat about leaving?\n\n\tSuppose, we have three people sitting. So, "0", "4" and "9" are occupied.\n\n\tNow, person "4" leaves.\n\n\tIt means that now, the intervals [0,4] and [4,9] are no longer valid because "4" is now empty.\n\t\n\tSo, we have to remove these intervals from our Max Heap, that is, \n\tremove the interval starting with "4" and the one ending with "4"\n\t\n\tAnd once we do that, it now means the interval [0,9] is again a valid interval so we should push it back in the max heap.\n\t\n\tAnd again, we have to calculate the closest distance by the same logic.\n\t\nAnd well, that\'s the whole approach explained.\n\nI hope I was able to explain the Max Heap solution in detail. Still, if you have any doubts, please comment below. I will try to answer if I can.\n\n```\nclass ExamRoom: \n\n def __init__(self, n: int):\n # Number of seats\n self.n = n\n \n # Max Heap\n self.maxHeap = [] \n\n def seat(self) -> int:\n \n # If heap is empty, then the whole row of seats is an available interval to us\n if not self.maxHeap:\n # left and right boundaries of this interval\n left,right = -1, self.n\n \n # Otherwise, we take the interval with the maximum distance between left and right boundaries\n # Note that here, boundaries simply mean at "left" and "right" indices, seats are occupied\n # And between these indices, there are available seats\n else:\n top = heappop(self.maxHeap)\n left,right = top[1], top[2]\n \n \n # Now that we have "left" and "right", let\'s decide at what seat person will sit\n \n # If all seats are empty, sit at the seat 0\n if left == -1: seat = 0\n \n # If a person is sitting at "left" index but there is no occupied seat on "right", \n # Sit at right most seat, that is, n - 1\n elif right == self.n: seat = self.n - 1\n \n # If seat "left" and "right" are occupied\n # Sit at "floor of (left + right / 2)" seat\n else: seat = (left + right) // 2\n \n # Now that we know at what "seat" person will sit, two new intervals will be created\n # First is from "left" to "seat" and the other is from "seat" to "right"\n # So we push these two intervals to the Max Heap \n # The deciding value based on which these intervals will be ordered is closest distance\n # So, for interval [left,seat] this distance is floor(seat - left / 2)\n # For the interval [seat,right] this distance is floor(right - seat / 2)\n \n # Closest Distance for the Left Interval => [left,seat]\n closestDistanceForLeftInterval = (seat - left) // 2\n \n # If "left" is -1 then the most optimal seat is seat "0"\n\t\t# And since interval is [left,seat] the closest distance will be "seat" only\n if left == -1: closestDistanceForLeftInterval = seat\n \n # Closest Distance for the Right Interval => [seat,right]\n closestDistanceForRightInterval = (right - seat) // 2\n \n # If "right" is "n" then the interval is [seat, n]\n # So, for this interval, the most optimal seat is the seat n - 1\n # So, the distance from most optimal seat to the "seat" will be "self.n - 1 - seat"\n if right == self.n: closestDistanceForRightInterval = self.n - 1 - seat\n\n # Push these two intervals in the maxHeap\n heappush(self.maxHeap, [-closestDistanceForLeftInterval, left, seat])\n heappush(self.maxHeap, [-closestDistanceForRightInterval, seat, right])\n \n # Return the seat number at which student will sit\n return seat\n \n\n def leave(self, p: int) -> None:\n \n # Now, just imagine what happens when a person leaves a seat\n # For example, if at any point, we have three person sitting in a row of 10 seats\n # Then seats "0", "4" and "9" will be occupied\n # And if person "4" leaves, it means, \n\t\t# we should now remove the intervals from maxHeap which either start with "4" or end with "4"\n # Because those intervals are no longer valid now that the seat "4" itself is empty\n \n intervalStartsAtP = None\n intervalEndsAtP = None\n \n # Let\'s just loop over the maxHeap since it is just a list internally\n for interval in self.maxHeap:\n # If this interval starts with "p"\n # This should be removed\n if interval[1] == p: intervalStartsAtP = interval\n \n # If this interval ends with "p"\n # This should be removed\n if interval[2] == p: intervalEndsAtP = interval\n \n # Remove the two\n self.maxHeap.remove(intervalStartsAtP)\n self.maxHeap.remove(intervalEndsAtP)\n \n # And now, the order is not proper so we have to manually call heapify here\n heapify(self.maxHeap)\n \n # Finally, since we remove the two intervals, it means a bigger interval is now available to us\n # Imagine that the two intervals we removed are [left, p] and [p, right]\n # So, it means [left,right] is now a new larger interval available to us, right?\n # So, put it in the heap\n left = intervalEndsAtP[1]\n right = intervalStartsAtP[2]\n \n # The same logic to get the closest distance as in seat() method\n closestDistance = (right - left) // 2\n \n # If "left" is -1 then the most optimal option is seat "0"\n # It means, if next person sits at seat "0", the closest seat will be at "right"\n # So closest distance here is "right"\n if left == -1: closestDistance = right\n \n # If "right" is "n" then the interval is [left, n]\n # So, for this interval, the most optimal seat is the seat n - 1\n # So, the distance from most optimal seat to the "left" will be "self.n - 1 - left"\n elif right == self.n: closestDistance = self.n - 1 - left\n \n # Push this interval in the maxHeap with its closest distance value\n heappush(self.maxHeap, [-closestDistance, left, right])\n``` | 18 | 0 | ['Heap (Priority Queue)', 'Python'] | 3 |
exam-room | Java TreeMap solution, O(logn) for each operation | java-treemap-solution-ologn-for-each-ope-at60 | The major idea is to store intervals into the TreeMap, every time when we add a seat, we get the longest interval and put a seat in the middle, then break the i | laughingwithwallace | NORMAL | 2018-06-17T04:32:43.056665+00:00 | 2018-10-08T15:54:29.866959+00:00 | 3,452 | false | The major idea is to store intervals into the TreeMap, every time when we add a seat, we get the longest interval and put a seat in the middle, then break the interval into two parts. Notice that for intervals with odd length, we simply trim it into even length using len -= len & 1.\n```\nclass ExamRoom {\n\n TreeMap<Integer, TreeSet<Integer>> map = new TreeMap<>();\n TreeSet<Integer> set = new TreeSet<>();\n int N;\n \n public ExamRoom(int N) {\n this.N = N;\n }\n \n public int seat() {\n int res = 0;\n if (set.size() == 0) {\n res = 0;\n } else {\n int first = set.first(), last = set.last();\n Integer max = map.isEmpty() ? null : map.lastKey();\n if (max == null || first >= max / 2 || N - 1 - last > max / 2) {\n if (first >= N - 1 - last) {\n addInterval(0, first);\n res = 0;\n } else {\n addInterval(last, N - 1 - last);\n res = N - 1;\n }\n } else {\n TreeSet<Integer> temp = map.get(max);\n int index = temp.first();\n int next = set.higher(index);\n int mid = (next + index) / 2;\n removeInterval(index, max);\n addInterval(index, mid - index);\n addInterval(mid, next - mid);\n res = mid;\n }\n }\n set.add(res);\n return res;\n }\n\t\t\n public void leave(int p) {\n Integer pre = set.lower(p);\n Integer next = set.higher(p);\n set.remove(p);\n if (next != null) {\n removeInterval(p, next - p);\n }\n if (pre != null) {\n removeInterval(pre, p - pre);\n if (next != null) {\n addInterval(pre, next - pre);\n }\n }\n }\n\t\t\n private void addInterval(int index, int len) {\n len -= len & 1; // trim to even number\n map.putIfAbsent(len, new TreeSet<>());\n map.get(len).add(index);\n }\n \n private void removeInterval(int index, int len) {\n len -= len & 1;\n Set<Integer> temp = map.get(len);\n if (temp == null) {\n return;\n }\n temp.remove(index);\n if (temp.size() == 0) {\n map.remove(len);\n }\n }\n}\n``` | 17 | 0 | [] | 3 |
