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group-sold-products-by-the-date | 2 Simple SQL Solutions: Group_Concat() and String_Agg() | 2-simple-sql-solutions-group_concat-and-vvn8i | \nSELECT \n sell_date,\n COUNT(product) AS num_sold,\n STRING_AGG(product, \',\') WITHIN GROUP (ORDER BY product ASC) AS products\nFROM (SELECT DISTINCT * FR | mirandanathan | NORMAL | 2021-02-20T19:11:14.699971+00:00 | 2021-02-20T19:11:35.272613+00:00 | 1,814 | false | ```\nSELECT \n sell_date,\n COUNT(product) AS num_sold,\n STRING_AGG(product, \',\') WITHIN GROUP (ORDER BY product ASC) AS products\nFROM (SELECT DISTINCT * FROM Activities) T\nGROUP BY sell_date\nORDER BY 1 ASC\n```\n\n```\nSELECT \n sell_date,\n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product ORDER BY product ASC) AS products\nFROM Activities\nGROUP BY sell_date\nORDER BY 1 ASC\n``` | 15 | 2 | [] | 3 |
group-sold-products-by-the-date | STRING_AGG + Explanation | string_agg-explanation-by-alirezah-ubkq | This query will show the disired result: \n select sell_date, count(distinct(product)) as num_sold,\n STRING_AGG(product,\',\') within group (order by product) | Alirezah | NORMAL | 2022-07-10T18:37:20.095072+00:00 | 2022-07-10T18:40:40.334501+00:00 | 2,341 | false | This query will show the disired result: \n select sell_date, count(distinct(product)) as num_sold,\n STRING_AGG(product,\',\') within group (order by product) as products\nfrom\n(SELECT DISTINCT sell_date, product FROM Activities) T\ngroup by sell_date\norder by sell_date\n\nthe reason for using this part ((SELECT DISTINCT sell_date, product FROM Activities) T) is that \nneed for bring distinct value on product name. (SQL server can not understand Distinct in the STRING_AGG, so need some query to send distinct value)\n\nin addition, postgres sql can understand Distinct in STRING_AGG, so you can use this query instead of top in postgrest sql:\n\n select sell_date, count(distinct(product)) as num_sold,\n STRING_AGG(distinct(product), \',\'\n order by product)\n as products\nfrom Activities\ngroup by sell_date\norder by sell_date\nAlso you can check this link: \nhttps://gregdodd.me/2021/08/24/distinct-list-in-string_agg/\n\nPlease UpVote if it was Helpful\n\n | 13 | 0 | ['MS SQL Server'] | 1 |
group-sold-products-by-the-date | Grouping With Aggregations 🏆 | grouping-with-aggregations-by-hridoy100-mhxc | My SQL\n\nFirst, we group the data by the sell_date column. This allows us to count the number of unique products sold on each sell date, which we store in the | hridoy100 | NORMAL | 2023-08-11T20:46:55.380098+00:00 | 2023-08-11T20:46:55.380124+00:00 | 1,533 | false | # My SQL\n\nFirst, we group the data by the `sell_date` column. This allows us to count the number of unique products sold on each sell date, which we store in the `num_sold` column.\n\nThe most challenging part is to sort and join all unique product names in each group to get the products column. We can use the `GROUP_CONCAT()` function to combine multiple values from multiple rows into a single string. The syntax of the `GROUP_CONCAT()` function is as follows:\n```\nGROUP_CONCAT(DISTINCT [column_name] [separator])\n```\nThe `DISTINCT` keyword ensures that only unique values are concatenated. The separator parameter specifies the character that should be used to separate the values.\n\n``` sql []\n# Write your MySQL query statement below\nSELECT sell_date, COUNT(DISTINCT(product)) as num_sold,\nGROUP_CONCAT(\n DISTINCT product \n ORDER BY product \n SEPARATOR \',\'\n) AS products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date ASC\n```\n\n# Pandas\n\nThe question asks us to group and summarize data by date. To do this, we first use the `groupby` function to group the `DataFrame` activities by date. This creates a new object called `groups`, which is a `DataFrameGroupBy` object.\n\nOnce we have the `DataFrameGroupBy` object, we can use the `agg()` function to perform aggregation operations on each group. The `agg()` function takes a list of aggregation tasks to perform. In this case, we are specifying two aggregation tasks:\n- Creating num_solid column with the number of unique products sold on each sell date. \n`num_sold=(\'product\', \'nunique\')`\n- Join all unique names within each group. \n`products=(\'product\', lambda x: \',\'.join(sorted(set(x))))`\n\n``` python3 []\nimport pandas as pd\n\ndef categorize_products(activities: pd.DataFrame) -> pd.DataFrame:\n groups = activities.groupby(\'sell_date\')\n\n stats = groups.agg(\n num_sold=(\'product\', \'nunique\'),\n products=(\'product\', lambda x: \',\'.join(sorted(set(x))))\n ).reset_index()\n stats.sort_values(\'sell_date\', inplace=True)\n return stats\n``` | 12 | 0 | ['Python', 'Python3', 'MySQL', 'Pandas'] | 0 |
group-sold-products-by-the-date | MySQL Easy Solution. | mysql-easy-solution-by-dilmurat-aliev-8k1t | Code\n\nSELECT \n sell_date, \n COUNT(DISTINCT(product)) AS num_sold, \n GROUP_CONCAT(DISTINCT(product)) AS products\nFROM Activities\nGROUP BY sell_date\n\n | dilmurat-aliev | NORMAL | 2023-07-07T11:00:13.327378+00:00 | 2023-07-07T11:01:36.748721+00:00 | 1,014 | false | # Code\n```\nSELECT \n sell_date, \n COUNT(DISTINCT(product)) AS num_sold, \n GROUP_CONCAT(DISTINCT(product)) AS products\nFROM Activities\nGROUP BY sell_date\n```\n\n | 12 | 0 | ['MySQL'] | 1 |
group-sold-products-by-the-date | MS SQL SERVER Simple Solution | ms-sql-server-simple-solution-by-nazir_c-72kh | \nselect sell_date, \n COUNT(product) as num_sold,\n STRING_AGG(product,\',\') WITHIN GROUP (ORDER BY product) as products from\n (select distinct sell | nazir_cinu | NORMAL | 2020-07-24T04:02:37.671223+00:00 | 2020-07-24T04:02:37.671260+00:00 | 3,115 | false | ```\nselect sell_date, \n COUNT(product) as num_sold,\n STRING_AGG(product,\',\') WITHIN GROUP (ORDER BY product) as products from\n (select distinct sell_date,product FROM Activities) Act\n GROUP BY sell_date\n``` | 11 | 1 | [] | 3 |
group-sold-products-by-the-date | Easy Solution | easy-solution-by-jahanviraycha-bgxn | SQL\n\nselect sell_date, count(distinct product) as num_sold,group_concat(distinct product) as products from Activities group by sell_date\n\n# Python\n\nimport | jahanviraycha | NORMAL | 2023-08-13T11:55:58.158026+00:00 | 2023-08-13T11:55:58.158052+00:00 | 1,960 | false | # SQL\n```\nselect sell_date, count(distinct product) as num_sold,group_concat(distinct product) as products from Activities group by sell_date\n```\n# Python\n```\nimport pandas as pd\n\ndef categorize_products(activities: pd.DataFrame) -> pd.DataFrame:\n return activities.groupby(\'sell_date\')[\'product\'].agg([(\'num_sold\',\'nunique\'),(\'products\',lambda x: \',\'.join(sorted(x.unique())))]).reset_index()\n\n``` | 10 | 0 | ['Pandas'] | 0 |
group-sold-products-by-the-date | MS SQL / Using STRING_AGG | ms-sql-using-string_agg-by-lilybone-5uwb | \nwith t as (\n select distinct * from Activities)\n\nselect \n sell_date\n ,count(1) as num_sold\n ,string_agg(product,\',\') within group (order | LilyBone | NORMAL | 2022-05-01T06:17:33.003223+00:00 | 2022-05-01T06:17:33.003263+00:00 | 1,769 | false | ```\nwith t as (\n select distinct * from Activities)\n\nselect \n sell_date\n ,count(1) as num_sold\n ,string_agg(product,\',\') within group (order by product) as products\nfrom t\ngroup by sell_date\norder by sell_date\n``` | 10 | 0 | ['MS SQL Server'] | 2 |
group-sold-products-by-the-date | Easy to understand Pandas solution | easy-to-understand-pandas-solution-by-hi-1fow | IntuitionI thought of how to use multiple aggregation function and save the values in different columns. I also needed to alphabetically sort the products as re | Himanshu_soni2023 | NORMAL | 2025-01-12T18:31:53.106437+00:00 | 2025-01-12T18:31:53.106437+00:00 | 942 | false | # Intuition
I thought of how to use multiple aggregation function and save the values in different columns. I also needed to alphabetically sort the products as required in the test cases. Meaning: the product starting with an alphabet which comes before any other should be the first.
# Approach
1. Group the dataframe.
2. use .agg function which will let us use multiple aggregation and other functions.
3. create new columns and feed values using aggregation functions on product column
4. .nunique() aggregation function will generate numerical count of unique products
5. I used lambda function to apply ','semicolon and join alphabetically sorted products
6. .set() was used to remove duplicate products (same products which are bought on the same day)
7. lastly, reset_index() function was used to convert the grouped df into a normal dataframe
# Complexity
- Time complexity:
### O(NLogN)
- Space complexity:
### O(N)
# Code
```pythondata []
import pandas as pd
def categorize_products(activities: pd.DataFrame) -> pd.DataFrame:
df = activities.groupby('sell_date').agg(
num_sold = ('product','nunique'),
products = ('product', lambda x: ','.join(sorted(set(x)))) # Sort and join unique product names
).reset_index()
return df
```
# This is my first solution I uploaded so please upvote me if you find it helpful. | 9 | 0 | ['Pandas'] | 0 |
group-sold-products-by-the-date | ✅ ✅ Best Solution Using Group By , Group_concat | best-solution-using-group-by-group_conca-3rem | Please Upvote if you Like it\n##### The GROUP BY statement groups rows that have the same values into summary rows.\n##### The GROUP BY statement is often used | vikramsinghgurjar | NORMAL | 2022-07-16T21:00:16.002290+00:00 | 2022-07-17T07:02:21.055963+00:00 | 1,322 | false | #### **Please Upvote if you Like it**\n##### The GROUP BY statement groups rows that have the same values into summary rows.\n##### The GROUP BY statement is often used with aggregate functions (COUNT(), MAX(), MIN(), SUM(), AVG()) to group the result-set by one or more columns.\n##### Another group method we use is **GROUP_CONCAT**, which concat items of grouped rows and joins them by using "," seperator. \n```\nSELECT sell_date,\n\t\tCOUNT(DISTINCT(product)) AS num_sold, \n\t\tGROUP_CONCAT(DISTINCT product ORDER BY product ASC) AS products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date ASC\n``` | 9 | 0 | [] | 0 |
group-sold-products-by-the-date | Simple and Easy Solution using group_concat | simple-and-easy-solution-using-group_con-0etk | \nSELECT \n sell_date,\n count(DISTINCT(product)) as num_sold,\n GROUP_CONCAT( DISTINCT product) as products\nfrom Activities \nGroup by sell_date;\n\n | pruthashouche | NORMAL | 2022-07-06T07:35:05.814784+00:00 | 2022-07-06T07:35:05.814823+00:00 | 706 | false | ```\nSELECT \n sell_date,\n count(DISTINCT(product)) as num_sold,\n GROUP_CONCAT( DISTINCT product) as products\nfrom Activities \nGroup by sell_date;\n```\n\n**Please UpVote if it was Helpful :)** | 8 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | SQL simple Query | sql-simple-query-by-deepakumar-developer-p7dz | \n# Code\n\nwith ProductCounts as (\n SELECT\n sell_date,\n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product ORDER BY product) AS | deepakumar-developer | NORMAL | 2023-12-25T12:47:20.939859+00:00 | 2023-12-25T12:47:20.939891+00:00 | 946 | false | \n# Code\n```\nwith ProductCounts as (\n SELECT\n sell_date,\n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product ORDER BY product) AS products\n FROM Activities\n GROUP BY sell_date\n)\nSELECT sell_date, num_sold, products\nFROM ProductCounts \nORDER BY sell_date;\n``` | 7 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | GROUP_CONCAT() explanation ✅ || MySQL✅ || Pandas || Beats 100% !! | group_concat-explanation-mysql-pandas-be-zko3 | Intuition\n Describe your first thoughts on how to solve this problem. \n- In order to agrregate the product into a single row grouped by sell_date, we use GROU | prathams29 | NORMAL | 2023-06-17T09:38:41.876407+00:00 | 2023-08-01T10:19:40.323922+00:00 | 747 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- In order to agrregate the product into a single row grouped by sell_date, we use **GROUP_CONCAT()** function.\n- We can also specify the sorting condition in GROUP_CONCAT() as \n`GROUP_CONCAT(DISTINCT product ORDER BY product) as products`\n\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT sell_date, COUNT(DISTINCT product) as num_sold, \n GROUP_CONCAT(DISTINCT product ORDER BY product) as products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date;\n```\n```\nimport pandas as pd\n\ndef group_sold_products_by_the_date(activities: pd.DataFrame) -> pd.DataFrame:\n df = activities.groupby(\'sell_date\')[\'product\'].agg([\n (\'num_sold\', \'nunique\'),\n (\'products\', lambda x: \',\'.join(sorted(x.unique())))]).reset_index()\n\n return df\n``` | 7 | 0 | ['MySQL', 'Pandas'] | 0 |
group-sold-products-by-the-date | SQL Solution using GROUP_CONCAT() | Definition and Syntax of GROUP_CONCAT() | sql-solution-using-group_concat-definiti-sp72 | --GROUP_CONCAT() function in MySQL is used to concatenate data from multiple rows into one field\n--Syntax: SELECT col1, col2, ..., colN\n GROUP_CONCAT ( | prag-sri | NORMAL | 2022-09-18T08:36:21.543578+00:00 | 2022-09-18T08:36:21.543617+00:00 | 1,327 | false | --GROUP_CONCAT() function in MySQL is used to concatenate data from multiple rows into one field\n--Syntax: SELECT col1, col2, ..., colN\n GROUP_CONCAT ( [DISTINCT] col_name1 \n [ORDER BY clause] [SEPARATOR str_val] ) \n FROM table_name GROUP BY col_name2;\n\n SELECT sell_date,\n COUNT(DISTINCT(product)) AS num_sold,\n GROUP_CONCAT(DISTINCT(product) SEPARATOR \',\') AS products\n FROM Activities\n GROUP BY sell_date; | 7 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | 1484. Group Sold Products By The Date | 1484-group-sold-products-by-the-date-by-ydkjk | ```\nSELECT sell_date, COUNT(DISTINCT(product)) as num_sold, \nGroup_Concat(distinct(product)\nORDER BY product) AS products\nFROM Activities \nGROUP BY sell_da | Spaulding_ | NORMAL | 2022-09-07T21:11:43.435018+00:00 | 2022-09-07T21:11:43.435054+00:00 | 986 | false | ```\nSELECT sell_date, COUNT(DISTINCT(product)) as num_sold, \nGroup_Concat(distinct(product)\nORDER BY product) AS products\nFROM Activities \nGROUP BY sell_date ORDER BY sell_date; | 7 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | SQL server easy solution | sql-server-easy-solution-by-karan-128-dhtt | Key points - \n1. Remove duplicates from the table using CTE\n2. Group by sell_date, count each sell date\n3. Use string_agg function and order within group by | karan-128 | NORMAL | 2022-08-29T19:29:44.793557+00:00 | 2022-08-29T19:29:44.793602+00:00 | 1,621 | false | **Key points - **\n1. Remove duplicates from the table using CTE\n2. Group by sell_date, count each sell date\n3. Use string_agg function and order within group by product.\n\n```\nwith cteactivity AS \n (SELECT DISTINCT *\n FROM activities)\nSELECT sell_date,\n\t\t count(product) AS num_sold,\n\t\t string_agg(product, \',\') within group (order by product) AS products\nFROM cteactivity\nGROUP BY sell_date\nORDER BY sell_date | 7 | 0 | ['MS SQL Server'] | 0 |
