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maximize-total-cost-of-alternating-subarrays | C++ solution with explanation | c-solution-with-explanation-by-roy258-h7g0 | 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 | roy258 | NORMAL | 2024-07-20T07:45:39.589326+00:00 | 2024-07-20T07:45:39.589349+00:00 | 1 | 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. $$O(n)$$ -->\nO(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Greedy Solution with Explanation | greedy-solution-with-explanation-by-umes-2by3 | Intuition\n Describe your first thoughts on how to solve this problem. \nwhenever a new_element is added in front of a subarray the sum of subarray(prev) become | umesh_346 | NORMAL | 2024-07-20T03:55:06.125562+00:00 | 2024-07-20T03:55:06.125587+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nwhenever a **new_element** is added in front of a subarray the sum of subarray(**prev**) becomes (**new_element**) - (**prev**) lets call it **temp**. \nif the **prev** is positive, solution maximizes if we split, \n**temp = new_element +... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | DP | dp-by-112115046-gx03 | \n\n# Code\n\nclass Solution:\n def maximumTotalCost(self, nums: List[int]) -> int:\n n = len(nums)\n @cache\n def dp(i, ok):\n | 112115046 | NORMAL | 2024-07-18T07:27:00.619582+00:00 | 2024-07-18T07:27:00.619613+00:00 | 2 | false | \n\n# Code\n```\nclass Solution:\n def maximumTotalCost(self, nums: List[int]) -> int:\n n = len(nums)\n @cache\n def dp(i, ok):\n if i == n: return 0\n if ok == 1:\n ans = nums[i] + dp(i+1, 0)\n else:\n ans = max(-nums[i] + dp(i+1, ... | 0 | 0 | ['Python3'] | 0 |
maximize-total-cost-of-alternating-subarrays | JAVA MEMO EASY SOLUTION USER FRIENDLY>>> | java-memo-easy-solution-user-friendly-by-4txp | Intuition\n Describe your first thoughts on how to solve this problem. \nIt is the simple DP Memoization Approach, Here in this case we can take or not-take the | ujjalmodak2000 | NORMAL | 2024-07-17T20:32:26.987364+00:00 | 2024-07-17T20:32:26.987385+00:00 | 7 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIt is the simple DP Memoization Approach, Here in this case we can take or not-take the next element of the array,If we don\'t take the next element,means it is not a subarray, so start from the next element to find a subarray. \n\n# Appr... | 0 | 0 | ['Dynamic Programming', 'Memoization', 'Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | Most Simple way to Understated | most-simple-way-to-understated-by-mentoe-taeg | \n\n# Approach\nrecursion : Take Not-take\n\n# Complexity\n- Time complexity : O(N)2 = O(N);\n\n- Space complexity:\nO(n2)\n\n# Code\n```\nclass Solution {\npub | Mentoes22 | NORMAL | 2024-07-17T12:41:52.817688+00:00 | 2024-07-17T12:41:52.817711+00:00 | 0 | false | \n\n# Approach\nrecursion : Take Not-take\n\n# Complexity\n- Time complexity : O(N)*2 = O(N);\n\n- Space complexity:\nO(n*2)\n\n# Code\n```\nclass Solution {\npublic:\n long long int helper(vector<int>& nums, int i, int state, vector<vector<long long int>>& dp) {\n if (i >= nums.size())\n return 0;... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | 5 Lines of code: simple DP with constant space. 0ms, beats 100% | 5-lines-of-code-simple-dp-with-constant-q90fk | Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n\nimpl Solution {\n pub fn maximum_total_cost(nums: Vec<i32>) -> i64 { \n | germanov_dev | NORMAL | 2024-07-14T17:59:05.261536+00:00 | 2024-07-14T18:07:38.607491+00:00 | 1 | false | # Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```\nimpl Solution {\n pub fn maximum_total_cost(nums: Vec<i32>) -> i64 { \n let (mut dp0, mut dp1) = (nums[0] as i64,nums[0] as i64);\n for idx in 1..nums.len() {\n (dp0, dp1) = (dp0.max(dp1) + nums[i... | 0 | 0 | ['Rust'] | 0 |