exam-room | Python 3 || w/ comments | python-3-w-comments-by-spaulding-6ub2 | \nclass ExamRoom:\n\n def __init__(self, n: int):\n self.n, self.room = n, []\n\n\n def seat(self) -> int:\n if not self.room: ans = 0 | Spaulding_ | NORMAL | 2023-01-27T20:42:10.629615+00:00 | 2024-08-04T23:54:25.114449+00:00 | 2,632 | false | ```\nclass ExamRoom:\n\n def __init__(self, n: int):\n self.n, self.room = n, []\n\n\n def seat(self) -> int:\n if not self.room: ans = 0 # sit at 0 if empty room \n\n else:\n dist, prev, ans = self.room[0], self.room[0], 0 # set best between door and first student \n\n for curr in self.room[1:]: # check between all pairs of students \n d = (curr - prev)//2 # to improve on current best\n\n if dist < d: dist, ans = d, (curr + prev)//2\n prev = curr\n\n if dist < self.n - prev-1: ans = self.n - 1 # finally, check whether last seat is best\n\n insort(self.room, ans) # sit down in best seat\n\n return ans\n\n def leave(self, p: int) -> None:\n self.room.remove(p)\n```\n[https://leetcode.com/problems/exam-room/submissions/1272745920/](https://leetcode.com/problems/exam-room/submissions/1272745920/)\n\n | 15 | 0 | ['Python3'] | 1 |
exam-room | C++ O(log N) for each operation, using BST(std::set) and hash table | c-olog-n-for-each-operation-using-bststd-1w0c | After inserted some students, the seats would be divided into many intervals.\nWe can sort these intervals, using a complex compare function.\nTo keep the sorte | zhoubowei | NORMAL | 2018-06-19T17:51:26.293727+00:00 | 2018-10-08T03:59:53.545733+00:00 | 2,471 | false | After inserted some students, the seats would be divided into many intervals.\nWe can sort these intervals, using a complex compare function.\nTo keep the sorted order after "insert" and "remove" operations, I use a BST to store these intervals.\n\nInsert operation:\n1. find the best interval (O(log n))\n2. get the middle point of the interval, split it into two intervals and insert them to the BST, remove old interval (O(log n))\n\nRemove operation (given a seat p):\n1. find the two intervals which are connected with p (O(1))\n2. delete the two intervals in the BST, and modify them in the hash table.(O(log n))\n\nTo find which interval to remove, I use a hash table, unordered_map<int, pair<set<Int, IntCmp>::iterator, set<Int, IntCmp>::iterator>>.\n\n```\nclass Int { // interval\npublic:\n Int(int a, int b, int n) {\n set(a, b, n);\n }\n void set(int a, int b, int n) {\n st = a;\n ed = b;\n if (a == INT_MIN) {\n mid = 0;\n if (b == INT_MAX) {\n dis = 0;\n return;\n }\n dis = b;\n return;\n }\n if (b == INT_MAX) {\n mid = n - 1;\n dis = mid - a;\n return;\n }\n\n mid = (a + b) >> 1;\n dis = min(mid - st, ed - mid);\n }\n int st;\n int ed;\n int mid;\n int dis;\n};\n\nstruct IntCmp {\n bool operator()(const Int& a, const Int& b) const { \n if (a.dis != b.dis) {\n return a.dis > b.dis;\n }\n if (a.mid != b.mid) {\n return a.mid < b.mid; \n }\n return (a.st == b.st) ? (a.ed < b.ed) : (a.st < b.st);\n }\n};\n\nclass ExamRoom {\npublic:\n int n;\n unordered_map<int, pair<set<Int, IntCmp>::iterator, set<Int, IntCmp>::iterator>> m;\n set<Int, IntCmp> s;\n \n ExamRoom(int N) {\n n = N;\n s.insert(Int(INT_MIN, INT_MAX, n));\n }\n \n int seat() {\n auto it = s.begin();\n auto it1 = s.insert(Int(it->st, it->mid, n));\n auto it2 = s.insert(Int(it->mid, it->ed, n));\n int pos = it->mid;\n m[pos] = make_pair(it1.first, it2.first);\n m[it->st].second = s.find(Int(it->st, it->mid, n));\n m[it->ed].first = s.find(Int(it->mid, it->ed, n));\n s.erase(it);\n return pos;\n }\n \n void leave(int p) {\n int st = m[p].first->st;\n int ed = m[p].second->ed;\n s.erase(m[p].first);\n s.erase(m[p].second);\n auto it = s.insert(Int(st, ed, n));\n m.erase(p);\n m[st].second = it.first;\n m[ed].first = it.first;\n }\n};\n\n/**\n * Your ExamRoom object will be instantiated and called as such:\n * ExamRoom obj = new ExamRoom(N);\n * int param_1 = obj.seat();\n * obj.leave(p);\n */\n``` | 15 | 0 | [] | 1 |
