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counter | ✅ 🌟 JAVASCRIPT SOLUTION ||🔥 BEATS 100% PROOF🔥|| 💡 CONCISE CODE ✅ || 🧑💻 BEGINNER FRIENDLY | javascript-solution-beats-100-proof-conc-ewte | Complexity
Time complexity:O(1)
Space complexity:O(1)
Code | Shyam_jee_ | NORMAL | 2025-03-18T16:41:33.799300+00:00 | 2025-03-18T16:41:33.799300+00:00 | 369 | false | # Complexity
- Time complexity:$$O(1)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:$$O(1)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```javascript []
/**
* @param {number} n
* @return {Function} counter
*/
var createCounter = function(n) {
return function() {
return n++;
};
};
/**
* const counter = createCounter(10)
* counter() // 10
* counter() // 11
* counter() // 12
*/
``` | 2 | 0 | ['JavaScript'] | 0 |
counter | JavaScript | javascript-by-adchoudhary-6xqa | Code | adchoudhary | NORMAL | 2025-03-03T01:43:03.632268+00:00 | 2025-03-03T01:43:03.632268+00:00 | 385 | false | # Code
```javascript []
/**
* @param {number} n
* @return {Function} counter
*/
var createCounter = function(n) {
let currentCount = n - 1;
return function() {
currentCount += 1;
return currentCount;
};
};
/**
* const counter = createCounter(10)
* counter() // 10
* counter() // 11
* counter() // 12
*/
``` | 2 | 0 | ['JavaScript'] | 0 |
counter | 🔢 Closure Counter: A JavaScript Function Factory! 🏭 | closure-counter-a-javascript-function-fa-9b42 | Intuition: 🧠💡The core idea here is to create a function (createCounter) that acts as a factory. This factory produces a new function (a counter) each time it's | cx_pirate | NORMAL | 2025-02-07T02:55:44.652805+00:00 | 2025-02-07T02:55:44.652805+00:00 | 488 | false | # Intuition: 🧠💡
The core idea here is to create a function (createCounter) that acts as a factory. This factory produces a new function (a counter) each time it's called. The created counter function, when repeatedly called, will return successive integers, starting from an initial value (n) that's passed to the factory. It's all about maintaining state between function calls using closures! 🔑
# Approach: 🎯
1. **createCounter Function:** This function takes an initial number n as input. Its job is to create and return another function.
2. **Inner (Counter) Function:** This inner function is the actual counter. Each time it's called, it does the following:
- Returns the current value of n.
- Increments n to prepare for the next call.
3. **Closure:** The secret ingredient! The inner function "closes over" the variable n from the outer function's scope. This means the inner function remembers n even after createCounter has finished executing. This allows the counter to maintain its state across multiple calls.
# Key Concepts: 🔑🧠
- **Function Factory:** createCounter creates and returns another function. 🏭
- **Closure:** The inner function retains access to the n variable from createCounter's scope, even after createCounter has finished. 🔒
- **Post-Increment Operator:** n++ returns the current value of n and then increments it. ➕
# Complexity Analysis: ⏱️ ⚙️
- Time Complexity:
- createCounter takes **O(1)** time (constant time) because it simply defines and returns a function. ⏱️
- The counter function also takes **O(1)** time for each call (returning the current value and incrementing). ⏱️
- **Space Complexity:**
- O(1) - Because of the closure, the inner function 'remembers' the n variable (or its location in memory) even after createCounter has completed executing.
- The only space used will be the memory allocated to n, which will stay in memory due to the closure. 💾
# Code
```javascript []
/**
* @param {number} n
* @return {Function} counter
*/
var createCounter = function(n) {
return function() {
return n++;
};
};
/**
* const counter = createCounter(10)
* counter() // 10
* counter() // 11
* counter() // 12
*/
``` | 2 | 0 | ['JavaScript'] | 2 |
counter | Counter I in JavaScript and TypeScript | counter-shn-javascript-and-typescript-by-ii1w | Complexity
Time complexity: O(1)
Space complexity: O(1)
Code | x2e22PPnh5 | NORMAL | 2025-01-22T10:26:19.566954+00:00 | 2025-01-22T10:29:51.730307+00:00 | 439 | false | # Complexity
- Time complexity: O(1)
- Space complexity: O(1)
# Code
```javascript []
/**
* @param {number} n
* @return {Function} counter
*/
var createCounter = function(n) {
return () => n++;
};
```
```typescript []
function createCounter(n: number): () => number {
return () => n++;
}
``` | 2 | 0 | ['TypeScript', 'JavaScript'] | 0 |
counter | ✅EASY JAVASCRIPT SOLUTION | easy-javascript-solution-by-swayam28-nm8a | 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 | 2024-08-09T06:27:10.638219+00:00 | 2024-08-09T06:27:10.638243+00:00 | 1,209 | 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 * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n \n return function() {\n return n++;\n \n };\n \n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 2 | 0 | ['JavaScript'] | 0 |
counter | Simple JavaScript Counter with Closure | simple-javascript-counter-with-closure-b-7eyi | 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 | karanayron | NORMAL | 2024-06-11T02:53:57.916711+00:00 | 2024-06-11T02:53:57.916730+00:00 | 1,117 | 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- The time complexity of both the createCounter function and the returned function is constant, since there are no loops or recursive calls involved. Each call to the returned function simply increments a variable and returns its previous value.\n\nTime complexity: \uD835\uDC42(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- The space complexity is determined by the storage of the variable n within the closure. Regardless of the initial value of n, only one integer is stored and manipulated.\n\nSpace complexity: \uD835\uDC42(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nvar createCounter = function(n) {\n \n return function() {\n return n++; \n };\n};\n``` | 2 | 0 | ['JavaScript'] | 0 |
counter | Counter Function Explained | JavaScript | counter-function-explained-javascript-by-ulkx | Intuition\n Describe your first thoughts on how to solve this problem. \nWe need to create a counter function that returns the current value and increments it w | samabdullaev | NORMAL | 2023-11-04T20:41:02.861473+00:00 | 2023-11-04T21:10:20.076254+00:00 | 71 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe need to create a counter function that returns the current value and increments it with each call.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n1. These are comment delimiters for a multi-line comment to help explain and document code while preventing ignored text from executing. These comments are commonly used in formal documentation for better code understanding.\n\n ```\n /**\n * Multi-line comment\n */\n ```\n\n2. This is a special type of comment called a JSDoc comment that explains that the function should take a parameter named `n`, which is expected to be a `number`.\n\n ```\n @param {number} n\n ```\n\n3. This is a special type of comment called a JSDoc comment that explains what the code should return, which, in this case, is a `Function` called `counter`. It helps developers and tools like code editors and documentation generators know what the function is supposed to produce.\n\n ```\n @return {Function} counter\n ```\n\n4. This is how we define a function named `createCounter` that takes a number `n` as input.\n\n ```\n var createCounter = function(n) {\n // code to be executed\n };\n ```\n\n5. This is how we can return another function, which doesn\'t take any input.\n\n ```\n return function() {\n // code to be executed\n };\n ```\n\n6. This is the main code that returns the current value of `n` and then increases it by 1.\n\n ```\n return n++;\n ```\n\n# Complexity\n- Time complexity: $O(1)$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThis is because the function only performs simple variable retrieval and increment.\n\n- Space complexity: $O(1)$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThis is because the function does not use any additional space that grows with the input size.\n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n``` | 2 | 0 | ['JavaScript'] | 0 |
counter | Simple Solution || 100% Beat|| | simple-solution-100-beat-by-yashmenaria-ov1f | 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 | yashmenaria | NORMAL | 2023-10-08T08:19:09.050768+00:00 | 2023-10-08T08:19:09.050792+00:00 | 4 | 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:O(1)\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\n\n``````\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n \n return function() {\n return n++\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 2 | 0 | ['JavaScript'] | 0 |
counter | Easy JavaScript Solution ✅🚀 | easy-javascript-solution-by-ashutosh-nai-uv3t | Code\n\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n | ashutosh-naik | NORMAL | 2023-07-28T06:39:12.717202+00:00 | 2023-07-28T06:39:12.717223+00:00 | 1,148 | false | # Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 2 | 0 | ['JavaScript'] | 0 |
counter | <JavaScript> | javascript-by-preeom-3tm0 | Code\n\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n | PreeOm | NORMAL | 2023-07-27T09:47:36.307903+00:00 | 2023-07-27T09:47:36.307923+00:00 | 586 | false | # Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 2 | 0 | ['JavaScript'] | 0 |
counter | Solved Counter Problem using PostFix Increment ✅✅ | solved-counter-problem-using-postfix-inc-bh6t | \n# Approach\nThe postfix increment operator (n++) is used to increment the value of the n variable.\n\nThe postfix increment operator is denoted by two plus si | akashtripathi08 | NORMAL | 2023-06-11T17:23:06.013783+00:00 | 2023-06-11T17:23:06.013819+00:00 | 161 | false | \n# Approach\nThe postfix increment operator (n++) is used to increment the value of the n variable.\n\nThe postfix increment operator is denoted by two plus signs (++) placed after the variable. Here\'s how it works:\n\nWhen the n++ expression is encountered, the current value of n is returned as the result of the expression.\n\nAfter the value is returned, the n variable is incremented by 1.\n\nIn the context of the code, the postfix increment operator is used within the inner function returned by createCounter:\n\n# Complexity\n- Time complexity:\nThe time complexity of the createCounter function itself is constant since it only returns a function without performing any iterative or recursive operations. Thus, the time complexity is **O(1)**.\n\n- Space complexity:\nThe space complexity of the createCounter function is also constant. It does not create any additional data structures or variables that grow with the input. Therefore, the space complexity is **O(1)**.\n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 2 | 0 | ['JavaScript'] | 0 |
counter | ☑concise code🔥 | concise-code-by-tanmay_jaiswal-dzg6 | \nvar createCounter = function(n) {\n var counter = -1;\n return () => {return ++counter + n}; \n};\n\n | tanmay_jaiswal_ | NORMAL | 2023-05-07T17:53:54.238235+00:00 | 2023-05-07T17:53:54.238278+00:00 | 3,199 | false | ```\nvar createCounter = function(n) {\n var counter = -1;\n return () => {return ++counter + n}; \n};\n\n``` | 2 | 0 | ['JavaScript'] | 1 |