group-sold-products-by-the-date | Oracle SQL Solution and Explanation. | oracle-sql-solution-and-explanation-by-r-gqsr | Solution & explanation -\nThe below code is not working but as per the oracle it should work.\nClick this link to check on oracle website.\n\n\n/* Write your PL | ruthvikc27-dev | NORMAL | 2022-07-06T13:25:43.359458+00:00 | 2022-07-06T19:12:25.273826+00:00 | 1,300 | false | # Solution & explanation -\nThe below code is not working but as per the oracle it should work.\nClick this [link](https://livesql.oracle.com/apex/livesql/file/content_HT1O85E4BHSBWN93G1B3M8SI2.html) to check on oracle website.\n\n```\n/* Write your PL/SQL query statement below */\n\nSELECT TO_CHAR(sell_date, \'YYYY-MM-DD\') AS "sell_date", \nCOUNT(DISTINCT(product)) AS "num_sold",\nLISTAGG(DISTINCT(product), \',\') WITHIN GROUP (ORDER BY product) AS "products"\nFROM activities \nGROUP BY sell_date \nORDER BY sell_date;\n```\n\nIn question description, it is mentioned as **"There is no primary key for this table, it may contain duplicates"** and it contains duplicate values. So just fetch unqiue values of product & sell_date columns from activities table instead of fetching directly from the activities table. Check the below code.\n```\n/* Write your PL/SQL query statement below */\n\nSELECT TO_CHAR(sell_date, \'YYYY-MM-DD\') AS "sell_date", \nCOUNT(DISTINCT(product)) AS "num_sold",\nLISTAGG(product, \',\') WITHIN GROUP (ORDER BY product) AS "products"\nFROM (\n SELECT DISTINCT product, sell_date \n FROM activities\n) \nGROUP BY sell_date \nORDER BY sell_date;\n```\n\nI hope it is helpful to you.\n | 7 | 0 | ['Oracle'] | 0 |
group-sold-products-by-the-date | Superb Logic MYSQL | superb-logic-mysql-by-ganjinaveen-37tu | \n# Logic is to use distinct and group_concat\n\nselect sell_date,\ncount(distinct product) as num_sold,\nGroup_concat(distinct product) as products from activi | GANJINAVEEN | NORMAL | 2023-04-17T12:02:27.404236+00:00 | 2023-04-17T12:02:27.404280+00:00 | 1,493 | false | \n# Logic is to use distinct and group_concat\n```\nselect sell_date,\ncount(distinct product) as num_sold,\nGroup_concat(distinct product) as products from activities\ngroup by sell_date;\n\n```\n# please upvote me it would encourage me alot\n | 6 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | Group concat | group-concat-by-florentina-wb43 | Code\n\n# Write your MySQL query statement below\nSELECT sell_date, COUNT(DISTINCT product) as num_sold, GROUP_CONCAT(\n DISTINCT product \n ORDER BY prod | florentina | NORMAL | 2023-02-05T09:53:44.945209+00:00 | 2023-02-05T09:53:44.945251+00:00 | 720 | false | # Code\n```\n# Write your MySQL query statement below\nSELECT sell_date, COUNT(DISTINCT product) as num_sold, GROUP_CONCAT(\n DISTINCT product \n ORDER BY product ASC \n separator \',\') AS Products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date;\n``` | 6 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | MySQL Easy Solution | mysql-easy-solution-by-roshanshetty2816-yqxz | \n# Write your MySQL query statement below\nSELECT sell_date,COUNT(DISTINCT(product)) AS num_sold,\nGROUP_CONCAT(DISTINCT product SEPARATOR "," ) AS products FR | roshanshetty2816 | NORMAL | 2022-09-03T08:20:59.768870+00:00 | 2022-09-03T08:20:59.768912+00:00 | 849 | false | ```\n# Write your MySQL query statement below\nSELECT sell_date,COUNT(DISTINCT(product)) AS num_sold,\nGROUP_CONCAT(DISTINCT product SEPARATOR "," ) AS products FROM Activities\nGroup by sell_date\nORDER BY sell_date;\n```\n# upvote if u find it useful | 6 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | sql very easy solution | sql-very-easy-solution-by-pratham_18-ccd8 | \nselect sell_date,\ncount(distinct(product)) num_sold,\ngroup_concat(distinct(product)) products\nfrom Activities\ngroup by sell_date\norder by sell_date; | Pratham_18 | NORMAL | 2022-05-28T08:20:05.355291+00:00 | 2022-05-28T08:20:05.355336+00:00 | 962 | false | ```\nselect sell_date,\ncount(distinct(product)) num_sold,\ngroup_concat(distinct(product)) products\nfrom Activities\ngroup by sell_date\norder by sell_date;``` | 6 | 0 | ['MySQL'] | 1 |
group-sold-products-by-the-date | simple solution | simple-solution-by-maboeths666-93f7 | SELECT sell_date, count(DISTINCT product) as num_sold, group_concat(DISTINCT product separator \',\') as products\nFROM Activities\nGROUP BY sell_date\nORDER BY | maboeths666 | NORMAL | 2022-01-21T22:33:55.665846+00:00 | 2022-01-21T22:33:55.665890+00:00 | 704 | false | SELECT sell_date, count(DISTINCT product) as num_sold, group_concat(DISTINCT product separator \',\') as products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date | 6 | 0 | [] | 0 |
group-sold-products-by-the-date | Easy MySQL Solution | easy-mysql-solution-by-saranpadmakar-8did | \nselect sell_date, count(distinct product) as num_sold , \ngroup_concat(distinct product order by product asc separator \',\') as products \nfrom | saranpadmakar | NORMAL | 2020-06-17T22:54:01.674313+00:00 | 2020-06-21T21:24:59.908761+00:00 | 2,810 | false | ```\nselect sell_date, count(distinct product) as num_sold , \ngroup_concat(distinct product order by product asc separator \',\') as products \nfrom activities \ngroup by sell_date\norder by sell_date\n``` | 6 | 0 | [] | 1 |
group-sold-products-by-the-date | ✅EASY MYSQL SOLUTION | easy-mysql-solution-by-swayam28-396n | 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 | swayam28 | NORMAL | 2024-08-23T08:55:35.070286+00:00 | 2024-08-23T08:55:35.070328+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\n```mysql []\n# Write your MySQL query statement below\nselect sell_date, count( DISTINCT product ) as num_sold ,\nGROUP_CONCAT( DISTINCT product order by product ASC separator \',\' ) as products\nFROM Activities GROUP BY sell_date order by sell_date ASC;\n``` | 5 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | MySQL Simple Clean Solution | mysql-simple-clean-solution-by-shree_gov-p7ii | Code\n\n# Write your MySQL query statement below\nSELECT sell_date,\n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product) AS products\nF | Shree_Govind_Jee | NORMAL | 2024-04-02T03:52:53.783866+00:00 | 2024-04-02T03:52:53.783908+00:00 | 839 | false | # Code\n```\n# Write your MySQL query statement below\nSELECT sell_date,\n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product) AS products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date\n``` | 5 | 0 | ['Database', 'MySQL'] | 2 |
group-sold-products-by-the-date | SQL simple query | sql-simple-query-by-deepakumar-developer-vy21 | \n\n# Code\n\nSELECT\n sell_date,\n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product ORDER BY product) AS products\nFROM Activities\nGROU | deepakumar-developer | NORMAL | 2023-12-25T13:00:26.479252+00:00 | 2023-12-25T13:00:26.479281+00:00 | 901 | false | \n\n# Code\n```\nSELECT\n sell_date,\n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product ORDER BY product) AS products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date;\n\n``` | 5 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | Group Sold Products By The Date | MySQL | group-sold-products-by-the-date-mysql-by-aykx | Intuition\n Describe your first thoughts on how to solve this problem. \nWe need to find the name and the number of different products sold on each date.\n\n# A | samabdullaev | NORMAL | 2023-10-25T21:46:19.563977+00:00 | 2023-10-25T21:46:19.563994+00:00 | 477 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe need to find the name and the number of different products sold on each date.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Select specific columns from the table.\n\n > `SELECT` \u2192 This command retrieves data from a database\n\n > `AS` \u2192 This command renames a column with an alias (temporary name). In most database languages, we can skip the `AS` keyword and get the same result\n\n\n2. Count how many different products were sold on each date.\n\n > `COUNT()` \u2192 This function returns the number of rows\n\n > `DISTINCT` \u2192 This statement returns only distinct (different) values\n\n3. Concatenate the product names in lexicographical order for each date.\n\n > `GROUP_CONCAT()` \u2192 This function concatenates and aggregates values from multiple rows within a specific column into a single string\n\n > `ORDER BY` \u2192 This command sorts the result-set in ascending (smallest to largest) order by default\n\n4. Group and arrange the results to show the number of products sold on each date in chronological order.\n\n > `GROUP BY` \u2192 This command groups rows that have the same values into summary rows, typically to perform aggregate functions on them\n\n# Complexity\n- Time complexity: $O(n log n)$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n`n` is the number of rows in the table. This is because the query involves sorting all the products in the worst case.\n\n# Code\n```\nSELECT \n sell_date, \n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product ORDER BY product) AS products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date;\n``` | 5 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | Pandas | SQL | Easy | Explained Step By Step | Group Sold Products By The Date | pandas-sql-easy-explained-step-by-step-g-e7oy | see the successfully Accepted Solution\n\n\nimport pandas as pd\n\ndef categorize_products(activities: pd.DataFrame) -> pd.DataFrame:\n \n grouped_table = | Khosiyat | NORMAL | 2023-09-19T23:01:06.233175+00:00 | 2023-10-01T17:39:55.576285+00:00 | 133 | false | [see the successfully Accepted Solution](https://leetcode.com/submissions/detail/1054010892/)\n\n```\nimport pandas as pd\n\ndef categorize_products(activities: pd.DataFrame) -> pd.DataFrame:\n \n grouped_table = activities.groupby(\'sell_date\')\n statistics_product =grouped_table.agg(\n num_sold=(\'product\', \'nunique\'),\n products=(\'product\', lambda x: \',\'.join(sorted(x.unique())))\n)\n \n statistics_product_index = statistics_product.reset_index()\n \n return statistics_product_index\n```\n\n\n\n**Intuition Explained**\nFirst, we group the \'activities\' DataFrame by the \'sell_date\' column. It effectively splits the DataFrame into multiple groups, each corresponding to a unique \'sell_date\' value.\n```\n grouped_table = activities.groupby(\'sell_date\')\n```\n\nThen we perform aggregation operations on each group created by the \'sell_date\'.\n```\n statistics_product =grouped_table.agg(\n```\n\nWe must calculate the number of unique products sold within each group. The result is a new column named \'num_sold\' in the statistics_product DataFrame, which represents the count of unique products sold on each \'sell_date\'.\n```\n num_sold=(\'product\', \'nunique\'),\n```\n\nWe must also create a new column named \'products\' in the statistics_product which gets the unique values of the \'product\' column within each group, sorts the unique product values. and joins the sorted unique product values into a single comma-separated string.\n```\n products=(\'product\', lambda x: \',\'.join(sorted(x.unique()))))\n```\n\nThen, we remove the current index and assigns a new default integer index to the DataFrame.\n```\n statistics_product_index = statistics_product.reset_index()\n```\n\n**SQL**\n\n[see the successfully Accepted Solution](https://leetcode.com/submissions/detail/1061707676/)\n\n```\nSELECT sell_date,\nCOUNT(DISTINCT product)\nAS num_sold,\n\nGROUP_CONCAT(DISTINCT product ORDER BY product ASC separator \',\')\nAS products\nFROM Activities\n\nGROUP BY sell_date\nORDER BY sell_Date ASC;\n```\n\n```\n-- Select the \'sell_date\', count of distinct \'product\', and a concatenated list of distinct \'product\' names\nSELECT sell_date,\n -- Count the distinct products sold on each \'sell_date\'\n COUNT(DISTINCT product) AS num_sold,\n\n -- Concatenate distinct product names into a comma-separated list, ordered alphabetically\n GROUP_CONCAT(DISTINCT product ORDER BY product ASC separator \',\') AS products\n\n-- Retrieve data from the \'Activities\' table\nFROM Activities\n\n-- Group the results by \'sell_date\' to aggregate data for each date\nGROUP BY sell_date\n\n-- Order the results by \'sell_date\' in ascending order\nORDER BY sell_Date ASC;\n```\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 | 5 | 0 | ['MySQL'] | 1 |
group-sold-products-by-the-date | SQL SERVER Solution | sql-server-solution-by-sharath_bayyaram-397l | it works well, and Don\'t forgot to upvote\n\nthis below line do magic\n\nstring_agg(product,\',\') within group (order by product asc) as products \n\n# Code\n | Sharath_Bayyaram | NORMAL | 2022-10-24T10:27:01.176689+00:00 | 2022-10-24T16:41:10.139524+00:00 | 831 | false | it works well, and **Don\'t forgot to upvote**\n\nthis below line do magic\n\n**string_agg(product,\',\') within group (order by product asc) as products **\n\n# Code\n```\n/* Write your T-SQL query statement below */\nselect sell_date, \n count(distinct product) as num_sold,\n string_agg(product,\',\') within group (order by product asc) as products\nfrom(select distinct sell_date,product from Activities)sq \ngroup by sell_date \norder by sell_date\n``` | 5 | 0 | ['MS SQL Server'] | 0 |