exam-room | [Python] O(log n) time for both seat() and leave() with a customized heap implementation | python-olog-n-time-for-both-seat-and-lea-6l3c | This solution is similar to https://leetcode.com/problems/exam-room/discuss/139941/Python-O(log-n)-time-for-both-seat()-and-leave()-with-heapq-and-dicts-Detaile | zhukov | NORMAL | 2018-06-24T08:43:04.742604+00:00 | 2018-10-10T11:53:06.481473+00:00 | 2,147 | false | This solution is similar to https://leetcode.com/problems/exam-room/discuss/139941/Python-O(log-n)-time-for-both-seat()-and-leave()-with-heapq-and-dicts-Detailed-explanation/147981. Many thanks to [@spatar](https://leetcode.com/spatar). \n\nThe main difference is that we will delete the segments from the heap when they become invalid due to the leave. Since the customized implementation of heap is not as optimized as heapq, the running time of this solution (116ms) is longer than the original heapq solution (72ms).\n\n```\nclass IndexedHeap:\n """This is a simple implementation of a min/max-heap.\n Note that we assume that each item of the heap is a list\n where its last element indicates the index of the item\n in the heap.\n """\n def __init__(self, reverse=False):\n # By default this heap is a min-heap. But if the flag \n # reverse is True, it will be a max-heap.\n self._reverse = reverse\n \n \n def _lt(self, x, y):\n if not self._reverse:\n return x < y\n else:\n return x > y\n \n \n def _le(self, x, y):\n if not self._reverse:\n return x <= y\n else:\n return x >= y\n \n \n def _percolate_up(self, heap):\n """Percolate the last item up to the appropriate \n position of the heap.\n """\n curr_index = len(heap) - 1\n while curr_index > 0:\n parent_index = (curr_index - 1) // 2\n if self._le(heap[parent_index], heap[curr_index]):\n break\n else:\n tmp = heap[parent_index]\n heap[parent_index] = heap[curr_index]\n heap[curr_index] = tmp \n # Swap back the indexes of the elements in the heap.\n heap[parent_index][-1] = parent_index\n heap[curr_index][-1] = curr_index\n \n curr_index = parent_index\n \n \n def _get_min_child(self, heap, item_index):\n """Get the index of the minimum child of an item.\n If the item doesn\'t have any child, we will return -1.\n """\n left_child = 2*item_index + 1\n if left_child > len(heap) - 1:\n return -1\n \n ret_index = left_child\n right_child = 2*item_index + 2\n if right_child < len(heap) and self._lt(heap[right_child], heap[left_child]):\n ret_index = right_child\n \n return ret_index\n \n \n def _percolate_down(self, heap, item_index):\n """Percolate the element at item_index down to the appropriate \n position of the heap.\n """\n curr_index = item_index\n min_child_index = self._get_min_child(heap, curr_index)\n while min_child_index != -1:\n if self._lt(heap[curr_index], heap[min_child_index]):\n break\n else:\n tmp = heap[min_child_index]\n heap[min_child_index] = heap[curr_index]\n heap[curr_index] = tmp\n # Swap back the indexes of the elements in the heap.\n heap[min_child_index][-1] = min_child_index\n heap[curr_index][-1] = curr_index\n \n curr_index = min_child_index\n min_child_index = self._get_min_child(heap, curr_index)\n \n \n def heaplify(self, heap_to_be):\n """Create a heap in-place from a list."""\n # Initialize the indexes of the elements in the heap.\n for index in range(len(heap_to_be)):\n heap_to_be[index][-1] = index \n \n curr_index = len(heap_to_be) // 2 - 1\n while curr_index >= 0:\n self._percolate_down(heap_to_be, curr_index)\n curr_index -= 1\n \n \n def heappush(self, heap, item):\n """Push one item into the heap."""\n item[-1] = len(heap)\n heap.append(item)\n # Percolate up the item from the end of the heap.