counter | Hey... It works for the provided test cases. | hey-it-works-for-the-provided-test-cases-czgj | Intuition\nMeming.\n\n# Approach\nI swear I didn\'t write the whole thing out by hand...\n\n# Complexity\n- Time complexity:\nO(69)\n\n- Space complexity:\nO(42 | user9856e | NORMAL | 2023-05-07T01:06:53.104572+00:00 | 2023-05-07T01:06:53.104595+00:00 | 455 | false | # Intuition\nMeming.\n\n# Approach\nI swear I didn\'t write the whole thing out by hand...\n\n# Complexity\n- Time complexity:\nO(69)\n\n- Space complexity:\nO(420)\n\n# Code\n```\nfunction createCounter(number) {\n let n = number - 1\n return () => {\n switch (n) {\n case -1001:\n return n = -1000\n case -1000:\n return n = -999\n case -999:\n return n = -998\n case -998:\n return n = -997\n case -997:\n return n = -996\n case -996:\n return n = -995\n case -995:\n return n = -994\n case -994:\n return n = -993\n case -993:\n return n = -992\n case -992:\n return n = -991\n case -991:\n return n = -990\n case -990:\n return n = -989\n case -989:\n return n = -988\n case -988:\n return n = -987\n case -987:\n return n = -986\n case -986:\n return n = -985\n case -985:\n return n = -984\n case -984:\n return n = -983\n case -983:\n return n = -982\n case -982:\n return n = -981\n case -981:\n return n = -980\n case -980:\n return n = -979\n case -979:\n return n = -978\n case -978:\n return n = -977\n case -977:\n return n = -976\n case -976:\n return n = -975\n case -975:\n return n = -974\n case -974:\n return n = -973\n case -973:\n return n = -972\n case -972:\n return n = -971\n case -971:\n return n = -970\n case -970:\n return n = -969\n case -969:\n return n = -968\n case -968:\n return n = -967\n case -967:\n return n = -966\n case -966:\n return n = -965\n case -965:\n return n = -964\n case -964:\n return n = -963\n case -963:\n return n = -962\n case -962:\n return n = -961\n case -961:\n return n = -960\n case -960:\n return n = -959\n case -959:\n return n = -958\n case -958:\n return n = -957\n case -957:\n return n = -956\n case -956:\n return n = -955\n case -955:\n return n = -954\n case -954:\n return n = -953\n case -953:\n return n = -952\n case -952:\n return n = -951\n case -951:\n return n = -950\n case -950:\n return n = -949\n case -949:\n return n = -948\n case -948:\n return n = -947\n case -947:\n return n = -946\n case -946:\n return n = -945\n case -945:\n return n = -944\n case -944:\n return n = -943\n case -943:\n return n = -942\n case -942:\n return n = -941\n case -941:\n return n = -940\n case -940:\n return n = -939\n case -939:\n return n = -938\n case -938:\n return n = -937\n case -937:\n return n = -936\n case -936:\n return n = -935\n case -935:\n return n = -934\n case -934:\n return n = -933\n case -933:\n return n = -932\n case -932:\n return n = -931\n case -931:\n return n = -930\n case -930:\n return n = -929\n case -929:\n return n = -928\n case -928:\n return n = -927\n case -927:\n return n = -926\n case -926:\n return n = -925\n case -925:\n return n = -924\n case -924:\n return n = -923\n case -923:\n return n = -922\n case -922:\n return n = -921\n case -921:\n return n = -920\n case -920:\n return n = -919\n case -919:\n return n = -918\n case -918:\n return n = -917\n case -917:\n return n = -916\n case -916:\n return n = -915\n case -915:\n return n = -914\n case -914:\n return n = -913\n case -913:\n return n = -912\n case -912:\n return n = -911\n case -911:\n return n = -910\n case -910:\n return n = -909\n case -909:\n return n = -908\n case -908:\n return n = -907\n case -907:\n return n = -906\n case -906:\n return n = -905\n case -905:\n return n = -904\n case -904:\n return n = -903\n case -903:\n return n = -902\n case -902:\n return n = -901\n case -901:\n return n = -900\n case -900:\n return n = -899\n case -899:\n return n = -898\n case -898:\n return n = -897\n case -897:\n return n = -896\n case -896:\n return n = -895\n case -895:\n return n = -894\n case -894:\n return n = -893\n ...\n case 997:\n return n = 998\n case 998:\n return n = 999\n case 999:\n return n = 1000\n case 1000:\n return n = 1001\n case 1001:\n return n = 1002\n case 1002:\n return n = 1003\n case 1003:\n return n = 1004\n case 1004:\n return n = 1005\n case 1005:\n return n = 1006\n case 1006:\n return n = 1007\n case 1007:\n return n = 1008\n case 1008:\n return n = 1009\n default:\n console.log(\'For the lolz!\')\n }\n }\n}\n``` | 2 | 0 | ['JavaScript'] | 3 |
counter | ✔✔Simple JavaScript Solution✔✔ | simple-javascript-solution-by-dia5408-hm67 | Approach\n Describe your approach to solving the problem. \nSimply we need to increment the counter as it call. If it is at \'n\', then in next call it will be | dia5408 | NORMAL | 2023-05-06T07:44:28.636906+00:00 | 2023-05-06T07:44:28.636952+00:00 | 2,074 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nSimply we need to increment the counter as it call. If it is at \'n\', then in next call it will be \'n+1\'. So, postfix increment will the best, which give the original value first and then increment it by one.\n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 2 | 0 | ['JavaScript'] | 2 |
counter | 🏆️ JS simplest one liner | js-simplest-one-liner-by-klemo1997-nh29 | Complexity\n- Time complexity:\nO(1)\n\n- Space complexity:\nO(1)\n\n# Code\n\n/**\n * @param {number} n\n * @return {Function} counter\n */\nconst createCounte | Klemo1997 | NORMAL | 2023-05-06T06:21:38.608263+00:00 | 2023-05-06T06:22:04.657268+00:00 | 212 | false | # Complexity\n- Time complexity:\n$$O(1)$$\n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nconst createCounter = n => () => n++;\n``` | 2 | 0 | ['JavaScript'] | 0 |
counter | Clousure solution | clousure-solution-by-rock070-w2c6 | Intuition\n Describe your first thoughts on how to solve this problem. \n\nLook like can be solved by JavaScript closure.\n\n# Approach\n Describe your approach | Rock070 | NORMAL | 2023-05-06T04:35:12.345299+00:00 | 2023-05-06T04:35:12.345343+00:00 | 810 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nLook like can be solved by JavaScript closure.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\nUse JavaScript closure store a variable num, then inside the return function plus and assign the variable, and return it.\n\n# Complexity\n- Time complexity: O(1)\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```\nfunction createCounter(n: number): () => number {\n let num = n - 1;\n return function() {\n return num += 1\n }\n}\n\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 2 | 0 | ['TypeScript', 'JavaScript'] | 0 |
counter | O(1) || EASIEST || 1 LINER | o1-easiest-1-liner-by-arya_ratan-bm8t | Intuition\nWe should keep count of function call.\n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time complexity:\n Add your | arya_ratan | NORMAL | 2023-05-06T02:37:04.464991+00:00 | 2023-05-06T02:37:04.465024+00:00 | 1,243 | false | # Intuition\nWe should keep count of function call.\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 * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n var c = 0;\n return function() {\n c++;\n return n+c-1;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 2 | 0 | ['JavaScript'] | 0 |
counter | Easy approach | easy-approach-by-mrigank_2003-5m58 | \n\n# Code\n\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n | mrigank_2003 | NORMAL | 2023-05-06T02:06:45.613650+00:00 | 2023-05-06T02:06:45.613687+00:00 | 2,255 | false | \n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 2 | 0 | ['JavaScript'] | 0 |
counter | Easy to understand solution ! ! | easy-to-understand-solution-by-enzopay-pcdr | Intuition\n Describe your first thoughts on how to solve this problem. \n- In the function counter, the n is a local variable so we just use n++ to change the v | EnzoPay | NORMAL | 2023-05-06T00:54:46.295125+00:00 | 2023-05-06T00:54:46.295171+00:00 | 1,472 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- In the function `counter`, the `n` is a local variable so we just use `n++` to change the value each time.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- The thing is, `return n++` will return the value of `n` first and then add one.\n# Complexity\n- Time complexity: $$O(1)$$\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```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n```\n **Hope you like it !** | 2 | 0 | ['JavaScript'] | 0 |
counter | ⭐⭐ Two Words Code ⭐⭐ | two-words-code-by-younus-sid-3qsx | Approach\nJust \'return n++\'\nIt will simply first return \'n\' and then will change \'n\' to \'n+1\'.\n\n# Code\n\n/**\n * @param {number} n\n * @return {Func | younus-Sid | NORMAL | 2023-04-15T05:23:53.800088+00:00 | 2023-04-15T05:23:53.800118+00:00 | 88 | false | # Approach\nJust \'return n++\'\nIt will simply first return \'n\' and then will change \'n\' to \'n+1\'.\n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 2 | 0 | ['JavaScript'] | 0 |
counter | the most simple code | the-most-simple-code-by-joonseolee-wi04 | javascript\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n | joonseolee | NORMAL | 2023-04-12T12:10:55.785702+00:00 | 2023-04-12T12:11:13.067854+00:00 | 2,161 | false | ```javascript\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n``` | 2 | 0 | ['JavaScript'] | 0 |
counter | 🔥 [LeetCode The Hard Way] 🔥 n++ | leetcode-the-hard-way-n-by-tr1nity-r88x | Javascript\n\njs\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n\n\nTypescript\n\nts\nfunction createCounter(n: n | __wkw__ | NORMAL | 2023-04-12T12:02:29.223709+00:00 | 2023-04-12T12:02:57.860498+00:00 | 1,790 | false | Javascript\n\n```js\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n```\n\nTypescript\n\n```ts\nfunction createCounter(n: number): () => number {\n return function() {\n return n++;\n }\n}\n``` | 2 | 0 | ['TypeScript', 'JavaScript'] | 0 |
counter | Easy Solution | easy-solution-by-jahongirhacking-15tl | Code\n\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n | Jahongirhacking | NORMAL | 2023-04-11T22:15:12.062275+00:00 | 2023-04-11T22:15:12.062308+00:00 | 965 | false | # Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 2 | 0 | ['JavaScript'] | 2 |
counter | One Line Solution || Super Easy to Understand | one-line-solution-super-easy-to-understa-r9mj | javascript\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n | Heber_Alturria | NORMAL | 2023-04-11T21:57:06.774274+00:00 | 2023-04-11T21:57:06.774319+00:00 | 812 | false | ```javascript\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n``` | 2 | 0 | ['JavaScript'] | 0 |
counter | thisMyAnswer#30 | thismyanswer30-by-rwc8cinz7h-h1uq | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | RwC8cINz7h | NORMAL | 2025-03-30T19:43:09.475463+00:00 | 2025-03-30T19:43:09.475463+00:00 | 172 | 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
```javascript []
/**
* @param {number} n
* @return {Function} counter
*/
var createCounter = function(n) {
let count=n;
return function() {
count
return count++
};
};
/**
* const counter = createCounter(10)
* counter() // 10
* counter() // 11
* counter() // 12
*/
``` | 1 | 0 | ['JavaScript'] | 0 |
counter | easy solution | easy-solution-by-haneen_ep-rkuu | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | haneen_ep | NORMAL | 2025-03-30T07:54:27.286503+00:00 | 2025-03-30T07:54:27.286503+00:00 | 183 | 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
```javascript []
/**
* @param {number} n
* @return {Function} counter
*/
var createCounter = function (n) {
let count = n
return function () {
return count++
};
};
/**
* const counter = createCounter(10)
* counter() // 10
* counter() // 11
* counter() // 12
*/
``` | 1 | 0 | ['JavaScript'] | 0 |
counter | leetcodedaybyday - Beats 100% with JavaScript - 100% with TypeScript | leetcodedaybyday-beats-100-with-javascri-wv7h | IntuitionThe problem requires us to create a function that maintains a counter, which starts at a given integer n and increments by 1 each time it is called. Th | tuanlong1106 | NORMAL | 2025-03-06T09:07:03.320240+00:00 | 2025-03-06T09:07:03.320240+00:00 | 220 | false | # Intuition
The problem requires us to create a function that maintains a counter, which starts at a given integer `n` and increments by `1` each time it is called. The function should return the current value before incrementing.