group-sold-products-by-the-date | ✅✅✅ Easiest and Understandable Solution | easiest-and-understandable-solution-by-l-fu6c | Write your MySQL query statement below\nSELECT \n sell_date,\n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product ORDER BY product SE | Lyra_ | NORMAL | 2022-09-12T04:07:42.431993+00:00 | 2022-09-12T04:07:42.432054+00:00 | 882 | false | # Write your MySQL query statement below\nSELECT \n sell_date,\n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product ORDER BY product SEPARATOR \',\') AS products\nFROM Activities\nGROUP BY sell_date | 5 | 0 | [] | 0 |
group-sold-products-by-the-date | Easy MySQL Query | Beginner friendly | easy-mysql-query-beginner-friendly-by-gp-3xlj | sql\nselect \n sell_date, \n count(distinct product) as num_sold, \n group_concat(distinct product) as products\nfrom activities\ngroup by sell_date\no | gparas | NORMAL | 2022-06-03T06:42:15.019827+00:00 | 2022-06-03T06:42:15.019880+00:00 | 469 | false | ```sql\nselect \n sell_date, \n count(distinct product) as num_sold, \n group_concat(distinct product) as products\nfrom activities\ngroup by sell_date\norder by sell_date;\n``` | 5 | 1 | [] | 0 |
group-sold-products-by-the-date | MySQL | Easy Solution | mysql-easy-solution-by-_dhruvbhatia-zyi7 | Pls Upvote if you like the solution!\n```\nSELECT sell_date,\n\t\tCOUNT(DISTINCT(product)) AS num_sold, \n\t\tGROUP_CONCAT(DISTINCT product ORDER BY product ASC | _dhruvbhatia_ | NORMAL | 2022-05-27T08:47:06.864755+00:00 | 2022-05-27T08:47:06.864798+00:00 | 447 | false | **Pls Upvote if you like the solution!**\n```\nSELECT sell_date,\n\t\tCOUNT(DISTINCT(product)) AS num_sold, \n\t\tGROUP_CONCAT(DISTINCT product ORDER BY product ASC SEPARATOR \',\') AS products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date ASC | 5 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | Simple MySQL Solution beats 100 % | simple-mysql-solution-beats-100-by-arkap-1llw | \nSELECT\n\tsell_date,\n\tCOUNT(DISTINCT (product)) AS num_sold, \n\tGROUP_CONCAT(DISTINCT(product)) AS products \nFROM Activities \nGROUP BY sell_date \nORDER | arkapravagupta2002 | NORMAL | 2022-05-22T09:33:06.008742+00:00 | 2022-05-26T03:17:50.625926+00:00 | 564 | false | ```\nSELECT\n\tsell_date,\n\tCOUNT(DISTINCT (product)) AS num_sold, \n\tGROUP_CONCAT(DISTINCT(product)) AS products \nFROM Activities \nGROUP BY sell_date \nORDER BY sell_date ASC;\n```\nEdit : As someone pointed out in the comments, we could use ```GROUP_CONCAT(DISTINCT(product) ORDER BY product ASC)``` to sort the products, but the above query works just as fine. | 5 | 0 | ['MySQL'] | 1 |
group-sold-products-by-the-date | quickest solution | quickest-solution-by-nirmalnk-s9hz | select sell_date, \ncount(distinct product) as num_sold, \ngroup_concat(distinct product) as products\nfrom Activities\ngroup by sell_date\norder by sell_date,p | nirmalnk | NORMAL | 2022-05-18T09:50:40.285536+00:00 | 2022-05-18T09:50:40.285560+00:00 | 417 | false | select sell_date, \ncount(distinct product) as num_sold, \ngroup_concat(distinct product) as products\nfrom Activities\ngroup by sell_date\norder by sell_date,products | 5 | 0 | ['MySQL'] | 2 |
group-sold-products-by-the-date | Oracle Solution - LISTAGG | oracle-solution-listagg-by-ppal87-ljkp | \nWITH cte as(\nselect distinct sell_date, product from Activities)\n\nselect TO_CHAR(sell_date,\'YYYY-MM-DD\') as sell_date, count(product) as num_sold,\nLISTA | ppal87 | NORMAL | 2020-06-18T13:45:38.075821+00:00 | 2020-06-18T13:45:38.075868+00:00 | 940 | false | ```\nWITH cte as(\nselect distinct sell_date, product from Activities)\n\nselect TO_CHAR(sell_date,\'YYYY-MM-DD\') as sell_date, count(product) as num_sold,\nLISTAGG(product, \',\') WITHIN GROUP (ORDER BY product) as products\nfrom cte\ngroup by sell_date\norder by sell_date\n``` | 5 | 0 | ['Oracle'] | 1 |
group-sold-products-by-the-date | Easy Beginner freindly solution || Deep Explaination. | easy-beginner-freindly-solution-deep-exp-oiqi | My solution aims to achieve the following tasks:\n\n1. Group by sell_date to get counts and concatenations of products sold on each date.\n2. Count distinct pro | Darshan_999 | NORMAL | 2024-07-17T06:31:14.685341+00:00 | 2024-07-17T06:31:14.685378+00:00 | 985 | false | My solution aims to achieve the following tasks:\n\n1. **Group by `sell_date`** to get counts and concatenations of products sold on each date.\n2. **Count distinct products** sold on each date.\n3. **Concatenate distinct product names** in ascending order, separated by commas.\n\nHere\'s a breakdown of the components and the complexities involved:\n\n## Intuition\nYou need to gather information about sales activities, specifically counting distinct products sold on each date and listing those products in a sorted, comma-separated string. This involves aggregating data by date and performing operations on grouped data.\n\n## Approach\n1. **Group By `sell_date`**: This ensures you perform operations on each date\'s sales activities separately.\n2. **Count Distinct Products**: Use `COUNT(DISTINCT product)` to get the number of unique products sold each day.\n3. **Concatenate Products**: Use `GROUP_CONCAT` to concatenate distinct product names, ordering them alphabetically.\n\n## Complexity\n### Time Complexity\n- **Grouping and Counting**: Grouping the data by `sell_date` and counting distinct products involves scanning through all records, which is \\(O(n)\\), where \\(n\\) is the number of rows in the `Activities` table.\n- **Concatenation and Sorting**: Concatenating and sorting products for each group is dependent on the number of distinct products per date. If \\(m\\) is the number of distinct products per date, the complexity for sorting and concatenation is \\(O(m \\log m)\\).\n\nOverall, the time complexity is \\(O(n + k \\cdot m \\log m)\\), where \\(k\\) is the number of distinct dates and \\(m\\) is the average number of distinct products per date.\n\n### Space Complexity\n- **Storage for Grouped Data**: Requires space for storing grouped results. If there are \\(k\\) distinct dates and each date has \\(m\\) distinct products on average, the space complexity is \\(O(k \\cdot m)\\).\n- **Intermediate Storage**: Space for intermediate results during concatenation and sorting, which is also \\(O(k \\cdot m)\\).\n\nOverall, the space complexity is \\(O(k \\cdot m)\\).\n\n## Code\n```sql\n-- Write your MySQL query statement below\nSELECT sell_date, \n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product ORDER BY product ASC SEPARATOR \',\') AS products\nFROM Activities \nGROUP BY sell_date \nORDER BY sell_date ASC;\n```\n\nThis query:\n1. Selects the `sell_date`.\n2. Counts the number of distinct `product` entries for each `sell_date`.\n3. Concatenates the distinct `product` entries in ascending order, separated by commas.\n4. Groups the results by `sell_date` and sorts the results in ascending order by `sell_date`.\n\nThis approach ensures efficient querying and clear results presentation for the sales activities on different dates. | 4 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | 100 % || INTUITIVE && WELL - EXPLAINED || EASY SQL QUERY || Using GROUP_CONCAT( ) || Beats 89.11 %.. | 100-intuitive-well-explained-easy-sql-qu-vzh5 | Intuition\n Describe your first thoughts on how to solve this problem. \nSELECT sell_date, --> show sell_date...\nCOUNT(DISTINCT product) AS num_sold, --> Using | ganpatinath07 | NORMAL | 2024-02-14T17:11:58.345562+00:00 | 2024-02-14T17:11:58.345593+00:00 | 1,180 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSELECT sell_date, --> **show** sell_date...\nCOUNT(DISTINCT product) AS num_sold, --> Using **aggregate** function **COUNT** to count product only once **alias** as num_sold...\nGROUP_CONCAT(DISTINCT product ORDER BY product) --> it helps to making a **list**...\nAS products --> **alias** as products...\nFROM Activities --> **Association** from **Activities** table...\nGROUP BY sell_date --> **Grouping** w.r.t. sell_date...\nORDER BY sell_date, product; --> Ordering w.r.t. sell_date & product.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUsing Fundamentals of **SQL**...\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nSELECT sell_date, \nCOUNT(DISTINCT product) AS num_sold, \nGROUP_CONCAT(DISTINCT product ORDER BY product) \nAS products \nFROM Activities \nGROUP BY sell_date \nORDER BY sell_date, product;\n``` | 4 | 0 | ['Database', 'MySQL'] | 0 |
group-sold-products-by-the-date | 💯SIMPLE | Explanation 2023🟢 | EASY💡 | simple-explanation-2023-easy-by-komronab-2wt6 | SELECT-> sell_date: This part of the query selects the sell_date column, which represents the date of sales activities.\n\nCOUNT-> (DISTINCT product) AS num_sol | komronabdulloev | NORMAL | 2023-10-15T08:24:20.008687+00:00 | 2023-10-15T08:24:20.008712+00:00 | 657 | false | `SELECT`-> sell_date: This part of the query selects the sell_date column, which represents the date of sales activities.\n\n`COUNT`-> (DISTINCT product) AS num_sold: This part counts the number of distinct products sold on each sell_date. The result is given the alias num_sold, which will represent the count of unique products sold on that particular date.\n\n`GROUP_CONCAT`-> (`DISTINCT` product `ORDER BY` product `ASC` SEPARATOR \',\') `AS` products: This part aggregates the distinct product names, ordered in ascending alphabetical order (`ORDER BY` product `ASC`), and separates them with a comma (\',\'). The result is given the alias products.\n\n`FROM`-> Activities: This specifies that the data is being retrieved from the Activities table, which likely contains records of sales activities with information about the sell date and the products sold.\n\n`GROUP BY`-> sell_date: This clause groups the results by the sell_date, meaning that it will aggregate the data based on the sell date. Each unique sell date will have its own row in the result set.\n\n`ORDER BY`-> sell_date `ASC`: Finally, the results are sorted in ascending order based on the sell_date, which means the dates will be listed in chronological order.\n\n---\n\n\n\n\n# Code\n```\nSELECT sell_date, COUNT(DISTINCT(product)) AS num_sold, GROUP_CONCAT(DISTINCT product \nORDER BY product \nASC SEPARATOR \',\') \nAS products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date \nASC\n``` | 4 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | MySQL Solution for Group Sold Products By The Date Problem | mysql-solution-for-group-sold-products-b-k9q4 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the solution is to group the activities by sell date and then calc | Aman_Raj_Sinha | NORMAL | 2023-06-05T12:38:09.007795+00:00 | 2023-06-05T12:38:09.007832+00:00 | 1,258 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the solution is to group the activities by sell date and then calculate the number of different products sold on each date. Additionally, we need to concatenate the names of the products and sort them lexicographically for each sell date.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach is to use the SQL query provided, which utilizes the GROUP BY clause to group the activities by sell date. Within each group, the COUNT(DISTINCT product) function is used to count the number of different products sold, while the GROUP_CONCAT(DISTINCT product ORDER BY product) function is used to concatenate and sort the product names lexicographically.\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 number of rows in the Activities table. Let\'s denote this number as n. The query performs a grouping operation, which typically has a time complexity of O(n log n) or O(n) depending on the database implementation. Additionally, the query involves sorting the product names lexicographically for each sell date, which also contributes to the overall time complexity.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of this solution depends on the number of distinct sell dates and the number of distinct products. Let\'s denote these numbers as d and p respectively. The space complexity is O(d + p), as it requires storing the distinct sell dates and the distinct product names in memory for grouping and concatenation operations.\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT sell_date, COUNT(DISTINCT product) AS num_sold, GROUP_CONCAT(DISTINCT product ORDER BY product) AS products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date;\n\n``` | 4 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | SQL Server CLEAN & EASY | sql-server-clean-easy-by-rhazem13-j7nd | \nWITH CTE AS (\n SELECT DISTINCT * FROM Activities\n)\n\nSELECT \nsell_date,\nCOUNT(product) AS num_sold,\nSTRING_AGG(product, \',\') WITHIN GROUP (ORDER BY p | rhazem13 | NORMAL | 2023-03-10T11:01:26.321865+00:00 | 2023-03-10T11:01:26.321905+00:00 | 1,671 | false | ```\nWITH CTE AS (\n SELECT DISTINCT * FROM Activities\n)\n\nSELECT \nsell_date,\nCOUNT(product) AS num_sold,\nSTRING_AGG(product, \',\') WITHIN GROUP (ORDER BY product ASC) AS products\nFROM CTE\nGROUP BY sell_date\nORDER BY 1 ASC\n``` | 4 | 0 | [] | 0 |
group-sold-products-by-the-date | SOLUTION WITH STRING_AGG FUNCTION ( SQL SERVER ) | solution-with-string_agg-function-sql-se-hg4h | 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 | 2023-03-05T14:46:17.054196+00:00 | 2023-03-05T14:46:17.054246+00:00 | 813 | 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\n-- /* SOLUTION 1: \r\nSELECT\r\n\tsell_date,\r\n\tCOUNT(product) num_sold,\r\n\tSTRING_AGG(product,\',\') WITHIN GROUP(ORDER BY product ) products \r\nFROM\r\n (\r\n\r\n SELECT\r\n DISTINCT\r\n sell_date,\r\n product\r\n FROM\r\n Activities\r\n )TBL\r\n\t\r\nGROUP BY\r\n\tsell_date\r\n--SOLUTION 1: */\r\n``` | 4 | 0 | ['MS SQL Server'] | 0 |
group-sold-products-by-the-date | simple solution | simple-solution-by-shulingchang-dn2l | \nSELECT sell_date, \nCOUNT(DISTINCT(product)) AS num_sold, \nGROUP_CONCAT(DISTINCT(product)) AS products \nFROM Activities \nGROUP BY sell_date \nORDER BY sell | shulingchang | NORMAL | 2022-08-18T04:11:45.973158+00:00 | 2022-08-18T04:11:45.973198+00:00 | 799 | false | ```\nSELECT sell_date, \nCOUNT(DISTINCT(product)) AS num_sold, \nGROUP_CONCAT(DISTINCT(product)) AS products \nFROM Activities \nGROUP BY sell_date \nORDER BY sell_date;\n``` | 4 | 0 | [] | 2 |