\n self._percolate_up(heap)\n \n \n def heappop(self, heap):\n """Pop the root (i.e., the minimum/maximum item) out of the heap."""\n if not heap:\n raise IndexError(\'empty heap\')\n \n ret = heap[0]\n heap[0] = heap[-1]\n # Reset the index of the root.\n heap[0][-1] = 0\n heap.pop()\n \n self._percolate_down(heap, 0)\n \n return ret\n \n \n def heapdelete(self, heap, item_index):\n """Delete the item at item_index from the heap."""\n if item_index < 0 or item_index >= len(heap):\n return\n \n # Move the last item to item_index.\n heap[item_index] = heap[-1]\n # Reset the index of the item at item_index.\n heap[item_index][-1] = item_index\n # Remove the last item from the heap.\n heap.pop()\n \n self._percolate_down(heap, item_index)\n \n return\n\n\nclass ExamRoom(object):\n def __init__(self, N):\n """\n :type N: int\n """\n self.N = N\n # This max-heap keeps a list of empty segments where the segments\n # are compared firstly with their lengths and then with its start indice\n # if their lengths are the same.\n self.heap = []\n # A dict from the segment start index to the segment\n self.avail_first = {}\n # A dict from the segment end index to the segment\n self.avail_last = {}\n # Create the min-heap helper.\n self.min_h = IndexedHeap()\n # Initially we have one big empty segment [0, N - 1].\n self.put_segment(0, self.N - 1)\n \n\n def put_segment(self, first, last):\n # Calculate the priority of the segment where the priority is the maximum \n # closest distance if the new student sits at one seat in the segment.\n if first == 0 or last == self.N - 1:\n # If the empty segment includes either 0 or N - 1\n priority = last - first\n else:\n priority = (last - first) // 2\n\n # segment[0] = priority;\n # segment[1] = the starting seat of the segment\n # segment[2] = the ending seat of the segment\n # segment[3] = the index of this segment in the heap.\n # Note that:\n # (1) Take the negative of priority because we want to use a min-heap.\n # (2) The segment index will be set in IndexedHeap.heappush().\n segment = [-priority, first, last, -1]\n\n self.avail_first[first] = segment\n self.avail_last[last] = segment\n\n self.min_h.heappush(self.heap, segment)\n\n def seat(self):\n """\n :rtype: int\n """\n # Get the segment with the maximum length.\n _, first, last, _ = self.min_h.heappop(self.heap)\n self.avail_first.pop(first)\n self.avail_last.pop(last)\n\n if first == 0:\n # Seat the student at 0.\n ret = 0\n if first != last:\n self.put_segment(first + 1, last)\n elif last == self.N - 1:\n # Seat the student at N - 1.\n ret = last\n if first != last:\n self.put_segment(first, last - 1)\n else:\n # Seat the student in the middle of the segment.\n ret = first + (last - first) // 2\n \n if ret > first:\n self.put_segment(first, ret - 1)\n if ret < last:\n self.put_segment(ret + 1, last)\n\n return ret\n\n def leave(self, p):\n """\n :type p: int\n :rtype: void\n """\n # Removing p at least leads to a hole at p.\n first = p\n last = p\n\n # Check if we need to delete the left and right segments \n # neighboring p.\n left = p - 1\n right = p + 1\n\n if left >= 0 and left in self.avail_last:\n # Delete the left segment if it exists. Note that we don\'t \n # delete the segment from the map self.avail_first since \n # its value will be overwritten in self.put_segment().\n segment_left = self.avail_last.pop(left)\n first = segment_left[1]\n # Instead of marking the segment as invalid, we delete \n # the segment from the heap.\n self.min_h.heapdelete(self.heap, segment_left[3])\n \n if right < self.N and right in self.avail_first:\n # Delete the right segment if it exists. Note that we don\'t \n # delete the segment from the map self.avail_last since \n # its value will be overwritten in self.put_segment(). \n segment_right = self.avail_first.pop(right)\n last = segment_right[2]\n # Instead of marking the segment as invalid, we delete \n # the segment from the heap.\n self.min_h.heapdelete(self.heap, segment_right[3])\n\n # Push the new segment into the heap.\n self.put_segment(first, last)\n``` | 12 | 0 | [] | 1 |