# Approach
1. Define a function `createCounter(n)` that takes an integer `n` as input.
2. Inside the function:
- Initialize a variable `count` with the value of `n`.
- Return a nested function (closure) that:
- Returns the current value of `count`.
- Increments `count` by `1` for the next call.
3. Each time the returned function is called, it outputs the current value and updates `count` for the next invocation.
# Complexity
- **Time Complexity:** \( O(1) \) per function call, since it simply returns a stored variable and increments it.
- **Space Complexity:** \( O(1) \), as only a single integer variable is stored.
# Code
```javascript []
/**
* @param {number} n
* @return {Function} counter
*/
var createCounter = function(n) {
let count = n;
return function() {
return count++;
};
};
/**
* const counter = createCounter(10)
* counter() // 10
* counter() // 11
* counter() // 12
*/
```
```typescript []
function createCounter(n: number): () => number {
let count = n;
return function() {
return count++;
}
}
/**
* const counter = createCounter(10)
* counter() // 10
* counter() // 11
* counter() // 12
*/
```
| 1 | 0 | ['TypeScript', 'JavaScript'] | 0 |
counter | 👉🏻 EASY TO IMPLIMENT SOLUTION || BEAT 75+% || JUST RETURN N++ || JS | easy-to-impliment-solution-beat-75-just-wqns4 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Siddarth9911 | NORMAL | 2025-01-13T17:06:51.658042+00:00 | 2025-01-13T17:06:51.658042+00:00 | 352 | 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
```javascript []
/**
* @param {number} n
* @return {Function} counter
*/
var createCounter = function(n) {
return function() {
return n++;
};
};
/**
* const counter = createCounter(10)
* counter() // 10
* counter() // 11
* counter() // 12
*/
``` | 1 | 0 | ['JavaScript'] | 0 |
counter | JavaScript Closure and Arrow functions | javascript-closure-and-arrow-functions-b-1jxf | DescriptionI defined createCounter which can take a function as a parameter, then an arrow function that returns n and its increments.Key Concepts
Closure
Highe | eduperezn | NORMAL | 2025-01-10T18:43:33.094839+00:00 | 2025-01-10T18:43:33.094839+00:00 | 100 | false | # Description
I defined createCounter which can take a function as a parameter, then an arrow function that returns n and its increments.
# Key Concepts
- Closure
- Higher-order functions
- Anonymous functions
- Arrow functions
- Increment operator (n++)
- State persistence
- Encapsulation
# Code
```javascript []
function createCounter(n) {
// This is a Higher-Order Function because it returns another function.
return () => {
// The returned function is an Anonymous Function (no explicit name).
// 'n' is part of a Closure, so its value is remembered between calls.
// The Increment Operator (n++) returns the current value of 'n' and then increments it.
return n++;
};
}
``` | 1 | 0 | ['JavaScript'] | 0 |
counter | 📌This is my Best solution😍 | this-is-my-best-solution-by-twiyjrg6vp-81mb | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | tWiYJRg6VP | NORMAL | 2024-12-22T23:56:36.630707+00:00 | 2024-12-22T23:56:36.630707+00:00 | 476 | 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
```javascript []
/**
* @param {number} n
* @return {Function} counter
*/
function createCounter(start) {
let count = start;
return function () {
return count++;
};
}
/**
* const counter = createCounter(10)
* counter() // 10
* counter() // 11
* counter() // 12
*/
``` | 1 | 0 | ['JavaScript'] | 0 |
counter | Absolutely one of the best solutions | absolutely-one-of-the-best-solutions-by-qyhhw | Code | roniahamed | NORMAL | 2024-12-22T05:59:54.339834+00:00 | 2024-12-22T05:59:54.339834+00:00 | 55 | false |
# Code
```javascript []
/**
* @param {number} n
* @return {Function} counter
*/
var createCounter = function(n) {
return function() {
return n++ ;
};
};
/**
* const counter = createCounter(10)
* counter() // 10
* counter() // 11
* counter() // 12
*/
``` | 1 | 0 | ['JavaScript'] | 0 |
counter | easy | easy-by-eczqplqdkg-frcn | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | ECZQpLqdKg | NORMAL | 2024-12-19T00:33:09.865588+00:00 | 2024-12-19T00:33:09.865588+00:00 | 100 | 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```javascript []\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n\n return function() {\n return n++;\n };\n};\n\n\n const counter = createCounter(10)\n counter() // 10\n counter() // 11\n counter() // 12\n\n``` | 1 | 0 | ['JavaScript'] | 0 |
counter | ☑️ Incrementing everytime function called. ☑️ | incrementing-everytime-function-called-b-8z1n | Code | Abdusalom_16 | NORMAL | 2024-12-16T05:44:31.232607+00:00 | 2024-12-16T05:44:31.232607+00:00 | 109 | false | # Code\n```javascript []\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n \n return function() {\n return n++;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 0 |
counter | Counter Function Using Closures in JavaScript | counter-function-using-closures-in-javas-g511 | Intuition\nThe problem involves generating a counter function that starts from a given integer n and increments its value each time the function is called. Usin | OmkarBhal | NORMAL | 2024-11-21T12:41:40.857865+00:00 | 2024-11-21T12:41:40.857902+00:00 | 221 | false | # Intuition\nThe problem involves generating a counter function that starts from a given integer n and increments its value each time the function is called. Using a closure is an ideal solution since closures allow functions to retain access to variables from their lexical scope, even after the outer function has executed. This way, the inner function can remember and update the value of n between calls.\n\n# Approach\nWe create a function createCounter that:\n\n\t1.\tAccepts an integer n as input.\n\t2.\tReturns an inner function (closure) that:\n\t\u2022\tReturns the current value of n.\n\t\u2022\tIncrements n for the next call.\n\nThe inner function \u201Cremembers\u201D the variable n from the outer function due to JavaScript closures. Each time the returned function is invoked, it increments n and provides the desired output.\n\n# Complexity\n- Time complexity:\n$$O(1)$$\nEach call to the counter function executes in constant time as it involves just returning and incrementing a variable.\n\n- Space complexity:\n$$O(1)$$\nThe space usage is constant, as we only maintain the variable n and the closure for the function.\n\n# Code\n```javascript []\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n // Define and return a function (closure) that captures `n`\n return function() {\n // Return the current value of `n` and then increment it by 1 for the next call\n return n++;\n };\n};\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 1 | ['JavaScript'] | 1 |
counter | One Liner💯💯 | one-liner-by-tanish-hrk-t1bc | Code\njavascript []\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n \n return function() {\n re | tanish-hrk | NORMAL | 2024-09-18T18:45:34.796046+00:00 | 2024-09-18T18:45:34.796081+00:00 | 56 | false | # Code\n```javascript []\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n \n return function() {\n return n++;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 1 |
counter | Simple Code🚀💯 || Easy to understand✅👍 | simple-code-easy-to-understand-by-jinees-bpia | \n# Approach\n Describe your approach to solving the problem. \n1.Closure:\n\n- The createCounter function returns an inner function that has access to the vari | jineesh_desai_09 | NORMAL | 2024-08-15T04:39:14.089685+00:00 | 2024-08-15T04:39:14.089718+00:00 | 107 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\n**1.Closure:**\n\n- The createCounter function returns an inner function that has access to the variable n from its outer lexical scope.\n- Each time the returned function is called, it returns the current value of n and then increments n by 1.\n- The use of closures allows the counter to retain its state between calls, enabling the counter to produce a sequence of increasing integers.\n\n**2.Example Usage:**\n\n- When createCounter(10) is called, it creates a counter that starts at 10.\n- The first call to counter() returns 10, the second call returns 11, and so on.\n# Complexity\n- **Time complexity:**\n The time complexity of your JavaScript createCounter function is **O(1)** (constant time)\n\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- **Space complexity:**\nThe space complexity of your JavaScript createCounter function is also **O(1)** (constant space).\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n**Notes:**\nThis implementation is a straightforward and effective way to generate a sequence of numbers that can be used in various contexts, such as iterating over an array or generating unique IDs.\n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n \n return function counter() {\n return n++;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n```\n**Don\u2019t forget to upvote if this solution helped! \uD83D\uDE0A** | 1 | 0 | ['JavaScript'] | 0 |
counter | Counter w/ TypeScript | counter-with-typescript-by-skywalkersam-zp5n | null | skywalkerSam | NORMAL | 2024-07-26T18:11:55.981956+00:00 | 2025-02-05T14:29:42.510988+00:00 | 9 | false |
```
function createCounter(n: number): () => number {
return function() {
return n++;
}
}
/**
* const counter = createCounter(10)
* counter() // 10
* counter() // 11
* counter() // 12
*/
``` | 1 | 0 | ['TypeScript'] | 0 |
counter | JavaScript solution | javascript-solution-by-siyadhri-u8yp | \n\n# Code\n\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n \n return function() {\n return n++; | siyadhri | NORMAL | 2024-06-09T14:27:35.181970+00:00 | 2024-06-09T14:27:35.181999+00:00 | 2 | false | \n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n \n return function() {\n return n++; \n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 0 |
counter | Closure Counter Without Increase | closure-counter-without-increase-by-isab-nax7 | 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 | isabellastor25 | NORMAL | 2024-06-01T02:39:46.853043+00:00 | 2024-06-01T02:39:46.853076+00:00 | 272 | 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 * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n let i = 0;\n return function() {\n n = n + i;\n if(i === 0){\n i = 1;\n }\n return n;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 1 |
counter | Kind of Jank | kind-of-jank-by-localnerdcoder-c7ka | 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 | LocalNerdCoder | NORMAL | 2024-03-14T14:53:18.767731+00:00 | 2024-03-14T14:53:18.767755+00:00 | 486 | 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 * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n \n return function() {\n current = n;\n current += 1;\n n = current;\n return current -= 1;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 1 |