group-sold-products-by-the-date | ✅[MySQL] Easy Solution using group_concat() | mysql-easy-solution-using-group_concat-b-3fzt | GROUP_CONCAT() function\n\nMySQL GROUP_CONCAT() function eturns a string with concatenated non-NULL value from a group.\nReturns NULL when there are no non-NULL | biT_Legion | NORMAL | 2022-08-07T10:58:53.623660+00:00 | 2022-08-07T10:58:53.623687+00:00 | 564 | false | ##### GROUP_CONCAT() function\n\nMySQL GROUP_CONCAT() function eturns a string with concatenated non-NULL value from a group.\nReturns NULL when there are no non-NULL values.\n\n\n```\nselect sell_date, \n\tcount(distinct(product)) as num_sold, \n\t\tgroup_concat(distinct(product)) as products\nfrom activities\ngroup by sell_date order by sell_date;\n``` | 4 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | simple readable group by solution | simple-readable-group-by-solution-by-ujj-gxfk | \n# Write your MySQL query statement below\nselect sell_date, count(distinct product) num_sold, \ngroup_concat(distinct product) as products from Activities gro | ujjwalavishek | NORMAL | 2022-07-26T06:17:46.771052+00:00 | 2022-07-26T06:17:46.771096+00:00 | 420 | false | ```\n# Write your MySQL query statement below\nselect sell_date, count(distinct product) num_sold, \ngroup_concat(distinct product) as products from Activities group by sell_date\n``` | 4 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | MySQL | Easy-Understanding | Beginner-Friendly | mysql-easy-understanding-beginner-friend-8v2t | \nselect sell_date,\n\t count(distinct product) as num_sold,\n group_concat(distinct product) as products\nfrom activities\ngroup by sell_date\norder | Saiko15 | NORMAL | 2022-07-12T09:03:36.825127+00:00 | 2022-07-12T09:03:36.825169+00:00 | 537 | false | ```\nselect sell_date,\n\t count(distinct product) as num_sold,\n group_concat(distinct product) as products\nfrom activities\ngroup by sell_date\norder by sell_date\n``` | 4 | 0 | ['MySQL'] | 1 |
group-sold-products-by-the-date | 1484. Group Sold Products By The Date | 1484-group-sold-products-by-the-date-by-4ki9n | \nselect sell_date,\ncount(distinct product)as num_sold,\ngroup_concat(distinct product order by product) as products\nfrom Activities \ngroup by sell_date\nord | rudra_0726 | NORMAL | 2022-07-11T03:19:39.437252+00:00 | 2022-07-11T03:19:39.437292+00:00 | 1,288 | false | ```\nselect sell_date,\ncount(distinct product)as num_sold,\ngroup_concat(distinct product order by product) as products\nfrom Activities \ngroup by sell_date\norder by sell_date;\n\n``` | 4 | 0 | ['MySQL', 'Oracle'] | 0 |
group-sold-products-by-the-date | GROUP_CONCAT ( ) FASTER THAN 95% | group_concat-faster-than-95-by-raghav_sh-sbr2 | \nSELECT sell_date, \nCOUNT(DISTINCT product) num_sold, \nGROUP_CONCAT(DISTINCT product) as products\nFROM Activities\nGROUP BY 1\n\n\nI am not sure if we have | Raghav_Shandilya | NORMAL | 2022-06-17T06:32:01.640433+00:00 | 2022-06-17T06:32:01.640462+00:00 | 336 | false | ```\nSELECT sell_date, \nCOUNT(DISTINCT product) num_sold, \nGROUP_CONCAT(DISTINCT product) as products\nFROM Activities\nGROUP BY 1\n```\n\nI am not sure if we have to use ORDER BY within GROUP_CONCAT as it returns the order lexicographically on its own. My assumption is that ---> ASC order is by default if you dont mention the ORDER BY in this clause. And also we dont need a SEPARATOR as the GROUP_CONCAT adds a \',\' between values by default. \n\nI can be wrong so please correct me if anyone knows more about this. | 4 | 0 | [] | 1 |
group-sold-products-by-the-date | [MySQL] BEATS 100.00% MEMORY/SPEED 0ms // APRIL 2022 | mysql-beats-10000-memoryspeed-0ms-april-l18y1 | \nSELECT sell_date,\n\t\tCOUNT(DISTINCT(product)) AS num_sold, \n\t\tGROUP_CONCAT(DISTINCT product ORDER BY product ASC SEPARATOR \',\') AS products\nFROM Activ | darian-catalin-cucer | NORMAL | 2022-04-23T09:27:32.489461+00:00 | 2022-04-23T09:27:32.489507+00:00 | 428 | false | ```\nSELECT sell_date,\n\t\tCOUNT(DISTINCT(product)) AS num_sold, \n\t\tGROUP_CONCAT(DISTINCT product ORDER BY product ASC SEPARATOR \',\') AS products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date ASC\n``` | 4 | 0 | ['MySQL'] | 1 |
group-sold-products-by-the-date | MYSQL Solution distict , group_concat ,group by | mysql-solution-distict-group_concat-grou-9blh | \nselect \n sell_date,\n count(distinct(product)) as num_sold,\n group_concat(distinct(product)) as products \nfrom Activities \n group by sell_dat | Sulaymon-Dev20 | NORMAL | 2022-04-23T08:40:34.523894+00:00 | 2022-04-23T08:41:08.727315+00:00 | 131 | false | ```\nselect \n sell_date,\n count(distinct(product)) as num_sold,\n group_concat(distinct(product)) as products \nfrom Activities \n group by sell_date \n``` | 4 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | Oracle | Listagg | group by | oracle-listagg-group-by-by-gowthamguttul-vmen | Please upvote, if it helps\n\n\nselect to_char(sell_date) sell_date,count(1)num_sold,LISTAGG(product,\',\') within group (order by product) products from (selec | GowthamGuttula | NORMAL | 2022-04-12T06:22:44.522459+00:00 | 2022-04-12T06:22:44.522508+00:00 | 892 | false | Please upvote, if it helps\n\n```\nselect to_char(sell_date) sell_date,count(1)num_sold,LISTAGG(product,\',\') within group (order by product) products from (select distinct * from Activities) group by sell_date order by sell_date\n``` | 4 | 0 | ['Oracle'] | 0 |
group-sold-products-by-the-date | MS SQL | ms-sql-by-siyu14-hdfb | Question\n\n1484. Group Sold Products By The Date\nEasy\n\nSQL Schema\nTable Activities:\n\n+-------------+---------+\n| Column Name | Type |\n+------------- | siyu14 | NORMAL | 2021-10-13T20:17:05.075210+00:00 | 2021-10-13T20:17:05.075254+00:00 | 984 | false | #### Question\n```\n1484. Group Sold Products By The Date\nEasy\n\nSQL Schema\nTable Activities:\n\n+-------------+---------+\n| Column Name | Type |\n+-------------+---------+\n| sell_date | date |\n| product | varchar |\n+-------------+---------+\nThere is no primary key for this table, it may contains duplicates.\nEach row of this table contains the product name and the date it was sold in a market.\n \n\nWrite an SQL query to find for each date, the number of distinct products sold and their names.\n\nThe sold-products names for each date should be sorted lexicographically. \n\nReturn the result table ordered by sell_date.\n\nThe query result format is in the following example.\n\nActivities table:\n+------------+-------------+\n| sell_date | product |\n+------------+-------------+\n| 2020-05-30 | Headphone |\n| 2020-06-01 | Pencil |\n| 2020-06-02 | Mask |\n| 2020-05-30 | Basketball |\n| 2020-06-01 | Bible |\n| 2020-06-02 | Mask |\n| 2020-05-30 | T-Shirt |\n+------------+-------------+\n\nResult table:\n+------------+----------+------------------------------+\n| sell_date | num_sold | products |\n+------------+----------+------------------------------+\n| 2020-05-30 | 3 | Basketball,Headphone,T-shirt |\n| 2020-06-01 | 2 | Bible,Pencil |\n| 2020-06-02 | 1 | Mask |\n+------------+----------+------------------------------+\nFor 2020-05-30, Sold items were (Headphone, Basketball, T-shirt), we sort them lexicographically and separate them by comma.\nFor 2020-06-01, Sold items were (Pencil, Bible), we sort them lexicographically and separate them by comma.\nFor 2020-06-02, Sold item is (Mask), we just return it.\n\n```\n\n\n#### Answer\n```SQL\n\n/* Write your T-SQL query statement below */\nselect sell_date, count(product) num_sold, STRING_AGG(product,\',\') WITHIN GROUP(order by product) as products from (select distinct * from Activities) as a \ngroup by sell_date\norder by sell_date\n\n``` | 4 | 0 | ['MS SQL Server'] | 0 |
group-sold-products-by-the-date | [MySQL] | mysql-by-ye15-qxnv | \n\nSELECT sell_date, COUNT(DISTINCT product) AS num_sold, GROUP_CONCAT(DISTINCT product ORDER BY product ASC) AS products\nFROM Activities\nGROUP BY sell_date\ | ye15 | NORMAL | 2021-01-22T02:29:19.924776+00:00 | 2021-01-22T02:29:19.924827+00:00 | 453 | false | \n```\nSELECT sell_date, COUNT(DISTINCT product) AS num_sold, GROUP_CONCAT(DISTINCT product ORDER BY product ASC) AS products\nFROM Activities\nGROUP BY sell_date\n``` | 4 | 0 | ['MySQL'] | 1 |
group-sold-products-by-the-date | Easy Oracle solution using LISTAGG | easy-oracle-solution-using-listagg-by-pr-1ufu | \nselect \n(to_char(sell_date,\'YYYY-MM-DD\')) as sell_date,\ncount(product) as num_sold,listagg(product,\',\') WITHIN GROUP (ORDER BY product) as products\nfro | praveenpurwar2004 | NORMAL | 2020-06-18T19:14:23.331351+00:00 | 2020-06-18T19:14:23.331475+00:00 | 264 | false | ```\nselect \n(to_char(sell_date,\'YYYY-MM-DD\')) as sell_date,\ncount(product) as num_sold,listagg(product,\',\') WITHIN GROUP (ORDER BY product) as products\nfrom (select distinct sell_date,product from Activities)\ngroup by sell_date\n``` | 4 | 0 | [] | 0 |
group-sold-products-by-the-date | [PostgreSQL] STRING_AGG | postgresql-string_agg-by-pbelskiy-yxwh | \nSELECT \n sell_date, \n COUNT(DISTINCT product) AS num_sold,\n STRING_AGG(DISTINCT product, \',\') AS products\nFROM Activities\nGROUP BY sell_date\nORDER | pbelskiy | NORMAL | 2024-06-05T19:16:34.283150+00:00 | 2024-06-05T19:16:34.283199+00:00 | 330 | false | ```\nSELECT \n sell_date, \n COUNT(DISTINCT product) AS num_sold,\n STRING_AGG(DISTINCT product, \',\') AS products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date\n\n``` | 3 | 0 | ['PostgreSQL'] | 0 |
group-sold-products-by-the-date | SQL: Daily Product Sales Summary | sql-daily-product-sales-summary-by-aleos-630r | \n# Code\n\nSELECT \n sell_date,\n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product ORDER BY product SEPARATOR \',\') AS products\n | aleos-dev | NORMAL | 2024-01-04T18:08:49.025599+00:00 | 2024-01-04T18:08:49.025625+00:00 | 869 | false | \n# Code\n```\nSELECT \n sell_date,\n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product ORDER BY product SEPARATOR \',\') AS products\nFROM activities\nGROUP BY sell_date\nORDER BY sell_date ASC;\n\n```\n\n##### GROUP_CONCAT Function\n```\nGROUP_CONCAT(column_name ORDER BY some_column SEPARATOR \'separator\')\n```\n- column_name: The column whose values you want to concatenate.\n- ORDER BY some_column: Optional clause to sort the values in a specific order.\n- SEPARATOR \'separator\': Optional clause to specify a string to separate the concatenated values. By default, it uses a comma (,).\n\n# Complexity\n- Time complexity: $$O(n log n)$$ due to the sorting required for both the GROUP BY and ORDER BY clauses, where n is the number of rows in the table.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(k)$$, where k is the number of distinct sell dates. \n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n\n | 3 | 0 | ['MySQL'] | 1 |
group-sold-products-by-the-date | T-SQL | Simple Solution | WITH & COUNT & STRING_AGG & WITHIN | t-sql-simple-solution-with-count-string_-p3b8 | Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(n)\n Add your space complexity here, e.g. O(n) \n\n# Co | ilyastuit | NORMAL | 2023-08-24T08:55:42.159449+00:00 | 2023-08-25T06:39:32.601236+00:00 | 1,445 | false | # Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```sql\nWITH t AS (SELECT DISTINCT * FROM Activities)\n\nSELECT \n t.sell_date,\n COUNT(t.product) AS num_sold,\n STRING_AGG(t.product, \',\') WITHIN GROUP (ORDER BY t.product) as products\nFROM t\nGROUP BY sell_date\nORDER BY sell_date\n```\n | 3 | 0 | ['Database', 'MS SQL Server'] | 2 |
group-sold-products-by-the-date | Using group by and agg functions | using-group-by-and-agg-functions-by-chv5-k9r1 | 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 | chv532 | NORMAL | 2023-08-22T16:19:03.926503+00:00 | 2023-08-22T16:19:03.926525+00:00 | 567 | 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 categorize_products(activities: pd.DataFrame) -> pd.DataFrame:\n act_df = activities.groupby(\'sell_date\')[\'product\'].agg([(\'num_sold\', \'nunique\'),\n (\'products\', lambda x: \',\'.join(sorted(x.unique())))]).reset_index()\n return act_df\n \n``` | 3 | 0 | ['Pandas'] | 0 |
group-sold-products-by-the-date | T-SQL STRING_AGG function used | t-sql-string_agg-function-used-by-user07-jinu | 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 | user0747y | NORMAL | 2023-05-12T05:02:51.336826+00:00 | 2023-05-12T05:02:51.336869+00:00 | 1,149 | 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 \n sell_date, \n count(distinct product) as num_sold, \n string_agg(product, \',\') WITHIN GROUP (ORDER BY product) as products\nfrom (select distinct sell_date, product from activities) ac\ngroup by ac.sell_date\n\n``` | 3 | 0 | ['MS SQL Server'] | 0 |
group-sold-products-by-the-date | MySQL Solution | mysql-solution-by-pranto1209-04jp | Code\n\n# Write your MySQL query statement below\nselect sell_date, count(distinct(product)) as num_sold, \ngroup_concat(distinct(product) order by product) as | pranto1209 | NORMAL | 2023-03-07T18:44:28.792882+00:00 | 2023-03-07T18:44:28.792924+00:00 | 1,838 | false | # Code\n```\n# Write your MySQL query statement below\nselect sell_date, count(distinct(product)) as num_sold, \ngroup_concat(distinct(product) order by product) as products\nfrom Activities group by sell_date order by sell_date;\n``` | 3 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | MySQL🐬89.2% FASTER🏎💨 | mysql892-faster-by-shubhamdraj-je2e | Approach\n Describe your approach to solving the problem. \nThe challange of this problem is how to aggregate the product names in one cell. So we use GROUP_CON | shubhamdraj | NORMAL | 2022-11-08T16:01:29.260819+00:00 | 2022-11-08T16:01:29.260859+00:00 | 720 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nThe challange of this problem is how to aggregate the product names in one cell. So we use `GROUP_CONCAT()` to aggregate distinct product names with a separator \',\' and also sort the product names. The rest is simple, we group by sell_date and COUNT DISTINCT product.\n\n# Code\n```\nSELECT \n sell_date, \n COUNT(DISTINCT product) as num_sold,\n GROUP_CONCAT(DISTINCT product ORDER BY product ASC SEPARATOR \',\') as products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date;\n```\n## Give it a **Upvote** If You Like My Explanation.\n### Have a Great Day/Night. | 3 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | just use group_concat | just-use-group_concat-by-takjaknigdy-trxf | \nselect \n\tsell_date, \n\tcount(distinct product) as num_sold, \n\tgroup_concat(distinct product) as products\nfrom Activities\ngroup by sell_date\norder by s | takjaknigdy | NORMAL | 2022-09-22T21:26:57.472648+00:00 | 2022-09-22T21:26:57.472690+00:00 | 287 | false | ```\nselect \n\tsell_date, \n\tcount(distinct product) as num_sold, \n\tgroup_concat(distinct product) as products\nfrom Activities\ngroup by sell_date\norder by sell_date\n``` | 3 | 0 | [] | 0 |