exam-room | Very simple C++ solution using set | very-simple-c-solution-using-set-by-pras-qh8v | \npublic:\n int last;\n set<int> s;\n ExamRoom(int n) {\n last = n-1;\n }\n \n int seat() {\n if(s.empty()){ //if the set is em | prashantarya10 | NORMAL | 2022-03-14T05:49:15.739499+00:00 | 2022-03-15T03:32:13.318876+00:00 | 1,235 | false | ```\npublic:\n int last;\n set<int> s;\n ExamRoom(int n) {\n last = n-1;\n }\n \n int seat() {\n if(s.empty()){ //if the set is empty, occupy the first seat\n s.insert(0);\n return 0;\n }\n \n //now we iterate over all the intervals\n //to check which interval produces the greatest difference\n \n int left=0, diff=INT_MIN, pos=-1;\n \n //if 0 is not in set then we have to check the interval [0, first occupied seat] also \n //if this interval produces the maximum difference then the student must occupy the 0th seat\n if(s.find(0)==s.end()){\n if(*s.begin()>diff){\n diff = *s.begin();\n pos = 0;\n }\n }\n \n //this loop checks the intervals like [1st occupied seat, 2nd occupied seat],\n //[2nd occupied seat, 3rd occuspied seat]...[second last occupied seat, last occupied seat]\n //if an interval in this loop produces the maximum difference then the student must occupy the seat at the mid of the interval\n for(int right: s){\n if((right-left)/2>diff){\n diff=(right-left)/2;\n pos = left + (right-left)/2;\n }\n left = right;\n }\n \n //if the n-1th seat is not occupied then we have to check the interval [last occupied seat, n-1] as well\n //if this interval produces the maximum difference then the student must occupy the (n-1)th seat\n if(s.find(last)==s.end()){\n if(last-left>diff){\n diff = last-left;\n pos = last;\n }\n }\n s.insert(pos);\n return pos;\n }\n \n void leave(int p) {\n s.erase(p);\n }\n};\n```\n | 11 | 0 | ['C', 'Ordered Set'] | 0 |
exam-room | Python & Javascript // short and simple solution // similar to prob. no. 849 | python-javascript-short-and-simple-solut-3oyv | Python\n\nclass ExamRoom:\n\n def __init__(self, N):\n self.seated, self.n = [], N - 1\n \n\n def seat(self):\n if not self.seated:\n | cenkay | NORMAL | 2018-06-17T04:02:23.526392+00:00 | 2018-06-17T04:02:23.526392+00:00 | 1,324 | false | * Python\n```\nclass ExamRoom:\n\n def __init__(self, N):\n self.seated, self.n = [], N - 1\n \n\n def seat(self):\n if not self.seated:\n self.seated += 0,\n return 0\n mx = ind = 0\n for i in range(1, len(self.seated)):\n l, r = self.seated[i - 1], self.seated[i]\n if (r - l) // 2 > mx:\n mx = (r - l) // 2\n ind = l + mx\n if self.seated[-1] != self.n and self.n - self.seated[-1] > mx:\n mx, ind = self.n - self.seated[-1], self.n\n if self.seated[0] >= mx:\n mx, ind = self.seated[0], 0\n self.seated.append(ind)\n self.seated.sort()\n return ind\n \n \n def leave(self, p):\n self.seated.remove(p)\n\n```\n* Javascript\n```\nvar ExamRoom = function(N) {\n this.seats = [], this.n = N - 1;\n};\n\n\nExamRoom.prototype.seat = function() {\n if (this.seats.length == 0) {\n this.seats = [0];\n return 0;\n }\n let mx = ind = 0, n = this.seats.length;\n for (let i = 1; i < n; i++) {\n l = this.seats[i - 1], r = this.seats[i];\n d = Math.floor((r - l) / 2);\n if (d > mx) {\n mx = d, ind = l + d;\n }\n }\n if (this.seats[n - 1] != this.n && this.n - this.seats[n - 1] > mx) {\n mx = this.n - this.seats[n - 1], ind = this.n;\n }\n if (this.seats[0] >= mx) {\n mx = this.seats[0], ind = 0;\n }\n this.seats.push(ind);\n this.seats.sort(function (a, b) {return a - b;});\n return ind;\n};\n\n\nExamRoom.prototype.leave = function(p) {\n this.seats.splice(this.seats.indexOf(p), 1);\n};\n``` | 9 | 0 | [] | 0 |
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