counter | Q2 Sol. (Just for my record) | q2-sol-just-for-my-record-by-lonetwyl-zq71 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nn = n + 1 \nWhen it cal | lonetwyl | NORMAL | 2024-02-27T09:31:54.198732+00:00 | 2024-02-27T09:31:54.198750+00:00 | 588 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nn = n + 1 \nWhen it calls the function, the number should increase by 1.\nThe result shows that the function is called first time, it would be the original number. Therefore, use "n-1" to correspoond to the output.\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 * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n n++;\n return n-1;\n \n };\n \n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 1 |
counter | very Easy code | very-easy-code-by-muhsin313-5lip | 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 | muhsin313 | NORMAL | 2024-02-27T01:52:36.012933+00:00 | 2024-02-27T01:52:36.012959+00:00 | 529 | 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 * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n \n return function() {\n return n++;\n };\n \n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 1 |
counter | 🚀 [ JS ] Solution | js-solution-by-maxi_s-c23v | 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 | maxi_s | NORMAL | 2024-02-26T16:19:57.406130+00:00 | 2024-02-26T16:19:57.406160+00:00 | 1,800 | 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 * @param {number} n\n * @return {Function} counter\n */\nconst createCounter = (n) => () => n++;\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 1 |
counter | [JS] Easy Solution | Clean Code | js-easy-solution-clean-code-by-lshigami-307r | 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 | lshigami | NORMAL | 2024-02-15T12:43:33.230000+00:00 | 2024-02-15T12:43:33.230033+00:00 | 854 | 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 * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n \n return function() {\n return n++; \n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 1 |
counter | Easy solution | easy-solution-by-truongtamthanh2004-wtyc | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | truongtamthanh2004 | NORMAL | 2024-02-08T07:38:49.421842+00:00 | 2024-02-08T07:38:49.421875+00:00 | 555 | 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 * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return () => n++;\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 1 |
counter | easy simple one liner code | easy-simple-one-liner-code-by-apophis29-85nv | Return an arrow function that +1 to n\n\n# Code\n\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n \n r | Apophis29 | NORMAL | 2024-01-12T15:51:18.697733+00:00 | 2024-01-12T15:51:18.697757+00:00 | 713 | false | Return an arrow function that +1 to n\n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n \n return ()=>n++ \n \n };\n\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 1 |
counter | One-liner simple solution | one-liner-simple-solution-by-shreya4dhin-2vye | Code\n\nlet createCounter = n => () => n++;\n | shreya4dhingra | NORMAL | 2023-12-02T07:15:17.601246+00:00 | 2023-12-02T07:15:17.601271+00:00 | 2 | false | # Code\n```\nlet createCounter = n => () => n++;\n``` | 1 | 0 | ['JavaScript'] | 1 |
counter | 2620. Counter: Easy Solution | 2620-counter-easy-solution-by-archana094-13xn | 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 | Archana0943 | NORMAL | 2023-10-30T16:51:18.459350+00:00 | 2023-10-30T16:51:18.459390+00:00 | 976 | 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 * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n //store the n value to the y var\n let y=n-1\n \n return function() {\n //return the value of y+1\n return y+=1\n\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 2 |
counter | Easy AF!! code | easy-af-code-by-ashirwad2573-m2cd | 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 | ashirwad2573 | NORMAL | 2023-08-30T04:56:09.636692+00:00 | 2023-08-30T04:56:09.636715+00:00 | 22 | 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 * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n let v=n;\n return function() {\n return (v++);\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 0 |
counter | Increment (++) operator solution | increment-operator-solution-by-user3087v-ijqc | Why the n++ operator works this way?\nIf used postfix, with operator after operand (for example, x++), the increment operator increments and returns the value b | user3087VQ | NORMAL | 2023-08-21T14:32:42.122172+00:00 | 2023-08-21T14:32:42.122192+00:00 | 3 | false | # Why the `n++` operator works this way?\nIf used postfix, with operator after operand (for example, `x++`), the increment operator increments and returns the value before incrementing.\n\nThat is why on the first call, the value of `n` is returned, and its value is incremented only after returning from the first call. On the second call we get the `n+1` value and so on.\n\nIf we want to return the incremented value on the first call - use the `++` operator before operand (for example, ++x). The increment operator increments and returns the value after incrementing. Then the first call will return `n+1` and not `n`\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 * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 0 |
counter | Easy & Simple 🚀 | easy-simple-by-axatsachani-pacj | Code\n\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function (n) {\n return function () {\n n++\n return | AxatSachani | NORMAL | 2023-08-15T16:35:01.789577+00:00 | 2024-07-13T13:29:34.211034+00:00 | 13 | false | # Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function (n) {\n return function () {\n n++\n return n - 1\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 0 |
counter | DAY 2 of JS challenge⏳Beats 97%! 1 liner solution for beginners to understand🚀 | day-2-of-js-challengebeats-97-1-liner-so-v6oi | Intuition\nWe use an arrow function for the increment of n to take place\n# Code\n\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCou | deleted_user | NORMAL | 2023-05-27T06:27:49.714948+00:00 | 2023-05-27T13:34:14.322363+00:00 | 1,748 | false | # Intuition\nWe use an arrow function for the increment of n to take place\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return ()=> n++\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 1 |
counter | Closures || ONE line✅|| JS || EASY || Beats 96% 😎😎✅✅☑☑✔ | closures-one-line-js-easy-beats-96-by-lu-m0ww | \n\n# Code\n\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++; \n | luffy233 | NORMAL | 2023-05-16T08:37:50.713238+00:00 | 2023-05-16T08:37:50.713280+00:00 | 474 | false | \n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++; \n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['Python', 'C++', 'Java', 'JavaScript', 'Erlang'] | 0 |
counter | Easy and simple approach JAVASCRIPT | easy-and-simple-approach-javascript-by-k-dg3s | \n\n# Code\n\n/**\n * @param {number} n\n * @return {Function} counter\n */\n \nvar createCounter = function(n) {\n var num = n-1;\n return function() {\n | krish2213 | NORMAL | 2023-05-11T14:24:50.345477+00:00 | 2023-05-11T14:24:50.345515+00:00 | 9 | false | \n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\n \nvar createCounter = function(n) {\n var num = n-1;\n return function() {\n num+=1;\n return(num);\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 0 |
counter | Very short solution TS | very-short-solution-ts-by-makarofalex-ajah | 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 | makarofalex | NORMAL | 2023-05-09T14:00:07.864893+00:00 | 2023-05-09T14:00:07.864931+00:00 | 120 | 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 const createCounter = (n: number): () => number => () => n++\n\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['TypeScript'] | 0 |
counter | Easy One Line | easy-one-line-by-bekki3-t8gy | return n post-incremented\n\n# Code\n\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() { | bekki3 | NORMAL | 2023-05-06T13:32:23.804052+00:00 | 2023-05-06T13:32:23.804088+00:00 | 530 | false | return n post-incremented\n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['C', 'C++', 'Java', 'Python3', 'JavaScript'] | 0 |
counter | [ JavaScript ] ✅✅ Simple JavaScript Solution | Typescript 🥳✌👍 | javascript-simple-javascript-solution-ty-2y1f | If You like the Solution, Don\'t Forget To UpVote Me, Please UpVote! \uD83D\uDD3C\uD83D\uDE4F\n# Runtime: 69 ms, faster than 6.77% of JavaScript online submissi | ashok_kumar_meghvanshi | NORMAL | 2023-05-06T05:02:37.713585+00:00 | 2023-05-06T05:19:14.205166+00:00 | 44 | false | # If You like the Solution, Don\'t Forget To UpVote Me, Please UpVote! \uD83D\uDD3C\uD83D\uDE4F\n# Runtime: 69 ms, faster than 6.77% of JavaScript online submissions for Counter.\n# Memory Usage: 41.9 MB, less than 56.40% of JavaScript online submissions for Counter.\n\n\tvar createCounter = function(n) {\n\t\treturn function() {\n\t\t\tlet result = n;\n\t\t\tn = n + 1;\n\t\t\treturn result;\n\t\t};\n\t};\n\n# Runtime: 62 ms, faster than 53.54% of TypeScript online submissions for Counter.\n# Memory Usage: 43.4 MB, less than 27.19% of TypeScript online submissions for Counter.\n\n\tfunction createCounter(n: number): () => number {\n\t\treturn function() {\n\t\t\tvar result = n++ ;\n\t\t\treturn result;\n\t\t}\n\t}\n# Thank You \uD83E\uDD73\u270C\uD83D\uDC4D | 1 | 0 | ['TypeScript', 'JavaScript'] | 0 |
counter | Day 2 => JS Challenge | day-2-js-challenge-by-divyanshu_singh_cs-5ssx | 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 | Divyanshu_singh_cs | NORMAL | 2023-05-06T03:59:49.071222+00:00 | 2023-05-06T03:59:49.071251+00:00 | 23 | 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 * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return ()=> n++\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 0 |
counter | ✌ Simple O(1) Solution ✌ | simple-o1-solution-by-silvermete0r-a7kv | Complexity\n- Time complexity:\nO(1)\n\n- Space complexity:\nO(1)\n\n# Code\nJavaScript []\nvar createCounter = function(n) {\n return function() {\n | silvermete0r | NORMAL | 2023-05-06T03:47:33.074938+00:00 | 2023-05-06T03:47:33.074981+00:00 | 7 | false | # Complexity\n- Time complexity:\n$$O(1)$$\n\n- Space complexity:\n$$O(1)$$\n\n# Code\n``` JavaScript []\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n``` | 1 | 0 | ['JavaScript'] | 0 |