group-sold-products-by-the-date | Group products using new STRING_AGG funtion | group-products-using-new-string_agg-funt-sljg | Upvote if you feel it easy :)\n\nThis solution uses latest function STRING_AGG introduced by sql server\n\nSELECT x.sell_date, SUM(x.num_sold) AS num_sold, STRI | user1652uo | NORMAL | 2022-09-10T17:25:02.068662+00:00 | 2022-09-15T16:56:10.295363+00:00 | 663 | false | **Upvote if you feel it easy :)**\n\nThis solution uses latest function STRING_AGG introduced by sql server\n```\nSELECT x.sell_date, SUM(x.num_sold) AS num_sold, STRING_AGG(x.product,\',\') as products\nFROM (\n\tSELECT top 100 percent sell_date, COUNT(DISTINCT product) as num_sold, product\n\tFROM Activities\n\tGROUP BY sell_date, product\n\tORDER BY sell_date, product\n) X\nGROUP BY x.sell_date\n``` | 3 | 0 | ['MS SQL Server'] | 0 |
group-sold-products-by-the-date | simple and easy | simple-and-easy-by-aman_verma-sqik | select sell_date,count(distinct product) num_sold,group_concat(distinct product) products from activities group by sell_date order by sell_date | aman_verma_ | NORMAL | 2022-09-08T06:54:53.652380+00:00 | 2022-09-08T06:54:53.652426+00:00 | 834 | false | select sell_date,count(distinct product) num_sold,group_concat(distinct product) products from activities group by sell_date order by sell_date | 3 | 0 | ['MySQL'] | 1 |
group-sold-products-by-the-date | Group_Concat function | Easy to Understand | MySQL Solution | group_concat-function-easy-to-understand-nuba | \nSELECT\n sell_date,\n COUNT(DISTINCT product) AS "num_sold",\n GROUP_CONCAT(\n DISTINCT product\n ORDER BY\n product\n ) | deleted_user | NORMAL | 2022-07-20T02:27:48.182731+00:00 | 2022-07-20T02:27:48.182760+00:00 | 418 | false | ```\nSELECT\n sell_date,\n COUNT(DISTINCT product) AS "num_sold",\n GROUP_CONCAT(\n DISTINCT product\n ORDER BY\n product\n ) AS "products"\nFROM\n activities\nGROUP BY\n sell_date;\n``` | 3 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | SQL Query using string_agg | sql-query-using-string_agg-by-lavinamall-xjtw | \nSELECT sell_date, \n COUNT(product) AS num_sold, \n STRING_AGG(product,\',\') WITHIN GROUP ( ORDER BY product ASC) AS products\nFROM (SELECT DISTINCT se | lavinamall | NORMAL | 2022-07-15T10:52:43.124836+00:00 | 2022-07-15T10:52:43.124873+00:00 | 709 | false | ```\nSELECT sell_date, \n COUNT(product) AS num_sold, \n STRING_AGG(product,\',\') WITHIN GROUP ( ORDER BY product ASC) AS products\nFROM (SELECT DISTINCT sell_date, product FROM Activities)A\nGROUP BY sell_date\n``` | 3 | 0 | ['MySQL', 'MS SQL Server'] | 0 |
group-sold-products-by-the-date | Group Sold Products By The Date | group-sold-products-by-the-date-by-kshit-ebz9 | \nSELECT sell_date,\nCOUNT(distinct(product)) AS num_sold,\nGROUP_CONCAT(DISTINCT(product) ORDER BY product SEPARATOR \',\') as products\nFROM Activities\nGROUP | kshitij_thakre | NORMAL | 2022-07-01T12:53:25.088593+00:00 | 2022-07-01T12:53:25.088644+00:00 | 387 | false | ```\nSELECT sell_date,\nCOUNT(distinct(product)) AS num_sold,\nGROUP_CONCAT(DISTINCT(product) ORDER BY product SEPARATOR \',\') as products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date,product;\n``` | 3 | 0 | ['MySQL'] | 1 |
group-sold-products-by-the-date | MySQl Solution | Group Sold Products By The Date | mysql-solution-group-sold-products-by-th-aqic | In this question , we need to group our data based on the selling date and we need to provide date and number of product sell at that day in the output :\n | manavi2000y | NORMAL | 2022-06-20T06:51:54.767445+00:00 | 2022-06-20T06:51:54.767485+00:00 | 393 | false | In this question , we need to group our data based on the selling date and we need to provide date and number of product sell at that day in the output :\n \n\t\t\t SELECT \n sell_date , \n COUNT(DISTINCT product) AS num_sold , \n FROM activities\n GROUP BY sell_date \n\t\n\nNow as we need to order the output table based on selling_date so , \n \n\t\t\t SELECT \n sell_date , \n COUNT(DISTINCT product) AS num_sold , \n FROM activities\n GROUP BY sell_date \n ORDER BY sell_date\n \nAt last we are left with one column where we need to display the name of products which were sold at that day , products need to be display by adding " , " between them for that we had used a fuction :\n\n**Group_concat( column name)** : Function is used to contcat the values and inclued \',\' between them . \n\t\n\t SELECT \n sell_date , \n COUNT(DISTINCT product) AS num_sold , \n GROUP_CONCAT(DISTINCT product) AS products\n FROM activities\n GROUP BY sell_date \n ORDER BY sell_date | 3 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | You got it | you-got-it-by-gogole-xyiw | select sell_date , Count(distinct product) \'num_sold\' ,\nGroup_concat(Distinct product order by product) \'products\'\nfrom activities group by sell_date; | GOGOLE | NORMAL | 2022-06-13T04:16:24.469972+00:00 | 2022-06-13T04:16:24.470012+00:00 | 250 | false | select sell_date , Count(distinct product) \'num_sold\' ,\nGroup_concat(Distinct product order by product) \'products\'\nfrom activities group by sell_date; | 3 | 0 | [] | 1 |
group-sold-products-by-the-date | concat 95% | concat-95-by-lightningmcqueenkachow-40bs | \nselect\n sell_date, \n count(distinct product) as num_sold, \n group_concat(distinct product order by product asc) as products \nfrom \n activitie | lightningMcqueenKachow | NORMAL | 2022-05-14T23:20:55.474016+00:00 | 2022-05-14T23:20:55.474048+00:00 | 203 | false | ```\nselect\n sell_date, \n count(distinct product) as num_sold, \n group_concat(distinct product order by product asc) as products \nfrom \n activities \ngroup by \n sell_date\norder by \n sell_date \n```\n\nif you like this, please upvote. | 3 | 0 | [] | 1 |
group-sold-products-by-the-date | TSQL STRING_AGG statement | tsql-string_agg-statement-by-yulinchao-fui6 | \nSELECT sell_date, COUNT(DISTINCT(product)) AS num_sold, \n STRING_AGG(product,\',\') WITHIN GROUP (ORDER BY product) AS products\nFROM \n(SELECT DISTINCT s | yulinchao | NORMAL | 2022-05-12T09:24:30.180756+00:00 | 2022-05-12T09:24:30.180797+00:00 | 146 | false | ```\nSELECT sell_date, COUNT(DISTINCT(product)) AS num_sold, \n STRING_AGG(product,\',\') WITHIN GROUP (ORDER BY product) AS products\nFROM \n(SELECT DISTINCT sell_date, product FROM Activities) T\nGROUP BY sell_date\n``` | 3 | 0 | [] | 0 |
group-sold-products-by-the-date | easy solution - distinct, group_concat, group by | easy-solution-distinct-group_concat-grou-mcdi | SELECT sell_date, count(distinct product) as num_sold, group_concat(distinct product) as products\nFROM activities\nGROUP BY sell_date; | anhdang1 | NORMAL | 2022-04-25T07:26:28.513573+00:00 | 2022-04-25T07:26:28.513612+00:00 | 355 | false | SELECT sell_date, count(distinct product) as num_sold, group_concat(distinct product) as products\nFROM activities\nGROUP BY sell_date; | 3 | 0 | [] | 0 |
group-sold-products-by-the-date | Using GROUP_CONCAT | Easy Understanding | SQL | using-group_concat-easy-understanding-sq-nhiu | The main problem here is to concatenate the distinct products that were sold on a particular date. \n\nWe can do that by using GROUP_CONCAT keyword.\nWe provide | raahulsaxena | NORMAL | 2022-04-11T14:29:59.102140+00:00 | 2022-04-11T14:29:59.102186+00:00 | 393 | false | The main problem here is to concatenate the distinct products that were sold on a particular date. \n\nWe can do that by using ```GROUP_CONCAT``` keyword.\nWe provide the `DISTINCT` keyword in both count and group_concat functions because we don\'t want any duplicate records to mess with our counts. (Question does mention that duplicate entries might be present).\n\n## Query:\n\n```sql\n\nselect sell_date, count(distinct product) as num_sold, group_concat(distinct product order by product) as products from Activities group by sell_date order by sell_date\n\n``` | 3 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | MySQL using COUNT, GROUP_CONTACT | mysql-using-count-group_contact-by-lahar-b6s1 | \nselect sell_date, count(distinct product) num_sold, \ngroup_concat(distinct product order by product) \nproducts from Activities group by sell_date;\n | lahari_bitra | NORMAL | 2022-03-25T11:23:11.883543+00:00 | 2022-03-25T11:23:11.883581+00:00 | 416 | false | ```\nselect sell_date, count(distinct product) num_sold, \ngroup_concat(distinct product order by product) \nproducts from Activities group by sell_date;\n``` | 3 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | Simple solution using group_concat in mysql | simple-solution-using-group_concat-in-my-dva7 | \nselect sell_date, \n count(distinct product) as num_sold,\n group_concat(distinct product order by product) as products\nfrom Activities\ngroup | ronakgadia | NORMAL | 2022-03-01T20:12:55.803413+00:00 | 2022-03-01T20:12:55.803444+00:00 | 332 | false | ```\nselect sell_date, \n count(distinct product) as num_sold,\n group_concat(distinct product order by product) as products\nfrom Activities\ngroup by sell_date\norder by sell_date\n``` | 3 | 0 | [] | 0 |
group-sold-products-by-the-date | MySQL | mysql-by-kool_panda-55ai | \nselect distinct sell_date, count(distinct product) as num_sold, \ngroup_concat(distinct product order by product) as products\nfrom activities \ngroup by sell | kool_panda | NORMAL | 2021-11-30T22:10:50.896441+00:00 | 2021-11-30T22:10:50.896475+00:00 | 564 | false | ```\nselect distinct sell_date, count(distinct product) as num_sold, \ngroup_concat(distinct product order by product) as products\nfrom activities \ngroup by sell_date\norder by sell_date\n``` | 3 | 0 | [] | 0 |
group-sold-products-by-the-date | 99.84% faster | 9984-faster-by-jonathanli42-i6hs | SELECT\n sell_date, \n\tCOUNT(DISTINCT product) AS num_sold, \n\tGROUP_CONCAT(DISTINCT product ORDER BY product ASC) AS products\nFROM\n Activities\nGROUP | jonathanli42 | NORMAL | 2020-10-24T05:19:21.646635+00:00 | 2020-10-24T05:19:21.646668+00:00 | 620 | false | SELECT\n sell_date, \n\tCOUNT(DISTINCT product) AS num_sold, \n\tGROUP_CONCAT(DISTINCT product ORDER BY product ASC) AS products\nFROM\n Activities\nGROUP BY 1\nORDER BY 1 ASC\n | 3 | 1 | [] | 1 |
group-sold-products-by-the-date | simple mysql answer | simple-mysql-answer-by-plh2-cz2u | answer 1\n> Runtime: 729 ms, faster than 25.00% of MySQL online submissions for Group Sold Products By The Date.\nMemory Usage: 0B, less than 100.00% of MySQL o | plh2 | NORMAL | 2020-07-05T08:24:34.022739+00:00 | 2020-07-05T08:27:38.020188+00:00 | 675 | false | # answer 1\n> Runtime: 729 ms, faster than 25.00% of MySQL online submissions for Group Sold Products By The Date.\nMemory Usage: 0B, less than 100.00% of MySQL online submissions for Group Sold Products By The Date.\n\n```sql\nselect \n\tsell_date, \n\tCOUNT(product) num_sold, \n\tgroup_concat(product order by product) products \nfrom (SELECT DISTINCT * FROM Activities) Activities\ngroup by sell_date\norder by sell_date\n```\n# answer 2\n> Runtime: 656 ms, faster than 35.06% of MySQL online submissions for Group Sold Products By The Date.\nMemory Usage: 0B, less than 100.00% of MySQL online submissions for Group Sold Products By The Date.\n\n```sql\nselect \n sell_date, COUNT(DISTINCT product) num_sold, \n group_concat(DISTINCT product) products \nfrom Activities\ngroup by sell_date\norder by sell_date\n``` | 3 | 0 | [] | 2 |
group-sold-products-by-the-date | MSSQL + XML PATH | mssql-xml-path-by-domicilio2114-h62s | \n1. Distinct Rows\n\t\tselect distinct * from activities\n\n\n2. XML Path\n\t\tselect \',\' + tb1.product from (select distinct * from activities) tb1 for XML | domicilio2114 | NORMAL | 2020-06-29T07:01:55.448875+00:00 | 2020-06-29T07:01:55.448924+00:00 | 332 | false | ```\n1. Distinct Rows\n\t\tselect distinct * from activities\n```\n```\n2. XML Path\n\t\tselect \',\' + tb1.product from (select distinct * from activities) tb1 for XML path(\'\')\n```\n```\n3. Add Stuff\n\t\tselect products = stuff((select \',\' + a.product from (select distinct * from activities) a for XML path(\'\')),1,1,\'\')\n```\n```\n4. Subquery\n\t\tselect sell_date, count(distinct product) \'num_sold\', \n\t\tproducts = stuff((select \',\' + a.product from (select distinct * from activities) a \n\t\twhere a.sell_date = aa.sell_date for XMl path(\'\')),1,1,\'\')\n\t\tfrom Activities aa group by aa.sell_date\n``` | 3 | 0 | [] | 2 |
group-sold-products-by-the-date | SQL Server | sql-server-by-adchoudhary-0785 | Code | adchoudhary | NORMAL | 2025-03-09T04:31:47.780617+00:00 | 2025-03-09T04:31:47.780617+00:00 | 473 | false | # Code
```mssql []
/* Write your T-SQL query statement below */
WITH t AS (SELECT DISTINCT * FROM Activities)
SELECT
t.sell_date,
COUNT(t.product) AS num_sold,
STRING_AGG(t.product, ',') WITHIN GROUP (ORDER BY t.product) as products
FROM t
GROUP BY sell_date
ORDER BY sell_date
``` | 2 | 0 | ['MS SQL Server'] | 0 |
group-sold-products-by-the-date | MySQL solution using - GROUP_CONCAT | mysql-solution-using-group_concat-by-swa-2lu2 | Code | swapit | NORMAL | 2025-02-28T11:03:55.984811+00:00 | 2025-02-28T11:03:55.984811+00:00 | 437 | false | # Code
```mysql []
SELECT sell_date, COUNT(DISTINCT(product)) AS num_sold, GROUP_CONCAT(DISTINCT product,'') AS products
FROM Activities
GROUP BY sell_date
``` | 2 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | Easy explanation | easy-explanation-by-pratyushpanda91-ggtk | ExplanationGrouping by sell_dateGROUP BY sell_date ensures we aggregate products per date.