counter | Easy code beats 90% | easy-code-beats-90-by-adicoder95-gqvv | Code\n\n// Define a function that takes an initial value `n` and returns another function\nfunction createCounter(n) {\n // Return the counter function that us | AdiCoder95 | NORMAL | 2023-04-14T04:30:04.795097+00:00 | 2023-04-14T04:30:04.795140+00:00 | 1,098 | false | # Code\n```\n// Define a function that takes an initial value `n` and returns another function\nfunction createCounter(n) {\n // Return the counter function that uses the initial value `n` and increments it every time it is called\n return function counter() {\n let result = n; // Store the current value of `n` in a variable `result`\n n++; // Increment the value of `n` for the next call to the function\n return result; // Return the stored value of `n` from the current call\n }\n}\n\n// Define a function that takes an initial value `n` and the number of times to call the counter function\nfunction getCounterSequence(n, numCalls) {\n const counter = createCounter(n); // Create a counter function using the initial value `n`\n const result = []; // Create an empty array to store the counter sequence\n for (let i = 0; i < numCalls; i++) { // Loop `numCalls` times\n result.push(counter()); // Call the counter function and push the result to the `result` array\n }\n return result; // Return the complete counter sequence\n}\n\n``` | 1 | 0 | ['JavaScript'] | 0 |
counter | Closure and Arrow function | closure-and-arrow-function-by-_kitish-r73d | Code\n\nconst createCounter = n => {\n let count = n, i = -1\n return () => count + ++i;\n};\n | _kitish | NORMAL | 2023-04-13T12:53:02.949213+00:00 | 2023-04-13T12:53:02.949262+00:00 | 5,392 | false | # Code\n```\nconst createCounter = n => {\n let count = n, i = -1\n return () => count + ++i;\n};\n``` | 1 | 0 | ['JavaScript'] | 1 |
counter | closure, inner function remembers the variable which is in it's birth scope | closure-inner-function-remembers-the-var-7a3x | here the inner function which is returned remembres the variable that was initialised in the function global scope that is the create counter function.\n\n# Cod | rajsiddi | NORMAL | 2023-04-12T07:09:20.865291+00:00 | 2023-04-15T15:44:45.900991+00:00 | 31 | false | here the inner function which is returned remembres the variable that was initialised in the function global scope that is the create counter function.\n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n let temp = n-1;\n return function() {\n \n return ++temp; \n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 0 |
counter | JS Easy solution :- Closure | js-easy-solution-closure-by-siteshmehta-syu1 | Approach\nA closure has been utilized in our code. This enables the retention of variable values and function declarations even after the function has finished | siteshmehta | NORMAL | 2023-04-12T05:50:03.834696+00:00 | 2023-04-12T05:50:52.249880+00:00 | 39 | false | # Approach\nA closure has been utilized in our code. This enables the retention of variable values and function declarations even after the function has finished executing.\n\n\n\n\n\n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n let counter = n; //created a counter variable to remember the value\n return function( ) {\n return counter++;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 0 |
counter | ✅✅ JavaScript || TypeScript Solution || Begineer Friendly ✅✅ | javascript-typescript-solution-begineer-d0n5t | \n\n# Approach\n Describe your approach to solving the problem. \nWe are getting n in the parent function as an argument. We have to return n and increase it by | 0xsonu | NORMAL | 2023-04-12T04:02:39.872666+00:00 | 2023-04-12T04:02:39.872702+00:00 | 72 | false | \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe are getting `n` in the parent function as an argument. We have to return n and increase it by one `n++`.\n\n# Code\n```TypeScript []\nfunction createCounter(n: number): () => number {\n return function() {\n return n++;\n }\n}\n\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n```\n\n```JavaScript []\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n return function() {\n return n++;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['TypeScript', 'JavaScript'] | 1 |
counter | Simple and Clean JavaScript solution | simple-and-clean-javascript-solution-by-7t4gk | \n\n# Code\n\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n let count = n;\n return function() {\n | sid_py | NORMAL | 2023-04-12T01:39:04.461161+00:00 | 2023-04-12T01:39:04.461211+00:00 | 198 | false | \n\n# Code\n```\n/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n let count = n;\n return function() {\n count++;\n return count - 1;\n };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */\n``` | 1 | 0 | ['JavaScript'] | 0 |
counter | another javascript | another-javascript-by-manminder11-wz5q | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | manminder11 | NORMAL | 2025-04-11T00:36:10.185265+00:00 | 2025-04-11T00:36:10.185265+00:00 | 1 | 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
```javascript []
let createCounter = function(n) {
return function counter() {
return n++
};
}
/**
* const counter = createCounter(10)
* counter() // 10
* counter() // 11
* counter() // 12
*/
``` | 0 | 0 | ['JavaScript'] | 0 |
longest-continuous-increasing-subsequence | [Java/C++]Clean solution | javacclean-solution-by-caihao0727mail-4rci | The idea is to use cnt to record the length of the current continuous increasing subsequence which ends with nums[i], and use res to record the maximum cnt.\n\n | caihao0727mail | NORMAL | 2017-09-10T03:16:27.752000+00:00 | 2018-08-26T23:44:47.535748+00:00 | 23,784 | false | The idea is to use ```cnt``` to record the length of the current continuous increasing subsequence which ends with ```nums[i]```, and use ```res``` to record the maximum ```cnt```.\n\nJava version:\n```\n public int findLengthOfLCIS(int[] nums) {\n int res = 0, cnt = 0;\n for(int i = 0; i < nums.length; i++){\n if(i == 0 || nums[i-1] < nums[i]) res = Math.max(res, ++cnt);\n else cnt = 1;\n }\n return res;\n }\n```\n\nC++ version:\n```\n int findLengthOfLCIS(vector<int>& nums) {\n int res = 0, cnt = 0;\n for(int i = 0; i < nums.size(); i++){\n if(i == 0 || nums[i-1] < nums[i]) res = max(res, ++cnt);\n else cnt = 1;\n }\n return res;\n }\n``` | 79 | 2 | [] | 15 |
longest-continuous-increasing-subsequence | Python DP solution | python-dp-solution-by-shijungg-3web | \nclass Solution(object):\n def findLengthOfLCIS(self, nums):\n if not nums:\n return 0\n dp = [1] * len(nums)\n for i in ran | shijungg | NORMAL | 2018-11-28T12:22:14.895945+00:00 | 2018-11-28T12:22:14.896004+00:00 | 4,165 | false | ```\nclass Solution(object):\n def findLengthOfLCIS(self, nums):\n if not nums:\n return 0\n dp = [1] * len(nums)\n for i in range(1, len(nums)):\n if nums[i] > nums[i-1]:\n dp[i] = dp[i - 1] + 1\n return max(dp)\n``` | 49 | 2 | [] | 5 |
longest-continuous-increasing-subsequence | Python Simple Solution | python-simple-solution-by-yangshun-bhjs | A continuous subsequence is essentially a subarray. Hence this question is asking for the longest increasing subarray and I have no idea why the question calls | yangshun | NORMAL | 2017-09-10T03:07:07.805000+00:00 | 2022-03-14T03:17:55.136204+00:00 | 8,997 | false | A continuous subsequence is essentially a subarray. Hence this question is asking for the longest increasing subarray and I have no idea why the question calls it continuous subsequence to confuse the readers. \n\nAnyway, we can make one pass of the array and keep track of the current streak of increasing elements, reset it when it does not increase.\n\n```\nclass Solution(object):\n def findLengthOfLCIS(self, nums):\n """\n :type nums: List[int]\n :rtype: int\n """\n # Time: O(n)\n # Space: O(1)\n max_len = i = 0\n while i < len(nums):\n curr = 1\n while i + 1 < len(nums) and nums[i] < nums[i + 1]:\n curr, i = curr + 1, i + 1\n max_len = max(max_len, curr)\n i += 1\n return max_len\n```\n\n**\uD83D\uDCAF Check out https://www.techinterviewhandbook.org for more tips and tricks by me to ace your coding interview \uD83D\uDCAF** | 32 | 1 | [] | 8 |
longest-continuous-increasing-subsequence | [C++] Simple solution | c-simple-solution-by-shubhambhatt-0aqy | Pls upvote if you find this helpful :)\nKeep the length of each increasing sequence in a variable and one global variable for updating it with all the lengths | shubhambhatt__ | NORMAL | 2020-06-04T08:39:30.889181+00:00 | 2020-06-04T08:39:30.889228+00:00 | 2,725 | false | ***Pls upvote if you find this helpful :)***\nKeep the length of each increasing sequence in a variable and one global variable for updating it with all the lengths of increasing sequences.\n```\nclass Solution {\npublic:\n int findLengthOfLCIS(vector<int>& nums) {\n if(nums.size()<=1)return nums.size();\n int answer=1,count=1;\n for(int i=0;i<nums.size()-1;i++){\n if(nums[i]<nums[i+1]){\n count++;\n answer=max(answer,count);\n }\n else{\n count=1;\n }\n }\n return answer;\n }\n};\n``` | 26 | 0 | ['C', 'C++'] | 1 |
longest-continuous-increasing-subsequence | Python O(n) Solution - O(1) Space Complexity | python-on-solution-o1-space-complexity-b-7mqd | \nclass Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n if not nums:\n return 0\n \n cur_len = 1\n m | xorobotxo | NORMAL | 2020-03-10T14:41:48.794141+00:00 | 2020-03-10T14:41:48.794176+00:00 | 2,967 | false | ```\nclass Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n if not nums:\n return 0\n \n cur_len = 1\n max_len = 1\n \n for i in range(1,len(nums)):\n if nums[i] > nums[i-1]:\n cur_len += 1\n else:\n max_len = max(max_len,cur_len)\n cur_len = 1\n \n return max(max_len,cur_len)\n``` | 21 | 0 | ['Python', 'Python3'] | 2 |
longest-continuous-increasing-subsequence | Java code---6 liner | java-code-6-liner-by-cbyitina-cvjy | \npublic int findLengthOfLCIS(int[] nums) {\n if(nums.length==0) return 0;\n int length=1,temp=1;\n for(int i=0; i<nums.length-1;i++) {\n | cbyitina | NORMAL | 2017-10-15T07:54:12.931000+00:00 | 2018-09-24T21:39:25.058046+00:00 | 5,441 | false | ```\npublic int findLengthOfLCIS(int[] nums) {\n if(nums.length==0) return 0;\n int length=1,temp=1;\n for(int i=0; i<nums.length-1;i++) {\n if(nums[i]<nums[i+1]) {temp++; length=Math.max(length,temp);}\n else temp=1; \n }\n return length;\n }\n``` | 21 | 2 | [] | 9 |
longest-continuous-increasing-subsequence | Java solution, DP | java-solution-dp-by-shawngao-xhrj | \nclass Solution {\n public int findLengthOfLCIS(int[] nums) {\n if (nums == null || nums.length == 0) return 0;\n int n = nums.length;\n | shawngao | NORMAL | 2017-09-10T03:14:20.426000+00:00 | 2018-10-02T22:57:03.991089+00:00 | 4,553 | false | ```\nclass Solution {\n public int findLengthOfLCIS(int[] nums) {\n if (nums == null || nums.length == 0) return 0;\n int n = nums.length;\n int[] dp = new int[n];\n \n int max = 1;\n dp[0] = 1;\n for (int i = 1; i < n; i++) {\n if (nums[i] > nums[i - 1]) {\n dp[i] = dp[i - 1] + 1;\n }\n else {\n dp[i] = 1;\n }\n max = Math.max(max, dp[i]);\n }\n \n return max;\n }\n}\n``` | 20 | 6 | [] | 4 |
longest-continuous-increasing-subsequence | Easy c++ 2 lines | easy-c-2-lines-by-coderaky-bf0g | Always set r and c as 1 cause subsequence length can\'t be smaller than 1.\n\nint findLengthOfLCIS(vector<int>& nums) {\n int r=1,c=1;\n for(int i | coderaky | NORMAL | 2021-11-14T04:08:08.283585+00:00 | 2021-11-14T04:08:08.283618+00:00 | 873 | false | Always set r and c as 1 cause subsequence length can\'t be smaller than 1.\n```\nint findLengthOfLCIS(vector<int>& nums) {\n int r=1,c=1;\n for(int i=1;i<nums.size();i++)\n r=max(r,nums[i]>nums[i-1]?++c:c=1);\n return r;\n }\n``` | 15 | 0 | [] | 0 |
longest-continuous-increasing-subsequence | [C++/Java] Clean Code - 3 liner [2 Pointers] | cjava-clean-code-3-liner-2-pointers-by-a-rfdt | C++ record length\n\nclass Solution {\npublic:\n int findLengthOfLCIS(vector<int>& a) {\n int mx = 0, len = 0;\n for (int i = 0; i < a.size(); | alexander | NORMAL | 2017-09-10T03:21:27.015000+00:00 | 2018-10-02T20:33:08.492935+00:00 | 3,145 | false | **C++ record length**\n```\nclass Solution {\npublic:\n int findLengthOfLCIS(vector<int>& a) {\n int mx = 0, len = 0;\n for (int i = 0; i < a.size(); i++) {\n if (i == 0 || a[i] <= a[i - 1]) len = 0;\n mx = max(mx, ++len);\n }\n return mx;\n }\n};\n```\n**C++ 2 pointer**\n```\nclass Solution {\npublic:\n int findLengthOfLCIS(vector<int>& a) {\n int mx = 0;\n for (int i = 0, j = 0; j < a.size(); j++) {\n if (j == 0 || a[j] <= a[j - 1]) i = j;\n mx = max(mx, j - (i - 1))\n }\n return mx;\n }\n};\n```\n3 liner\n```\nclass Solution {\npublic:\n int findLengthOfLCIS(vector<int>& a) {\n int mx = 0;\n for (int i = 0, j = 0; j < a.size(); i = (j == 0 || a[j] <= a[j - 1]) ? j : i, mx = max(mx, j - (i - 1)), j++) { }\n return mx;\n }\n};\n```\n**Java - 2 pointer**\n```\nclass Solution {\n public int findLengthOfLCIS(int[] a) {\n int mx = 0;\n for (int i = 0, j = 0; j < a.length; i = (j == 0 || a[j] <= a[j - 1]) ? j : i, mx = Math.max(mx, j - i + 1), j++) { }\n return mx;\n }\n}\n```\n\n**Java length variable**\n```\nclass Solution {\n public int findLengthOfLCIS(int[] a) {\n int mx = 0, len = 0;\n for (int i = 0; i < a.length; i++) {\n if (i == 0 || a[i] <= a[i - 1]) len = 0;\n mx = Math.max(mx, ++len);\n }\n return mx;\n }\n}\n``` | 14 | 3 | [] | 1 |
longest-continuous-increasing-subsequence | Java | 100% faster | simple - O(n) time and O(1) space - simple and only few lines | java-100-faster-simple-on-time-and-o1-sp-5hct | \npublic int findLengthOfLCIS(int[] nums) \n{\n\tint maximum = 1;\n\tint currentMax = 1;\n\tfor(int i = 1; i < nums.length; i++)\n\t{\n\t\tcurrentMax = nums[i] | C0deHacker | NORMAL | 2020-12-20T21:21:16.873826+00:00 | 2020-12-20T21:21:16.873868+00:00 | 1,158 | false | ```\npublic int findLengthOfLCIS(int[] nums) \n{\n\tint maximum = 1;\n\tint currentMax = 1;\n\tfor(int i = 1; i < nums.length; i++)\n\t{\n\t\tcurrentMax = nums[i] > nums[i - 1] ? currentMax + 1 : 1; \n\t\tmaximum = Math.max(maximum, currentMax);\n\t}\n\n\treturn nums.length == 0 ? 0 : maximum;\n}\n``` | 8 | 0 | ['Java'] | 3 |
longest-continuous-increasing-subsequence | 1 ms Java Solution | 1-ms-java-solution-by-janhvi__28-xp98 | \nclass Solution {\n public int findLengthOfLCIS(int[] nums) {\n if(nums.length==0) return 0;\n int length=1,temp=1;\n for(int i=0; i<nu | Janhvi__28 | NORMAL | 2022-08-07T18:15:13.796383+00:00 | 2022-08-07T18:15:13.796413+00:00 | 1,154 | false | ```\nclass Solution {\n public int findLengthOfLCIS(int[] nums) {\n if(nums.length==0) return 0;\n int length=1,temp=1;\n for(int i=0; i<nums.length-1;i++) {\n if(nums[i]<nums[i+1]) {temp++; length=Math.max(length,temp);}\n else temp=1; \n }\n return length;\n }\n}\n``` | 7 | 0 | [] | 0 |
longest-continuous-increasing-subsequence | JavaScript Clean 2 Sliding-Window Approaches | javascript-clean-2-sliding-window-approa-2q1c | Solution 1\n\n\nvar findLengthOfLCIS = function(nums) {\n let len = 1, maxLen = 0;\n \n for(let i = 0; i < nums.length; i++) {\n if(nums[i] < nu | control_the_narrative | NORMAL | 2020-06-04T17:19:21.125439+00:00 | 2020-06-04T17:24:10.279296+00:00 | 893 | false | ## Solution 1\n\n```\nvar findLengthOfLCIS = function(nums) {\n let len = 1, maxLen = 0;\n \n for(let i = 0; i < nums.length; i++) {\n if(nums[i] < nums[i+1]) len++;\n else len = 1;\n maxLen = Math.max(len, maxLen);\n }\n return maxLen; \n};\n```\n\n## Solution 2\n\n```javascript\nvar findLengthOfLCIS = function(nums) {\n if(nums.length < 2) return nums.length;\n let left = 0, right = 1, maxLen = 0;\n \n while(right < nums.length) {\n if(nums[right-1] >= nums[right]) left = right;\n right++;\n maxLen = Math.max(right - left, maxLen);\n }\n return maxLen \n};\n``` | 7 | 1 | ['Sliding Window', 'JavaScript'] | 2 |
longest-continuous-increasing-subsequence | Python 3 Three Solutions Using Stack and Sliding Window and DP | python-3-three-solutions-using-stack-and-xoju | Using Stack\npython\nclass Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n if not nums:\n return 0\n if len(nums)= | deepak_goswami | NORMAL | 2020-03-16T12:16:33.089833+00:00 | 2020-03-16T12:16:55.342713+00:00 | 520 | false | **Using Stack**\n```python\nclass Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n if not nums:\n return 0\n if len(nums)==1:\n return 1\n st = []\n max_len = float(\'-inf\')\n for i in range(len(nums)):\n if not st:\n st.append(nums[i])\n elif st and st[-1]<nums[i]:\n st.append(nums[i])\n elif st and st[-1]>=nums[i]:\n st = []\n st.append(nums[i])\n max_len = max(max_len,len(st))\n return max_len\n```\n\n**Sliding Window**\n```python \nclass Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n if len(nums)==1:\n return 1\n ans = 0 \n ind = 0\n for i in range(1,len(nums)):\n if nums[i-1]>=nums[i]:\n ind = i\n ans = max(ans,i-ind+1)\n return ans\n```\n\n\n**Dp Approach**\n```python\nclass Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n if not nums:\n return 0\n dp = [1] * len(nums)\n for i in range(1, len(nums)):\n if nums[i] > nums[i-1]:\n dp[i] = dp[i - 1] + 1\n return max(dp)\n\t\t\n | 7 | 1 | [] | 0 |
longest-continuous-increasing-subsequence | Simple Java ☕ Solution using 2 pointer approach | Time: Beats 100% of the Solutions ✅ | simple-java-solution-using-2-pointer-app-sh3t | Intuition\n Describe your first thoughts on how to solve this problem. \nWe are to find the longest continuous increasing subsequence(LCIS), for that, what I th | hetpatelcse | NORMAL | 2023-01-30T15:17:18.989145+00:00 | 2023-01-30T15:20:14.156711+00:00 | 1,563 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe are to find the longest continuous increasing subsequence(LCIS), for that, what I thought was that, if we are checking the length from an element at any index, and if we reach a point that now our subsequence is not increasing, then we need not to check the length of longest cont. inc. subsequence.\n\nFor eg:\n[1,3,5,4,***2,3,4,5***]\n\nhere 2,3,4,5 is our LCIS, what I mean to say in above paragraph was, when we check the LCIS from element at index 0, it ends at element at index 2, now we need not to look for LCIS from element at index 1 and 2, as there length will always be lesser than that of element at index 0.\n\nI checked the length of each such part of array and returned the maximum one.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe will use 2 pointers to solve this problem, we will keep the track of previous element and j will keep track of current element. \n\nWe will compare nums[i] and nums[j] if nums[j]>nums[i], then currentCount(variable to store the length of current LCIS) will be incremented. \n\nIf nums[j]>nums[i] is false, means this was the end of current LCIS and we now need to check the next LCIS means, we will now chance i to j, increament j and make currentCount to 1.\n\nIn every iteration we are also updating maxCount if currentCount gets greater than it. \n\nAt last we are returning the maxCount.\n\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```\nclass Solution \n{\n public int findLengthOfLCIS(int[] nums) \n {\n int maxCount = 1;\n int currentCount = 1;\n int i = 0 ;\n int j = 1;\n while(j<nums.length)\n {\n if(nums[j]>nums[i])\n {\n currentCount++;\n i++;\n j++;\n }\n else\n {\n i = j;\n j++;\n currentCount = 1;\n }\n if(maxCount<currentCount)\n {\n maxCount = currentCount;\n }\n \n } \n return maxCount; \n }\n}\n```\n\n\n****Do Upvote if found useful.**** \u2B06\uFE0F | 6 | 0 | ['Two Pointers', 'Java'] | 1 |
longest-continuous-increasing-subsequence | Two python solutions using dp and a straightforward soln | two-python-solutions-using-dp-and-a-stra-03mn | DP solution.\nclass Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n dp=[1]*len(nums)\n for i in range(1,len(nums)):\n | guneet100 | NORMAL | 2022-08-18T17:37:30.558682+00:00 | 2022-08-18T17:37:30.558720+00:00 | 1,323 | false | 1. DP solution.\n```class Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n dp=[1]*len(nums)\n for i in range(1,len(nums)):\n if nums[i]>nums[i-1]:\n dp[i]+=dp[i-1]\n return max(dp)\n ```\n2\n```\nclass Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n counter=1\n temp=1\n for i in range(0,len(nums)-1):\n if nums[i]<nums[i+1]:\n temp+=1\n if temp>counter:\n counter=temp\n else:\n temp=1\n return counter | 6 | 0 | ['Python', 'Python3'] | 3 |