Counting Distinct Products Sold Per DateCOUNT(DISTINCT product) AS nu | pratyushpanda91 | NORMAL | 2025-02-17T16:31:54.256604+00:00 | 2025-02-17T16:31:54.256604+00:00 | 664 | false | # Explanation
Grouping by sell_date
GROUP BY sell_date ensures we aggregate products per date.
Counting Distinct Products Sold Per Date
COUNT(DISTINCT product) AS num_sold counts the unique products sold for each date.
Concatenating Product Names in Lexicographical Order
GROUP_CONCAT(DISTINCT product ORDER BY product SEPARATOR ',') AS products
This ensures the product names are sorted lexicographically and separated by commas.
Ordering Results by Date
ORDER BY sell_date ensures the results are sorted in ascending order of sell_date.
# Code
```mysql []
# Write your MySQL query statement below
SELECT sell_date,
COUNT(DISTINCT product) AS num_sold,
GROUP_CONCAT(DISTINCT product ORDER BY product SEPARATOR ',') AS products
FROM Activities
GROUP BY sell_date
ORDER BY sell_date;
``` | 2 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | Group_concat | group_concat-by-aadithya18-5ado | IntuitionActually, I am also new to this. Seen a hint from the discussion section and found out there is group_concat available in sql.Code | aadithya18 | NORMAL | 2025-02-10T20:58:27.927582+00:00 | 2025-02-10T20:58:27.927582+00:00 | 375 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Actually, I am also new to this. Seen a hint from the discussion section and found out there is group_concat available in sql.
# Code
```mysql []
# Write your MySQL query statement below
Select
a.sell_date, Count(distinct(a.product)) as num_sold,
GROUP_CONCAT(DISTINCT a.product order by a.product) as products
From
Activities a
GROUP BY
sell_date
ORDER BY
sell_date;
``` | 2 | 0 | ['Database', 'MySQL'] | 0 |
group-sold-products-by-the-date | ✅✅100% working solution 🔥🔥using GROUP_CONCAT || | 100-working-solution-using-group_concat-lfhpo | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | ritikg4360 | NORMAL | 2025-02-01T10:50:58.860397+00:00 | 2025-02-01T10:50:58.860397+00:00 | 456 | 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 []
select sell_date, count(distinct product) as num_sold,
GROUP_CONCAT(distinct product order by product separator ',') as products
from Activities
group by sell_date
order by sell_date ;
``` | 2 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | 👉🏻EASY TO UNDERSTAND SOLUTION || FAST || COUNT() || GROUP_CONCAT() || MySQL | easy-to-understand-solution-fast-count-g-rfp0 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Siddarth9911 | NORMAL | 2025-01-04T18:23:29.447505+00:00 | 2025-01-04T18:23:29.447505+00:00 | 447 | 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 []
SELECT sell_date, COUNT(DISTINCT product) AS num_sold, GROUP_CONCAT(DISTINCT product ORDER BY product ASC separator ',') AS products
FROM Activities
GROUP BY sell_date
ORDER BY sell_date ASC;
``` | 2 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | EASY WAY USING GROUP_CONCAT | easy-way-using-group_concat-by-siddu_pat-83ll | \n# Code\nmysql []\nselect sell_date,\ncount(distinct product)as num_sold,\ngroup_concat(distinct product order by product) as products\nfrom Activities \ngroup | siddu_patil005 | NORMAL | 2024-11-29T13:34:40.858690+00:00 | 2024-11-29T13:34:40.858713+00:00 | 400 | false | \n# Code\n```mysql []\nselect sell_date,\ncount(distinct product)as num_sold,\ngroup_concat(distinct product order by product) as products\nfrom Activities \ngroup by sell_date\norder by sell_date;\n``` | 2 | 0 | ['MySQL'] | 0 |
group-sold-products-by-the-date | 📊 Efficient SQL Solution for Grouped Product Count and Concatenation 🏆 (Beats 91% in Speed!) | efficient-sql-solution-for-grouped-produ-nb3u | \uD83D\uDCCA SQL Solution for Grouping and Counting Sold Products \uD83C\uDFC6\n\n### Intuition\n Describe your first thoughts on how to solve this problem. \nT | rshikharev | NORMAL | 2024-11-07T15:54:28.822560+00:00 | 2024-11-07T15:54:28.822594+00:00 | 572 | false | ## \uD83D\uDCCA SQL Solution for Grouping and Counting Sold Products \uD83C\uDFC6\n\n### Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to find the number of unique products sold for each date and list them in lexicographical order.\n\n### Approach\n<!-- Describe your approach to solving the problem. -->\n1. Use `GROUP BY` to group entries by each `sell_date`.\n2. Apply `COUNT(DISTINCT product)` to count unique products sold for each date.\n3. Use `GROUP_CONCAT` with `DISTINCT` and `ORDER BY` to concatenate product names in lexicographical order.\n\n### Complexity\n- **Time complexity:** O(n), where n is the number of entries in the `Activities` table.\n- **Space complexity:** O(n), for storing the grouped results.\n\n# Code\n```mysql []\n# Write your MySQL query statement below\nSELECT \n sell_date,\n COUNT(DISTINCT product) AS num_sold,\n GROUP_CONCAT(DISTINCT product ORDER BY product ASC) AS products\nFROM Activities\nGROUP BY sell_date\nORDER BY sell_date;\n```\n\nThis query effectively groups and counts unique products for each date, ensuring product names are listed alphabetically.\n\nI welcome any feedback or suggestions for optimizing the code! | 2 | 0 | ['Database', 'MySQL'] | 0 |
complement-of-base-10-integer | Solution | solution-by-deleted_user-b37f | C++ []\nclass Solution {\n public:\n int bitwiseComplement(int N) {\n int mask = 1;\n\n while (mask < N)\n mask = (mask << 1) + 1;\n\n return mas | deleted_user | NORMAL | 2023-05-20T07:49:01.127884+00:00 | 2023-05-20T08:13:42.329372+00:00 | 7,395 | false | ```C++ []\nclass Solution {\n public:\n int bitwiseComplement(int N) {\n int mask = 1;\n\n while (mask < N)\n mask = (mask << 1) + 1;\n\n return mask ^ N;\n }\n};\n```\n\n```Python3 []\nclass Solution:\n def bitwiseComplement(self, n: int) -> int:\n cnt=0\n ans=0\n if n==0:\n return 1\n while n>0:\n if n&1:\n cnt+=1\n else:\n ans =ans +(2**cnt)\n cnt+=1\n n=n>>1\n return ans\n```\n\n```Java []\nclass Solution {\n public int bitwiseComplement(int n) {\n if(n == 0) return 1;\n int res = 0;\n int fac = 1;\n \n while(n != 0){\n res += fac * (n % 2 == 0 ? 1 : 0);\n fac *= 2;\n n /= 2;\n }\n return res;\n }\n}\n``` | 466 | 0 | ['C++', 'Java', 'Python3'] | 4 |
complement-of-base-10-integer | [Java/C++/Python] Find 111.....1111 >= N | javacpython-find-1111111-n-by-lee215-vgm0 | Hints\n1. what is the relationship between input and output\n2. input + output = 111....11 in binary format\n3. Is there any corner case?\n4. 0 is a corner case | lee215 | NORMAL | 2019-03-17T04:02:27.659592+00:00 | 2019-07-07T12:38:35.215299+00:00 | 20,691 | false | ## **Hints**\n1. what is the relationship between input and output\n2. input + output = 111....11 in binary format\n3. Is there any corner case?\n4. 0 is a corner case expecting 1, output > input\n\n\n## **Intuition**\nLet\'s find the first number `X` that `X = 1111....1 > N`\nAnd also, it has to be noticed that,\n`N = 0` is a corner case expecting`1` as result.\n\n<br>\n\n## **Solution 1**:\n`N + bitwiseComplement(N) = 11....11 = X`\nThen `bitwiseComplement(N) = X - N`\n\n**Java:**\n```\n public int bitwiseComplement(int N) {\n int X = 1;\n while (N > X) X = X * 2 + 1;\n return X - N;\n }\n```\n\n**C++:**\n```\n int bitwiseComplement(int N) {\n int X = 1;\n while (N > X) X = X * 2 + 1;\n return X - N;\n }\n```\n\n**Python:**\n```\n def bitwiseComplement(self, N):\n X = 1\n while N > X: X = X * 2 + 1\n return X - N\n```\n<br>\n\n## **Solution 2**:\n`N ^ bitwiseComplement(N) = 11....11 = X`\n`bitwiseComplement(N) = N ^ X`\n\n**Java:**\n```\n public int bitwiseComplement(int N) {\n int X = 1;\n while (N > X) X = X * 2 + 1;\n return N ^ X;\n }\n```\n\n**C++:**\n```\n int bitwiseComplement(int N) {\n int X = 1;\n while (N > X) X = X * 2 + 1;\n return N ^ X;\n }\n```\n\n**Python:**\n```\n def bitwiseComplement(self, N):\n X = 1\n while N > X: X = X * 2 + 1;\n return N ^ X\n```\n<br>\n\n## **Complexity**\n`O(logN)` Time\n`O(1)` Space\n<br>\n\n## Python 1-lines\nUse `bin`\n```\n def bitwiseComplement(self, N):\n return (1 << len(bin(N)) >> 2) - N - 1\n```\nUse `translate`\n```\n def bitwiseComplement(self, N):\n return int(bin(N)[2:].translate(string.maketrans(\'01\', \'10\')), 2)\n``` | 321 | 4 | [] | 21 |
complement-of-base-10-integer | C++ EASY TO SOLVE || Beginner friendly with detailed explanation and dry run | c-easy-to-solve-beginner-friendly-with-d-dx7a | Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation i.e if it\'s 101 =Complement=> | Cosmic_Phantom | NORMAL | 2022-01-04T02:30:48.071484+00:00 | 2024-08-23T02:25:07.500595+00:00 | 7,364 | false | Given a positive integer, output its **complement number**. The complement strategy is to flip the bits of its binary representation i.e if it\'s **101** =Complement=> **010**\n\n**Intuition:-**\nThere is not much of a intuition in this question as the question is loud and clear about it\'s use of bit manipulation .\n\nTo make things a bit more interesting let\'s do this question with and wihout bit manipulation \n***\n***\n**Using Bit Manipulations:-**\n***\n**Bit masking can be summarized with this image (^_^)**\n \nThus, we can conclude that masking means to keep/change/remove a desired part of information.\n***\n**Dry run of bitmasking:-**\n\n***\n**Code[Using bitmasking]:-**\n```\nclass Solution {\npublic:\n int bitwiseComplement(int num) {\n\t\t //base case\n if(num == 0) return 1;\n unsigned mask = ~0;\n while( mask & num ) mask = mask << 1;\n return ~num ^ mask;\n }\n};\n```\n**Time Complexity:** *`O(log(num))`*\n**Space Complexity:** *`O(1)`*\n***\n***\n**Using XOR**\nBasic idea is to find the smallest power of 2 that is larger than the `input number num`, and output the difference between `powerof2s` and `num` . \n\n**Dry run of XOR:-**\n```\nFor example :-\nInput: num = 5(101) ,\nThus the smallest power of 2 (and just larger than 5) is 8 (1000)\nThe output should be 8 - 5 - 1 = 2 (010).\n```\n**Code [using XOR]:-**\n```\nclass Solution {\npublic:\n int bitwiseComplement(int num) {\n int powerof2s = 2, temp = num;\n \n while(temp>>1) {\n temp >>= 1;\n powerof2s <<= 1;\n }\n \n return powerof2s - num - 1;\n }\n};\n```\n\n***\n***\n**Without using Bit manipulation:-**\n***\n**Algorithm:-**\n1. At start convert convert the decimal number to its binary representation \n2. Then start overriding the bits with 1 as 0 and 0 with 1 .\n3. Then again start converting the binary representation back to decimal representation .\n\n**Code:-**\n```\nclass Solution {\npublic:\n int bitwiseComplement(int num) {\n //base case\n if(num==0) return 1;\n vector<int> temp; \n\t\t// convert to binary representation\n while( num != 0 ){\n temp.push_back( num % 2 );\n num /= 2;\n } \n\t\t// make complement\n for(int i=0; i<temp.size(); i++){\n if( temp[i] == 1 ) temp[i] = 0;\n else if( temp[i] == 0 ) temp[i] = 1;\n } int res = 0;\n\t\t//Again convert them back to decimal representation \n for(int i=temp.size()-1; i>-1; i--) res = res * 2 + temp[i];\n return res;\n }\n};\n```\n**An optmization of the above code:- #Credit goes to @bakar7**\n```\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n if (n == 0) {\n return 1;\n }\n\n int answer = 0;\n int power = 1;\n while (n) {\n answer += ((1 - (n % 2)) * power);\n n /= 2;\n power *= 2;\n }\n\n return answer;\n }\n};\n```\n\n\n\n | 120 | 50 | ['Bit Manipulation', 'C', 'Bitmask', 'C++'] | 9 |
complement-of-base-10-integer | C++ 2 lines, XOR mask | c-2-lines-xor-mask-by-votrubac-8g5q | We construct a XOR mask from all ones until it\'s larger or equal than the number. For example, for 100101, the mask will be 111111. And for 1111, the mask will | votrubac | NORMAL | 2019-03-17T04:00:28.157284+00:00 | 2019-03-17T04:00:28.157319+00:00 | 10,004 | false | We construct a XOR mask from all ones until it\'s larger or equal than the number. For example, for ```100101```, the mask will be ```111111```. And for ```1111```, the mask will be ```1111```.\n\nThe result is the XOR operation of the number and the mask.\n```\nint bitwiseComplement(int N, int c = 1) {\n while (c < N) c = (c << 1) + 1;\n return N ^ c;\n}\n``` | 118 | 6 | [] | 9 |