longest-continuous-increasing-subsequence | 674: Solution with step by step explanation | 674-solution-with-step-by-step-explanati-kjn2 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n1. We handle the base cases where the input list is empty or has only one | Marlen09 | NORMAL | 2023-03-19T18:04:01.823554+00:00 | 2023-03-19T18:04:01.823599+00:00 | 874 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. We handle the base cases where the input list is empty or has only one element. In both cases, the length of the longest continuous in1.creasing subsequence is the length of the input list itself, so we simply return that length.\n\n2. We initialize two variables cur_len and max_len to keep track of the length of the current increasing subsequence and the maximum length seen so far, respectively. We set both variables to 1 because the first element of the input list is always part of a subsequence of length 1.\n\n3. We iterate through the input list starting from the second element. For each element, we check if it is greater than the previous element. If it is, then it is part of the current increasing subsequence, so we increase the length of the subsequence by 1 (cur_len += 1). We also update the maximum length seen so far (max_len = max(max_len, cur_len)) if necessary. If the current element is not greater than the previous element, then it is the start of a new increasing subsequence, so we reset the length of the subsequence to 1 (cur_len = 1).\n\n4. After iterating through the entire input list, we return the maximum length of any increasing subsequence seen (return max_len).\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```\nclass Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n # Base case: empty list or single element list\n if len(nums) == 0 or len(nums) == 1:\n return len(nums)\n \n # Initialize variables to keep track of current length and max length\n cur_len = 1 # current length of increasing subsequence\n max_len = 1 # maximum length of increasing subsequence\n \n # Iterate through the list starting from the second element\n for i in range(1, len(nums)):\n # If the current element is greater than the previous element, it is part of the increasing subsequence\n if nums[i] > nums[i-1]:\n cur_len += 1 # increase the length of the subsequence\n max_len = max(max_len, cur_len) # update the maximum length if necessary\n else:\n cur_len = 1 # reset the length of the subsequence if the current element is not greater than the previous element\n \n return max_len\n\n``` | 5 | 0 | ['Array', 'Python', 'Python3'] | 0 |
longest-continuous-increasing-subsequence | Python Easy Solution without Sliding Window, O(n), 54ms, beats 93.23% with Explanation | python-easy-solution-without-sliding-win-0i8l | Explanation\n1. Edge case if there is only one element in nums\n2. add a counter and a max counter (in case the maximum is at beginning)\n3. iterate through num | alexl5311 | NORMAL | 2022-05-07T04:00:20.616319+00:00 | 2022-05-07T04:35:18.187564+00:00 | 520 | false | **Explanation**\n1. Edge case if there is only one element in nums\n2. add a counter and a max counter (in case the maximum is at beginning)\n3. iterate through nums:\n\t4. if it\'s the first element, continue (no previous, will evoke error)\n\t5. each time the element is greater than the previous one (increasing), add 1 to counter\n\t6. if not, check if the counter is the largest one yet (if it\'s largest subarray yet)\n\t\t7. if yes, update max counter\n\t\t8. reset counter to 1\n7. check if last counter is max (edge case)\n8. return max counter\n\n```\nclass Solution(object):\n def findLengthOfLCIS(self, nums):\n """\n :type nums: List[int]\n :rtype: int\n """\n if len(nums) == 1:\n return 1\n \n counter = 1\n max_counter = 0\n for i in range(len(nums)):\n if i == 0:\n continue\n else: \n if nums[i] > nums[i-1]:\n counter += 1\n else:\n if counter > max_counter:\n max_counter = counter\n counter = 1\n if counter > max_counter:\n max_counter = counter\n \n return max_counter\n```\n\n**If you liked this, please upvote to support me!** | 5 | 0 | ['Python'] | 0 |
longest-continuous-increasing-subsequence | Easy Approach || Python3 | easy-approach-python3-by-harxsan-gc06 | Approach\n Simple Intuition\n# Complexity\n- Time complexity:\n O(N)\n- Space complexity:\n O(1)\n# Code\npython3 []\nclass Solution:\n def findLeng | harxsan | NORMAL | 2024-08-29T09:18:37.676851+00:00 | 2024-08-29T09:18:37.676890+00:00 | 521 | false | # Approach\n Simple Intuition\n# Complexity\n- Time complexity:\n O(N)\n- Space complexity:\n O(1)\n# Code\n```python3 []\nclass Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n cnt = 1\n max_cnt = 0\n if len(nums) == 1:\n return 1\n for i in range(1,len(nums)):\n if nums[i] > nums[i -1]:\n cnt += 1\n else:\n cnt = 1\n if cnt > max_cnt:\n max_cnt = cnt\n \n return max_cnt\n``` | 4 | 0 | ['Python3'] | 1 |
longest-continuous-increasing-subsequence | simple and easy understand solution | simple-and-easy-understand-solution-by-s-fdi6 | if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n\nclass Solution {\npublic:\n int findLengthOfLCIS(vector<int>& nums) \n {\n int mx | shishirRsiam | NORMAL | 2024-04-07T03:57:07.459715+00:00 | 2024-04-07T03:57:07.459743+00:00 | 1,137 | false | # if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n```\nclass Solution {\npublic:\n int findLengthOfLCIS(vector<int>& nums) \n {\n int mx = 1, cur = 1, n = nums.size();\n for(int i=1;i<n;i++)\n {\n if(nums[i] > nums[i-1]) cur++;\n else \n {\n mx = max(mx, cur);\n cur = 1;\n }\n }\n return max(mx, cur);\n }\n};\n``` | 4 | 1 | ['Array', 'C', 'C++', 'Java', 'C#'] | 3 |
longest-continuous-increasing-subsequence | 🔥Beats 99.71%🔥✅Easy (Java/C++/Python) Solution With Detailed Explanation✅ | beats-9971easy-javacpython-solution-with-jno5 | Intuition\nThe method findLengthOfLCIS employs a single pass through the array to identify and measure the lengths of continuous increasing subsequences, updati | suyogshete04 | NORMAL | 2024-01-26T04:48:52.833857+00:00 | 2024-01-26T04:48:52.833894+00:00 | 1,365 | false | # Intuition\nThe method `findLengthOfLCIS` employs a single pass through the array to identify and measure the lengths of continuous increasing subsequences, updating a count for each subsequence. It maintains and updates the maximum length (`ans`) of these subsequences encountered so far. Whenever a non-increasing pair is found, the count is reset, ensuring only continuous increasing subsequences are considered.\n\n\n\n# Approach\n1. **Variable Initialization**: \n - `n`: Stores the length of the input array `nums`.\n - `ans`: Initially set to -1, this variable will hold the maximum length of an increasing subsequence found during the iteration. The initial value of -1 is a placeholder and will be updated during the iteration.\n - `count`: Initialized to 1, this variable keeps track of the current length of an increasing subsequence.\n\n2. **Iterating Through the Array**: \n - The loop iterates over the array elements, except for the last element, as the comparison is always made with the next element.\n - `for (int i = 0; i < n - 1; i++)`: This loop iterates through the array elements.\n\n3. **Identifying Increasing Subsequences**: \n - `if (nums[i] < nums[i + 1])`: If the current element is less than the next element, it indicates that the subsequence is still increasing, so `count` is incremented.\n - `else`: If the current element is not less than the next element, it means the current increasing subsequence has ended. In this case, `ans` is updated to be the maximum of `ans` and `count`, and then `count` is reset to 1 to start counting the length of a new subsequence.\n\n4. **Final Check and Update**: \n - After the loop, there\'s a final update to `ans`: `ans = Math.max(ans, count);`. This is necessary because the last increasing subsequence might be the longest, but it wouldn\'t be checked inside the loop as the loop ends before the last element.\n\n5. **Return the Result**: \n - The method returns `ans`, which contains the length of the longest continuous increasing subsequence found in the array.\n\n# Complexity\n\n1. **Time Complexity: O(n)**\n - The method iterates through the array `nums` exactly once. During this iteration, it performs constant-time operations for each element, such as comparison and updating the count and maximum length values.\n - Since the number of operations is proportional to the length of the array, the time complexity is linear, or `O(n)`, where `n` is the number of elements in the array.\n\n2. **Space Complexity: O(1)**\n - The space complexity is constant because the method uses a fixed number of variables (`n`, `ans`, `count`) regardless of the size of the input array.\n - No additional data structures that grow with the input size are used, so the amount of space required remains the same, leading to a space complexity of `O(1)`.\n\n# Code\n```Java []\npublic class Solution {\n public int findLengthOfLCIS(int[] nums) {\n int n = nums.length;\n int ans = -1;\n int count = 1;\n for (int i = 0; i < n - 1; i++) {\n if (nums[i] < nums[i + 1]) {\n count++;\n } else {\n ans = Math.max(ans, count);\n count = 1;\n }\n }\n\n ans = Math.max(ans, count);\n\n return ans;\n }\n}\n\n```\n```C++ []\nclass Solution {\npublic:\n int findLengthOfLCIS(vector<int>& nums) {\n int n = nums.size();\n int ans = -1;\n int count = 1;\n for (int i = 0; i < n - 1; i++) {\n if (nums[i] < nums[i + 1])\n count++;\n else {\n ans = max(ans, count);\n count = 1;\n }\n }\n\n ans = max(ans, count);\n\n return ans;\n }\n};\n```\n```Python []\nfrom typing import List\n\nclass Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n if not nums:\n return 0\n\n ans = 1\n count = 1\n for i in range(1, len(nums)):\n if nums[i] > nums[i - 1]:\n count += 1\n ans = max(ans, count)\n else:\n count = 1\n\n return ans\n\n``` | 3 | 0 | ['C++', 'Java', 'Python3'] | 0 |
longest-continuous-increasing-subsequence | Beginners Python Code | beginners-python-code-by-geethika1829-f7l6 | Code\n\nclass Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n n = len(nums)\n ans = 1\n curr_len = 1\n for i in | geethika1829 | NORMAL | 2023-09-05T05:48:18.066665+00:00 | 2023-09-05T05:48:18.066698+00:00 | 418 | false | # Code\n```\nclass Solution:\n def findLengthOfLCIS(self, nums: List[int]) -> int:\n n = len(nums)\n ans = 1\n curr_len = 1\n for i in range(1,n):\n if nums[i] > nums[i-1]:\n curr_len = curr_len + 1\n ans = max(ans,curr_len)\n else:\n curr_len = 1\n return ans\n \n``` | 3 | 0 | ['Python3'] | 1 |
longest-continuous-increasing-subsequence | Very easy and simple solution in JavaScript!!! WoW. Just watch it. :) | very-easy-and-simple-solution-in-javascr-oiem | \n# Code\n\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar findLengthOfLCIS = function(nums) {\n let max = 0, curr = 0\n for(let i = 0 ; i | AzamatAbduvohidov | NORMAL | 2023-05-13T13:51:12.905377+00:00 | 2023-05-13T13:51:12.905428+00:00 | 592 | false | \n# Code\n```\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar findLengthOfLCIS = function(nums) {\n let max = 0, curr = 0\n for(let i = 0 ; i < nums.length; i++) {\n if(nums[i] < nums[i + 1]) {\n curr++\n max = Math.max(max, curr)\n } else {\n curr = 0\n }\n }\n return max > 0 ? max + 1 : 1\n};\n``` | 3 | 0 | ['JavaScript'] | 0 |
longest-continuous-increasing-subsequence | C++ Easy Solution | c-easy-solution-by-anandmohit852-kjo7 | \n\n# Code\n\nclass Solution {\npublic:\n int findLengthOfLCIS(vector<int>& nums) {\n int maxi=1;\n int c=1;\n for(int i=0;i<nums.size() | anandmohit852 | NORMAL | 2023-02-17T14:00:31.389354+00:00 | 2023-02-17T14:00:31.389424+00:00 | 647 | false | \n\n# Code\n```\nclass Solution {\npublic:\n int findLengthOfLCIS(vector<int>& nums) {\n int maxi=1;\n int c=1;\n for(int i=0;i<nums.size()-1;i++){\n if(nums[i]<nums[i+1]){\n c++;\n }\n else{\n c=1;\n }\n if(maxi<c){\n maxi=c;\n }\n }\n return maxi;\n }\n};\n``` | 3 | 0 | ['C++'] | 0 |
longest-continuous-increasing-subsequence | JS faster than 100% O(n) | js-faster-than-100-on-by-kunkka1996-7c8c | \n\n\nvar findLengthOfLCIS = function(nums) {\n let output = 0;\n let count = 0;\n\n for (let i = 0; i < nums.length; i++) {\n if (!count || num | kunkka1996 | NORMAL | 2022-08-27T03:10:01.906523+00:00 | 2022-08-27T03:10:01.906564+00:00 | 663 | false | \n\n```\nvar findLengthOfLCIS = function(nums) {\n let output = 0;\n let count = 0;\n\n for (let i = 0; i < nums.length; i++) {\n if (!count || nums[i] > nums[i - 1]) {\n count++;\n } else {\n output = Math.max(output, count);\n count = 1;\n }\n }\n \n return Math.max(output, count);\n};\n``` | 3 | 0 | ['JavaScript'] | 2 |
longest-continuous-increasing-subsequence | Easy Java Solution || 1ms & faster than 100% | easy-java-solution-1ms-faster-than-100-b-p4l0 | \nclass Solution {\n public int findLengthOfLCIS(int[] nums) {\n int max = 1;\n int count = 1;\n \n for(int i = 1 ; i < nums.leng | vmanishk | NORMAL | 2022-08-09T12:07:06.074277+00:00 | 2022-08-09T12:07:06.074330+00:00 | 493 | false | ```\nclass Solution {\n public int findLengthOfLCIS(int[] nums) {\n int max = 1;\n int count = 1;\n \n for(int i = 1 ; i < nums.length; i++) {\n if (nums[i - 1] < nums[i]) {\n count++;\n } else {\n count = 1;\n }\n if (max < count) {\n max = count;\n }\n }\n return max;\n }\n}\n``` | 3 | 0 | ['Java'] | 0 |
longest-continuous-increasing-subsequence | Python O(n) greedy | python-on-greedy-by-ypmagic2-2z9i | \ndef findLengthOfLCIS(self, nums: List[int]) -> int:\n max_length = 0\n cur_length = 0\n cur_max = float(\'-inf\')\n for num in num | ypmagic2 | NORMAL | 2020-07-24T06:44:00.413776+00:00 | 2020-07-24T06:44:00.413837+00:00 | 372 | false | ```\ndef findLengthOfLCIS(self, nums: List[int]) -> int:\n max_length = 0\n cur_length = 0\n cur_max = float(\'-inf\')\n for num in nums:\n if num > cur_max:\n cur_length += 1\n cur_max = num\n else:\n max_length = max(max_length, cur_length)\n cur_length = 1\n cur_max = num\n return max(max_length, cur_length)\n``` | 3 | 0 | ['Greedy', 'Python'] | 1 |
longest-continuous-increasing-subsequence | Javascript O(n) | javascript-on-by-fbecker11-8ep8 | \n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar findLengthOfLCIS = function(nums) {\n let max = 0;\n let curr = 0;\n let prev = -Infinity;\n | fbecker11 | NORMAL | 2020-07-02T01:51:02.779448+00:00 | 2020-07-02T01:51:02.779486+00:00 | 289 | false | ```\n/**\n * @param {number[]} nums\n * @return {number}\n */\nvar findLengthOfLCIS = function(nums) {\n let max = 0;\n let curr = 0;\n let prev = -Infinity;\n for (let n of nums){\n if(n > prev){\n max = Math.max(max, ++curr) \n }else{\n curr = 1;\n }\n prev = n\n }\n return max\n};\n``` | 3 | 0 | ['JavaScript'] | 1 |
longest-continuous-increasing-subsequence | Straight-forward and fast Python | straight-forward-and-fast-python-by-he00-n7f4 | ```\nclass Solution(object):\n def findLengthOfLCIS(self, nums):\n """\n :type nums: List[int]\n :rtype: int\n """\n if le | he000193 | NORMAL | 2019-11-18T19:21:13.665534+00:00 | 2019-11-18T19:21:13.665588+00:00 | 355 | false | ```\nclass Solution(object):\n def findLengthOfLCIS(self, nums):\n """\n :type nums: List[int]\n :rtype: int\n """\n if len(nums) < 2:\n return len(nums)\n tmp, max_l = 1, 0\n for i in range(len(nums) - 1): \n if nums[i] < nums[i + 1]:\n tmp += 1\n else:\n max_l = max(max_l, tmp)\n tmp = 1\n return max(max_l, tmp) | 3 | 0 | ['Python'] | 0 |
longest-continuous-increasing-subsequence | dp is powerfull | dp-is-powerfull-by-lin2017gg-gta9 | \nclass Solution {\n public int findLengthOfLCIS(int[] nums) {\n if(nums.length == 0) return 0;\n int[] dp = new int[nums.length];\n dp[ | lin2017gg | NORMAL | 2018-04-20T00:02:26.938086+00:00 | 2018-04-20T00:02:26.938086+00:00 | 296 | false | ```\nclass Solution {\n public int findLengthOfLCIS(int[] nums) {\n if(nums.length == 0) return 0;\n int[] dp = new int[nums.length];\n dp[0] = 1;\n for(int i = 1;i < nums.length;++i) {\n if(nums[i] > nums[i - 1]) {\n dp[i] = dp[i - 1] + 1;\n } else {\n dp[i] = 1;\n }\n }\n int max = Integer.MIN_VALUE;\n for(int i = 0;i < nums.length;++i) {\n if(dp[i] > max) {\n max = dp[i];\n }\n }\n return max;\n }\n}\n\n``` | 3 | 1 | [] | 0 |
longest-continuous-increasing-subsequence | JAVA ✅|| BEGINNER FRIENDLY 🧠|| 100% BEATS 🚀 | java-beginner-friendly-100-beats-by-gopa-vivn | IntuitionWe need to find the longest continuous increasing subsequence (LCIS) in a given array. The idea is to keep track of how long the current increasing sub | GopalDev | NORMAL | 2025-02-13T03:35:04.188071+00:00 | 2025-02-13T03:35:04.188071+00:00 | 160 | false | ### Intuition
We need to find the longest continuous increasing subsequence (LCIS) in a given array. The idea is to keep track of how long the current increasing subsequence is while iterating through the array. If the current element is smaller than the next one, the subsequence continues. If not, the sequence ends, and we check if it was the longest sequence so far. This ensures that we capture the longest LCIS by the end of the iteration.
### Approach
1. **Initialization**:
- Start by setting a variable (`count`) to 1, representing the length of the current increasing subsequence, as any individual element is considered an increasing subsequence of length 1.
- Another variable (`max_length`) is used to store the longest increasing subsequence found so far.
2. **Iterate through the Array**:
- For each element, check if it is smaller than the next element.
- If it is, increment the current sequence's length (`count`).
- If the sequence breaks (i.e., the next element is smaller or equal), compare the current sequence length (`count`) with `max_length`. If `count` is greater, update `max_length`. Then reset `count` to 1 for the next potential subsequence.
3. **Final Check**:
- After iterating through the array, perform a final check to account for the case where the longest increasing subsequence is at the end of the array.
### Complexity
- **Time Complexity**:
$$O(n)$$
- **Space Complexity**:
$$O(1)$$
# Code
```java []
class Solution {
public int findLengthOfLCIS(int[] nums) {
int ans=-1;
int count=1;
for(int i=0; i<nums.length-1; i++){
if(nums[i]<nums[i+1]){
count++;
}else{
ans= Math.max(ans , count);
count=1;
}
}
return Math.max(ans,count);
}
}
``` | 2 | 0 | ['Java'] | 0 |
longest-continuous-increasing-subsequence | Beginner's logic (0 ms) | beginners-logic-0-ms-by-dinhvanphuc-1f3z | IntuitionApproachUse max function to find max length when encounter a smaller number, remember that the initial value of count is always 1 because we are compar | DinhVanPhuc | NORMAL | 2025-01-09T07:52:17.945653+00:00 | 2025-01-09T07:52:17.945653+00:00 | 241 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
Use max function to find max length when encounter a smaller number, remember that the initial value of count is always 1 because we are comparing the number after, not the number earlier (check testcases to understand)
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
int findMaxLength = INT_MIN;
int count = 1, size = nums.size();
if (size == 1) return 1;
for (int i = 1; i < size; i++) {
if (nums[i] < nums[i - 1]) {
if (count > 1) {
findMaxLength = max(findMaxLength, count);
count = 1;
}
} else if (nums[i] > nums[i - 1]) count++;
else {
if (count > 1) {
findMaxLength = max(findMaxLength, count);
count = 1;
}
}
} return max(findMaxLength, count);
}
};
``` | 2 | 0 | ['C++'] | 0 |
longest-continuous-increasing-subsequence | 1ms 99.95% in java very easy code | 1ms-9995-in-java-very-easy-code-by-galan-bz25 | Code | Galani_jenis | NORMAL | 2024-12-19T13:45:29.758271+00:00 | 2024-12-19T13:45:29.758271+00:00 | 377 | false | 
# Code
```java []
class Solution {
public int findLengthOfLCIS(int[] nums) {
int ans = -1;
int count = 1;
for (int i = 0; i < nums.length - 1; i++) {
if (nums[i] < nums[i + 1]) {
count++;
} else {
ans = Math.max(ans, count);
count = 1;
}
}
return Math.max(ans, count);
}
}
``` | 2 | 0 | ['Java'] | 0 |
longest-continuous-increasing-subsequence | 674. Longest Continuous Increasing Subsequence | 674-longest-continuous-increasing-subseq-iq2m | Intuition\n Describe your first thoughts on how to solve this problem. \nSo it is so simple naive approach which make you so clear and understand\n# Approach\n | saumyap271 | NORMAL | 2024-12-08T02:37:04.837017+00:00 | 2024-12-08T02:37:04.837042+00:00 | 156 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSo it is so simple naive approach which make you so clear and understand\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```cpp []\nclass Solution {\npublic:\n int findLengthOfLCIS(vector<int>& nums) {\n int n=nums.size();\n int count=1, maxcount=1;\n for(int i=1; i<n; i++){\n if(nums[i]>nums[i-1]){\n count++;\n }\n else{\n count=1;\n }\n maxcount=max(count,maxcount);\n }\n return maxcount;\n }\n};\n``` | 2 | 0 | ['C++'] | 0 |
longest-continuous-increasing-subsequence | Beats 100% | beats-100-by-sahilsaw-4u6h | Intuition\n Describe your first thoughts on how to solve this problem. \nGoing forward, keep track of how many values satisfy the condition arr[i] > arr[i+1]. T | SahilSaw | NORMAL | 2024-10-05T19:05:56.116255+00:00 | 2024-10-05T19:05:56.116273+00:00 | 68 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nGoing forward, keep track of how many values satisfy the condition arr[i] > arr[i+1]. The moment this condition breaks, reset your count and store the maximum value in a variable. Print it at the end.\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```cpp []\nclass Solution {\npublic:\n int findLengthOfLCIS(vector<int>& arr) {\n int size=1,ans=1;\n arr.push_back(arr[arr.size()-1]-1);\n for(int i=1;i<arr.size();i++)\n {\n if(arr[i]>arr[i-1])\n {\n size++;\n ans=max(ans,size);\n }\n else{\n size=1;\n }\n }\n return ans;\n }\n};\n``` | 2 | 0 | ['C++'] | 0 |
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