complement-of-base-10-integer | Well Explained JAVA & C++ || 2 Approaches || Easy for mind to Accept it | well-explained-java-c-2-approaches-easy-8e2cz | Intitution\n\nWe have to convert 5 -----> 101 -----> 010 ------> 2\nSteps: 1 2 3\n\n\n\nMethod - 1 :\nApproach to use:\n\n | hi-malik | NORMAL | 2022-01-04T01:46:11.509527+00:00 | 2022-01-04T05:23:12.006494+00:00 | 6,511 | false | **Intitution**\n```\nWe have to convert 5 -----> 101 -----> 010 ------> 2\nSteps: 1 2 3\n```\n\n\n**Method - 1 :**\n**Approach to use:**\n```\nNow we take modulo of 5 i.e. % 2 ----> 1 ^\n\t\t\t\t\t\t\t\t2 ----> 0 |\n\t\t\t\t\t\t\t\t1 ----> 1 |\n\t\tand finally it become\'s 0 \n\nWhat I did is we take modulo of 5 i.e. 5 % 2 = 1 ;then divide by 2 i.e 5 / 2 = 2;\nagain modulo of 2 i.e 2 % 2 = 0 ;then again we divide by 2 2 / 2 = 1;\n\n\ngoing from down to top, we will get 101, now by compliment we can convert 101 to 010\n\nNow for 010, taking least significant digit. we go from right to left\ni.e. 0 * 2^0\n +1 * 2^1\n +0 * 2^2\n ---------\n = 2\n```\n**Java**\n```\nclass Solution {\n public int bitwiseComplement(int n) {\n if(n == 0) return 1; // Checking for base case\n int res = 0;\n int fac = 1; // keep for 2 basically\n \n while(n != 0){\n // first we need to check what is our bit in 2 by taking modulo\n res += fac * (n % 2 == 0 ? 1 : 0);\n // res is the number convert back to decimal + factor * n % 2 if comes 0 then we take 1 otherwise 0 this is our complement\n \n fac *= 2;\n n /= 2;\n }\n return res;\n }\n}\n```\n**C++**\n```\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n if(n == 0) return 1; // Checking for base case\n int res = 0;\n int fac = 1; // keep for 2 basically\n \n while(n != 0){\n // first we need to check what is our bit in 2 by taking modulo\n res += fac * (n % 2 == 0 ? 1 : 0);\n // res is the number convert back to decimal + factor * n % 2 if comes 0 then we take 1 otherwise 0 this is our complement\n \n fac *= 2;\n n /= 2;\n }\n return res;\n }\n};\n```\nANALYSIS:-\n* **Time Complexity :-** BigO(logN)\n\n* **Space Complexity :-** BigO(1)\n\n**Method - 2 :**\n**Approach to use:**\n```\n5 -> 101 ^ 010 -> 111 // taking XOR of it\'s no and complement we get 111\n\n10 - > 1010 ^ 0101 -> 1111\n\nSo, looking at above example we can say that,\n A ^ B = C\nthen, \n A ^ A ^ B = A ^ C\nbut A ^ A is 0;\ntherefore, 0 ^ B is B;\nHence we can say B = A ^ C\n\nSo, these property of XOR apply in our case\n\nLet say our no is "N" and no we have to find is "X"\nlet say N ^ X gives [111....] something,\nwhere 111 is length of our N\nand X = [1111..] length is same as binary representation of N\n\nSo, from this the formula we get is :\n -----------------\n X = [1111...] ^ N\n -----------------\n \nBut here thing is how we will get [1111...], in java we have method by taking Integer.toBinaryString(). \nAfter taking length we will do first bitwise left shift of 1\ni.e 1 << 3 gives 1000\nand if i subtract -1\ngives 111\n\nthis is what i need in case of N = 5, where the length of my binary representation is 3\n\nSo, that\'s how we gonna do it.\n```\n**JAVA**\n```\nclass Solution {\n public int bitwiseComplement(int n) {\n return n == 0 ? 1 : ((1 << Integer.toBinaryString(n).length()) - 1) ^ n;\n }\n}\n```\n**C++**\n```\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n return n>0? (int)((1ll<<((int)log2(n)+1))-1)^n : 1;\n }\n};\n\n```\nANALYSIS:-\n* **Time Complexity :-** BigO(logN)\n\n* **Space Complexity :-** BigO(1)\n\n*Guys, If you downvote. Please do let me know in comment section. What\'s wrong with the code or explanation. So, that I can correct and improve it. \nThanks (:* | 83 | 13 | ['C', 'Java'] | 14 |
complement-of-base-10-integer | 4 approaches: bitwise operation, math formular, one naive simulation, and bonus | 4-approaches-bitwise-operation-math-form-ns9g | Find the smallest binary number c that is all 1s, (e.g. \u201C111\u201D, \u201C11111\u201D) that is greater or equal to N.\nans = C ^ N or C \u2013 N\n\nMore de | codedayday | NORMAL | 2020-05-04T15:19:22.775885+00:00 | 2020-10-05T14:23:48.800864+00:00 | 4,489 | false | Find the smallest binary number c that is all 1s, (e.g. \u201C111\u201D, \u201C11111\u201D) that is greater or equal to N.\nans = C ^ N or C \u2013 N\n\nMore details:\n^: bitwise XOR operator\n**Table 1. The truth table of bitwise XOR operator:**\n**1^0 = 1\n1^1 = 0**\n0^0 = 0\n0^1 = 1\n**The upper two rows of Table 1 is used in the problem.**\n\n\n\nTime complexity: O(log(n))\nSpace complexity: O(1)\n\nSolution 1: bit shift based solution\n```\nclass Solution {// bit shift based solution\npublic:\n int bitwiseComplement(int N) {\n int c = 1; // c = pow(2, x) \u2013 1; c is the smallest number >= N\n while (c < N) \n c = (c << 1) | 1;\n //c = c*2 + 1; same as above\n return N ^ c; \n //return c - N; // also ok \n }\n};\n```\n\nSolution 2: math based solution: log2, pow\n\n```\nclass Solution { //math based solution: log2, pow\npublic:\n int bitwiseComplement(int N) { \n if(N == 0) return 1;\n int exponenet = (int)log2(N)+1;\n int flipper = (int)pow(2, exponenet)-1; \n return N ^ flipper;\n }\n};\n```\n\nWhat, the above two solution is still too anti-human?!\nOf course, it is.\nIf you are still not comfortable with the above two methods due to too much specific knowledge about bit operation,\nyou might find the last solution might be more natural :\n\nSolution 3: naive simulation: find each single digit, then flip it\n```\nclass Solution { // BEST1: find each single digit, then flip it\npublic:\n int findComplement(int num) {\n int res = 0, ind = 0;\n while(num > 0){\n //find single digit from low to high significance\n int digit = num % 2; \n int flippedDigit = (digit == 1? 0 : 1); // flip digit\n res += (flippedDigit << ind);\n num /= 2; \n ind++;\n }\n return res;\n }\n};\n\n```\nThis is a bonus solution: \n\nSolution 4: Get all-1 mask by spreading 1 from most significant to the rest\n```\nclass Solution { // Get all-1 mask by spreading 1 from most significant to the rest:\npublic://Time/Space: O(1); O(1)\n int findComplement(int num) {\n\t if(num ==0) return 1;\n int mask = num;\n mask |= mask >> 1;\n mask |= mask >> 2;\n mask |= mask >> 4;\n mask |= mask >> 8;\n mask |= mask >> 16;\n return num ^ mask;\n }\n};\n\n```\nExplanation: \n\nTo understand the solution, we need to go backwards. The aim is to xor the given number with a mask. The mask should contain all 1s in its rightmost bits. However, the number of rightmost bits is important. In the case of 5(101), for example, the number of rightmost bits must be 3, since 5 uses 3 rightmost bits. The mask must be 111 for 5(101). When we xor 111 with 101, we will get 010. As another example, the mask will be 111111 for 38(100110)\n\nSo the problem boils down to generating the mask. Let\'s think about 8-bit numbers. We can later extend the same logic to 32-bit integers. I will count the bits from right to left, starting with 1.\n\nThe largest positive numbers represented by 8 bits will set 7th bit to be 1.\nIn binary format, they should be like:\n01ABCDEF\nwhere A,B,C,D,E,and F might be 0 or 1 and doesn\'t matter.\n\nThe first operation, mask |= mask>>1; will set the 6th bit. So, mask will become (011bcdef). Once again, we do not care the actual value of c,d,e,f.\nNow, we know that the 7th and 6th bits are set, we can safely shift the mask to right by not 1 but 2 bits. mask |= mask>>2; will now set the 5th and 4th bits. \nBy the same reason, we can now shift the mask to right by not 1, 2 or 3 but 4 bits. \nThat is the threshold for 8-bit numbers. We do not need to shift more.\n\nThe following is a detailed illustration with an example:\n\n```\nNote1: 8-bit: 2^7-1, the large number is: 127\nNote2: the first few positive numbers in 8-bit, must be in the format like: 01ABCDEF, where A,B,C,D,E,F belongs to [0,1]\nFor example: \n127 is: 01111111\n126 is: 01111110\n125 is: 01111101\n..\n\n\nLet\'s assume: num = 64 to make our point:\nmask = num; // this ensure the highest significance digit is same as that of num.\n\n01ABCDEF (mask, we do not the value of A,B,C,D,E,F) \n\nIllustration of operation: mask |= mask >> 1;\n01ABCDEF\n 01ABCDE\n---------- OR\n011?????\n\nIllustration of operation: mask |= mask >> 2;\n011?????\n 011???\n---------- OR\n01111???\n\nIllustration of operation: mask |= mask >> 4;\n01111???\n 0111\n---------- OR\n01111111\n\nWe found our goal: \n01111111\n\n```\n\n\n\n\nFor 32-bit integers, the shift operations should continue until reaching to 16. For 64-bit longs, that will be 32.\n\n\nReference/Credit:\nhttps://leetcode.com/problems/number-complement/discuss/96103/maybe-fewest-operations\n | 79 | 2 | ['Simulation'] | 3 |
complement-of-base-10-integer | ☕️ [C++,Java,Python ] | Easy Solution | MASK+XOR | 100% | cjavapython-easy-solution-maskxor-100-by-rjlm | C++ | Mask + XOR\nLogic: In order to find the complement of an integer we have to flip all the 0\'s to 1\'sand all the 1\'s to 0\'s. We know that XOR means one | coderaky | NORMAL | 2022-01-04T02:03:06.823748+00:00 | 2022-01-04T14:41:26.260540+00:00 | 2,437 | false | # C++ | Mask + XOR\n**Logic:** In order to find the complement of an integer we have to flip all the ``0\'s`` to ``1\'s ``and all the ``1\'s`` to ``0\'s``. We know that XOR means one or the other is true but not both. So if we XOR our input with a mask ``11111...``, flipping will happen and we will get our the desired result (complement of that number).\n\n``N ^ bitwiseComplement(N) = 11....11 = Mask``\n``bitwiseComplement(N) = N ^ Mask ``\n\n**Approach:** we will create a mask of all ones until it\'s larger or equal to the number. For example, for ``1010``, the mask will be ``1111``. Once the mask is created, we will XOR it with Number to get the complement.\n```\n 1010\n^1111\n------\n 0101\n------\n```\n\n**We have 2 Method based on ways of creating mask:**\n\n**Method 1:** \n\nCreating mask by using n<<2+1:\n```\n(number) => (how-number-is-derived)=>binary-string\nn = 1 => 1 => 1\nn = 3 => (2*1 + 1) => 11\nn = 7 => (3*2 + 1) => 111\nn = 15 => (7*2 + 1) => 1111\n```\n\n*Complexity*\nTime: O(logn) i.e, O(no of digits in n)\nSpace: O(1)\n\n*Code*\n<iframe src="https://leetcode.com/playground/YqGUiZuv/shared" frameBorder="0" width="400" height="300"></iframe>\n\n**Method 2:**\nCreate mask by bits count:\n```\nfor example, if we take 1010\n(int)(log2(n)+1=> total bits in n =>4\n(int) pow(2, (int)(log2(n)+1) )-1=> binary representation consist of (no of bits) ones=>1111\n```\n\n*Complexity*\nTime: O(log N) since log function has time complexity of log N. \nSpace: O(1)\n\n*Code:*\n```\nint bitwiseComplement(int n) {\n return n ? n ^ (int) pow(2, (int)(log2(n)+1) )-1 : 1;\n}\n```\n\nPlease Upvote! if find helpful \nGuys, if you downvote the post make sure to leave a reason so that I can correct those points, suggestion are welcomed :)\nFind me on: https://coderaky.com\n | 69 | 42 | ['Bit Manipulation'] | 6 |
complement-of-base-10-integer | ✅ [C++/Python/Java] 1 Line || O(1) || Log and Bit OP || Detailed Explanation | cpythonjava-1-line-o1-log-and-bit-op-det-4j6q | PLEASE UPVOTE if you like \uD83D\uDE01 If you have any question, feel free to ask. \n step 1: get the mask of all 1s with the number of binary digits equal to t | linfq | NORMAL | 2022-01-04T05:10:13.069090+00:00 | 2022-01-04T06:30:43.305079+00:00 | 3,090 | false | **PLEASE UPVOTE if you like** \uD83D\uDE01 **If you have any question, feel free to ask.** \n* step 1: get the `mask` of all 1s with the number of binary digits equal to the number of binary digits of n\n\t* for example\n\t\t* `n = 5 = 0b101` so the number of binary digits is `3` and we get `mask = 0b111 = 7`\n\t\t* `n = 8 = 0b1000` so the number of binary digits is `4` and we get `mask = 0b1111 = 15`\n\t\t* `n = 0 = 0b0` so the number of binary digits is `1` and we get `mask = 0b1 = 1`\n\t* `mask = (2 << int(log2(n))) - 1`\n\t\t* for example: `n = 5` => `int(log2(n)) = 2` => `mask = (2 << 2) - 1 = (0b10 << 2) - 1 = 0b1000 - 1 = 0b111 = 7`\n\t\t* **Note `n = 0` is an exception, we just add `max(n, 1)` to deal with it.**\n* step 2: `return mask - n` or `return mask ^ n` or `return mask & (~n)`\n\t* since mask is all-one like `0b11...1` and with same number of binary digits of `n`, `mask - n` is the complement of n\n\t* for example\n\t\t* `n = 5 = 0b101` => `mask = 0b111 = 7` => `mask - n = 0b111 - 0b101 = 0b010 = 2`\n\t\t* `n = 8 = 0b1000` => `mask = 0b1111 = 15` => `mask - n = 0b1111 - 0b1000 = 0b111 = 7`\n\n@pgthebigshot reminded me that `mask - n` also works, really smart!\n\n\n**C++**\n```\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n return ((2 << int(log2(max(n, 1)))) - 1) - n;\n\t\t// return ((2 << int(log2(max(n, 1)))) - 1) ^ n; // also work\n // return ((2 << int(log2(max(n, 1)))) - 1) & (~n); // also work\n }\n};\n```\n**Java**\n```\nclass Solution {\n public int bitwiseComplement(int n) {\n\t\treturn ((2 << (int)(Math.log(Math.max(n, 1)) / Math.log(2))) - 1) - n;\n // return ((2 << (int)(Math.log(Math.max(n, 1)) / Math.log(2))) - 1) ^ n; // also work\n // return ((2 << (int)(Math.log(Math.max(n, 1)) / Math.log(2))) - 1) & (~n); // also work\n }\n}\n```\n**Python**\n```\nclass Solution(object):\n def bitwiseComplement(self, n):\n\t\treturn ((2 << int(math.log(max(n, 1), 2))) - 1) - n\n # return ((2 << int(math.log(max(n, 1), 2))) - 1) ^ n # also work\n # return ((2 << int(math.log(max(n, 1), 2))) - 1) & (~n) # also work\n```\n\n**PLEASE UPVOTE if you like** \uD83D\uDE01 **If you have any question, feel free to ask.** | 57 | 28 | ['C', 'Python', 'Java'] | 9 |
complement-of-base-10-integer | ♾️ C++ | ✅100% Beats | ✅Full Explanation | 1009. Complement of Base 10 Integer | c-100-beats-full-explanation-1009-comple-124u | \n\n# \u2B06\uFE0FPlease Upvote If You Like the Solution :)\n\n# Approach\n1. We know that we need to complement. For that we need to change 0 to 1 and 1 to 0.\ | AnantKumawat22 | NORMAL | 2023-05-26T11:00:03.224713+00:00 | 2023-05-26T13:59:52.147595+00:00 | 2,869 | false | \n\n# \u2B06\uFE0FPlease Upvote If You Like the Solution :)\n\n# Approach\n1. We know that we need to complement. For that we need to change `0` to `1` and `1` to `0`.\n\n2. We need to think, what can we use here from our bitwise operators `(i.e &, |, ^, ~)`.\n\n3. Suppose we have,\n\n```\nEx. n = 10, binary-> 1010\n 1 0 1 0\n x x x x\n -----------\n 0 1 0 1\n\n```\n4. Think if we `xor` the `n` with `1111 (mask)`, we will get our desired answer. Like, \n\n```\n 1 0 1 0\n 1 1 1 1 ==> mask\n ----------- ==> XOR\n 0 1 0 1\n```\n\n5. Now we just need to make `mask`, for that we can take `1` and do `(left shit + add one)` to make it `1111`. Like, \n\n```\nmask -> 0 0 0 1 \n 0 0 1 0 ==> left shift (mask << 1)\n 0 0 1 1 ==> add 1 (mask + 1)\n```\n\n6. we know that if we do `&` of `n` with `1111` then we will get that `n` itself.\n\n7. so we will do the `5th step` until `((mask & n) != n)`.\n\n8. At last we can able to make `1111 (mask)` and now return `(mask ^ n)`, see `4th step`. \n\n# Complexity\n- Time complexity: O(ceil(log(n)))\n- Space complexity: O(1)\n\n**Why Time complexity: O(ceil(log(n)))???**\n\n1. we are running the while loop for the number of bit of n. \n n = 2^x ==> (x = number of bits).\n Ex. n = 10, binary -> 1 0 1 0 (4 bits)\n 10 = 2^x\n log2(10) = x\n ceil (log2(10)) = x, \n\n2. log2(10) == 3.322, but we represent 10 in 4 bits so take ceil value.\n\n# Code\n```\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n int mask = 1;\n \n while((mask & n) != n){\n mask = mask << 1;\n mask += 1;\n }\n return mask ^ n;\n }\n};\n``` | 38 | 0 | ['Bit Manipulation', 'C', 'Bitmask', 'C++'] | 4 |
complement-of-base-10-integer | C++ 0ms solution using bitwise operation | c-0ms-solution-using-bitwise-operation-b-eklw | The idea is to find the bit length of N, construct a string of 1\'s with the same length, and XOR with N.\n\nFor example, 5 in binary is 101. Notice that 101 ^ | vanesschanppi | NORMAL | 2020-05-05T03:54:47.577522+00:00 | 2020-05-10T02:07:09.681309+00:00 | 2,128 | false | The idea is to find the bit length of N, construct a string of 1\'s with the same length, and XOR with N.\n\nFor example, 5 in binary is 101. Notice that 101 ^ 111 = 010 (2 in decimal).\n\n int bitwiseComplement(int N) {\n int comp = 1;\n while (comp < N) \n comp = (comp << 1) | 1;\n return N ^ comp;\n }\n\t | 35 | 1 | [] | 5 |
complement-of-base-10-integer | [Python] Oneliner explained | python-oneliner-explained-by-dbabichev-vlvp | What we actually need to find in this problem is length of our number in bits: if we have 10 = 1010, then complementary is 5 = 101. Note, that 5 + 10 = 15 = 2^4 | dbabichev | NORMAL | 2020-10-05T09:12:35.316946+00:00 | 2020-10-05T09:12:35.316990+00:00 | 1,624 | false | What we actually need to find in this problem is length of our number in bits: if we have `10 = 1010`, then complementary is `5 = 101`. Note, that `5 + 10 = 15 = 2^4 - 1`. So, let us just find length of our number in bits and subtract `N-1`.\n\n**Complexity** time complexity is `O(log N)` to find its length; space complexity is also `O(log N)` to keep binary representation of our number.\n\n```\nclass Solution:\n def bitwiseComplement(self, N):\n return (1<<(len(bin(N)) - 2)) - 1 - N\n```\n\nIf you have any questions, feel free to ask. If you like solution and explanations, please **Upvote!** | 29 | 4 | [] | 2 |
complement-of-base-10-integer | [JAVA] Detailed Explanation of Intution| BIT MASKING|BIT FLIP | 5 Approach | java-detailed-explanation-of-intution-bi-a9ir | Intution: Since we have to find the complement of a given decimal number in binary form, we will use the complement property which is N + Compliment(N) = 111... | abhi9720 | NORMAL | 2022-01-04T03:39:16.146367+00:00 | 2022-01-04T10:13:18.104859+00:00 | 942 | false | *Intution:* Since we have to find the complement of a given decimal number in binary form, we will use the complement property which is N + Compliment(N) = 111....111.\nFoe Example : \n* N = 5 => N = 101 , Compliment(N) = 010 => 101 + 010 = 111\n* N = 8 => N = 1000 , Compliment(N) = 0111 => 1000 + 0111 = 1111\n* N = 17 => N = 10001, Compliment(N) = 01110 => 10001 + 01110 = 11111 and so on....\n\nSo we need to find out a number just greater than N having all 1 bits and then subtract N from it to get the complement.\nExample : for N=5(101) => 111 is the number just greater than 5 having all 1s. So, 111 - 101 = 010 (Compliment)\n\n***Approach1: (1ms)***\n*Explanation:* let n = 5 and i=0, j=0\n* Iteration1 : (i=0 < n) => i = 0 + (2^0) = 1 (1) , j = 1\n* Iteration2 : (i=1 < n) => i = 1 + (2^1) = 3 (11) , j = 2\n* Iteration3 : (i=3 < n) => i = 3 + (2^2) = 7 (111) , j = 3\n* Iteration4 : (i=7 > n) => loop breaks\nCompliment = i -n => 111 - 101 = 010\n```\n\nclass Solution {\n public int bitwiseComplement(int n) {\n \n if(n==0) return 1;\n \n int i = 0;\n int j = 0;\n \n while(i<n){\n i += Math.pow(2,j);\n j++;\n }\n \n return i-n;\n }\n}\n```\n\n\n***Approach2: (0ms)***\n*Explanation:* Just replace the arithmetical operation into bit manipulation as i += Math.pow(2^j) can be replaced with i<<n | 1. Let n = 5 and i=0, j=0\n* Iteration1 : (i=0 < n) => i =0<<1 | 1 = 1 | 1 = 1\n* Iteration2 : (i=1 < n) => i = 1<<1 | 1 = 10 | 1 = 11 = 3\n* Iteration3 : (i=3 < n) => i = 3<<1 | 1 = 11<<1 | 1 = 110 | 1 = 111 = 7\n* Iteration4 : (i=7 > n) => loop breaks\nCompliment = i -n => 111 - 101 = 010\n\n```\nclass Solution {\n public int bitwiseComplement(int n) {\n \n if(n==0) return 1;\n \n int i = 0;\n \n while(i<n){\n i = i<<1 | 1;\n }\n \n return i-n;\n }\n}\n```\n\n\n***Approach3: (0ms)***\n*Explanation:* let n = 10 and i=1, j=0\n* Iteration1 : (i=1 < n) => i = 1 * 2 +1 = 3 (11)\n* Iteration2 : (i=3 < n) => i = 3 * 2 + 1 = 7 (111) \n* Iteration3 : (i=7 < n) => i = 7 * 2 + 1 = 15 (1111)\n* Iteration4 : (i=15 > n) => loop breaks\nCompliment = i -n => 1111 - 1010 = 0101 => 101 = 5\n\n```\nclass Solution {\n public int bitwiseComplement(int n) {\n \n int i = 1;\n \n while(i<n){\n i = i*2 + 1;\n }\n \n return i-n;\n }\n}\n```\n\n\n\n*In every approach at last I am calculating compliment using i - n, which can also be done using **XOR* as *i ^n will give the compliment*:\n* i = 111 , n = 101 (5) => i ^ n = 111 ^ 101 = 010 (compliment)\n* i = 1111, n = 1010 (10) => i ^ n = 1111 ^ 1010 = 0101 (compliment)\n\n\n***Approach4: (0ms)*** : Using Xor Property\n In complement of n , every bit is opposite to n\n* so at every place we have odd number of 1 so xor will be 1 at every place\n* n^bitwiseComplement(n) = 111....1\n\n```\nMath:\na xor b = c\n(a xor b) xor a = c xor a\n(a xor a) xor b = c xor a\n0 xor b = c xor a\nb = c xor a\n```\n\n* n^111....1 = bitwiseComplement(n)\n* FInd the upper bound Number which has all 1\'s, Number just greater than n, will all bit 1\n```\n Suppose N= 5\n X = 2+1 = 3 , 6+1 = 7 \n X>N , X^N = 111^101 = 010\n```\n \n```\nclass Solution {\n public int bitwiseComplement(int n) {\t \n int X = 1; \n while(X<n) X= X*2+1;\n return X^n; \n }\n}\n```\n\n\n***Approach5: (0ms)*** : Using Masking\nConsider N = 10 ( 1010 )\nWe will start processing from right most bit (LSB) , Using **i** Variable we will we track bit , if it is set then we will unset it or vice-versa.\n\n```\ni=0 , intial bit position\nif n&1 == 0 means last bit is unset , so set it \n result += 1<<i;\nand if set then it will set to unset \n\n\t i++;// move to next bit position\n\tn >>= 1; // now move to second last bit or right shift to process next bit\n```\nfor 0 index , it is 0 bit set to 1\nfor 2 index bit , it is 0 bit set to 1\n**result = 1 + 100 = 0101**\n\n\n```\nclass Solution {\n public int bitwiseComplement(int n) {\n \n if(n==0) return 1;\n \n int result = 0;\n int i = 0; \n while(n>0){\n if((n&1) == 0){\n result += 1<<i;\n }\n i++;\n n >>= 1;\n }\n \n return result;\n }\n}\n```\n \n\n | 21 | 8 | ['Bit Manipulation', 'Bitmask', 'Java'] | 1 |
complement-of-base-10-integer | Java 0ms 32mb... beats 100% | java-0ms-32mb-beats-100-by-msadler-66mi | Intuition:\nWe want to XOR N\'s binary with an array of 1\'s of the same length.\nApproach:\nFind out how long N is (via while loop.... 2^y=N).\nThen we subtrac | msadler | NORMAL | 2019-03-19T20:08:28.372736+00:00 | 2019-05-07T23:51:13.870383+00:00 | 2,507 | false | **Intuition:**\nWe want to XOR N\'s binary with an array of 1\'s of the same length.\n**Approach:**\nFind out how long N is (via while loop.... 2^y=N).\nThen we subtract 1 from our number, and we will get an array of 1\'s in binary.\n(ie. if x=2^8, then x-1 in binary is "1111 1111").\n\n\n\n```\nclass Solution {\n public int bitwiseComplement(int N) {\n if (N == 0) return 1;\n if (N == 1) return 0;\n int x = 1;\n while(x<= N){\n x = x << 1; // equialently written as x*=2;\n }\n return N ^ (x-1);\n }\n}\n``` | 18 | 1 | [] | 6 |
complement-of-base-10-integer | Python 4 lines: faster than 99%, memory less than 100% | python-4-lines-faster-than-99-memory-les-hn24 | \nclass Solution:\n def bitwiseComplement(self, N: int) -> int:\n P = 2\n while P <= N:\n P *= 2\n return P - 1 - N\n\nIf P i | chenyian | NORMAL | 2019-05-14T01:58:29.385585+00:00 | 2020-06-21T02:28:27.028516+00:00 | 1,817 | false | ```\nclass Solution:\n def bitwiseComplement(self, N: int) -> int:\n P = 2\n while P <= N:\n P *= 2\n return P - 1 - N\n```\nIf P is the smallest power of 2 larger than N, then N and its binary complement M satisfies N+M=P-1. | 17 | 0 | [] | 5 |
complement-of-base-10-integer | C++ | 100% faster | Beginner friendly | Simple Math | c-100-faster-beginner-friendly-simple-ma-ft8z | \nint bitwiseComplement(int n) {\n int i;\n for(i=0;i<=n;)\n {\n i=(i*2)+1;\n if(n<=i)\n break;\n | shyamal122 | NORMAL | 2022-01-04T11:04:10.846331+00:00 | 2022-01-04T11:04:10.846371+00:00 | 759 | false | ```\nint bitwiseComplement(int n) {\n int i;\n for(i=0;i<=n;)\n {\n i=(i*2)+1;\n if(n<=i)\n break;\n }\n return i-n; \n }\n```\t\n\n\n**If you have some doubts feel free to ask me.**\nIt\'s my first post on leetcode.If you like it please upvote. | 13 | 1 | ['Math', 'C'] | 1 |
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