maximize-total-cost-of-alternating-subarrays | Simple 2D dp | simple-2d-dp-by-tayal-np1h | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | tayal | NORMAL | 2024-07-13T08:22:05.471103+00:00 | 2024-07-13T08:22:05.471119+00:00 | 0 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | JAVA - 100% Faster - (2D) DP - (Positive - Negative) Approach | java-100-faster-2d-dp-positive-negative-vrg2i | \n# Code\n\nclass Solution {\n public long maximumTotalCost(int[] nums) {\n \n int n = nums.length;\n long add =0, sub=0;\n if(n= | AbhirMhjn | NORMAL | 2024-07-13T03:41:33.832133+00:00 | 2024-07-13T03:41:33.832156+00:00 | 1 | false | \n# Code\n```\nclass Solution {\n public long maximumTotalCost(int[] nums) {\n \n int n = nums.length;\n long add =0, sub=0;\n if(n==1){\n return nums[0];\n }\n // long[][] dp = new long[n][2];\n \n // dp[0][0] = nums[0];\n // dp[0][1] = nums[... | 0 | 0 | ['Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | Clean solution | clean-solution-by-hawtinzeng-e2c1 | Intuition\n Describe your first thoughts on how to solve this problem. \ndon\'t rely on the test set, maybe the test set will give you some misunderstanding.\n\ | HawtinZeng | NORMAL | 2024-07-09T01:12:46.047864+00:00 | 2024-07-09T01:12:46.047882+00:00 | 1 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\ndon\'t rely on the test set, maybe the test set will give you some misunderstanding.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$... | 0 | 0 | ['TypeScript'] | 0 |
maximize-total-cost-of-alternating-subarrays | Python || DP || Binary length Subarray | python-dp-binary-length-subarray-by-in_s-peku | Consider Subarray of length 1 or 2 to maximazie your answer.\nTC : O(n)\n\nCode:\n\n\nclass Solution:\n def maximumTotalCost(self, nums: List[int]) -> int:\n | iN_siDious | NORMAL | 2024-07-04T18:12:18.002937+00:00 | 2024-07-04T18:12:18.002974+00:00 | 1 | false | Consider Subarray of length 1 or 2 to maximazie your answer.\nTC : O(n)\n\nCode:\n\n```\nclass Solution:\n def maximumTotalCost(self, nums: List[int]) -> int:\n n=len(nums)\n @cache\n def dp(idx):\n if idx>=n: return 0\n #take one length subarray\n ans=dp(idx+1)+... | 0 | 0 | ['Dynamic Programming', 'Python3'] | 0 |
maximize-total-cost-of-alternating-subarrays | Clean java solution | DP easy to understand | clean-java-solution-dp-easy-to-understan-ev0j | Code\n\nclass Solution {\n Map<String, Long> memo;\n public long maximumTotalCost(int[] nums) {\n memo = new HashMap<>();\n return dfs(nums, | SG-C | NORMAL | 2024-07-04T12:22:43.379979+00:00 | 2024-07-04T12:22:43.380009+00:00 | 5 | false | # Code\n```\nclass Solution {\n Map<String, Long> memo;\n public long maximumTotalCost(int[] nums) {\n memo = new HashMap<>();\n return dfs(nums, 0, true);\n }\n private long dfs(int[] nums, int i, boolean sign){\n if(i == nums.length)\n return 0;\n\n String state = i ... | 0 | 0 | ['Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | Recursion + Memoization | recursion-memoization-by-surendrapokala1-ocnd | 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 | surendrapokala111 | NORMAL | 2024-07-04T09:38:25.439949+00:00 | 2024-07-04T09:38:25.439984+00:00 | 7 | 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. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Elegant DP Solution with Clear Mathematical & Constructive Proof | elegant-dp-solution-with-clear-mathemati-ulel | Intuition\nThe problem requires splitting an array into subarrays to maximize the total cost, where the cost of a subarray is defined in an alternating addition | Sambosa123 | NORMAL | 2024-07-04T04:03:26.068081+00:00 | 2024-09-26T23:19:35.056721+00:00 | 7 | false | # Intuition\nThe problem requires splitting an array into subarrays to maximize the total cost, where the cost of a subarray is defined in an alternating addition-subtraction manner. This can be approached using dynamic programming by considering only subarrays of length 1 or 2, simplifying the calculation of costs.\n\... | 0 | 0 | ['Array', 'Dynamic Programming', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Easy DP Solution | easy-dp-solution-by-chinna_dubba-16u4 | 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 | chinna_dubba | NORMAL | 2024-07-03T13:48:47.147762+00:00 | 2024-07-03T13:48:47.147800+00:00 | 2 | 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. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | Beginner Friendly : Easy Solution - Recursive + DP approach | beginner-friendly-easy-solution-recursiv-hqna | \n\n# Code 1 - Recursive( WITH TLE)\n\nclass Solution:\n def solve(self,ind,flag,n,nums):\n if ind == n:\n return 0\n if flag == 0:\ | astralamind | NORMAL | 2024-07-02T18:35:25.674616+00:00 | 2024-07-02T18:35:25.674653+00:00 | 4 | false | \n\n# Code 1 - Recursive( WITH TLE)\n```\nclass Solution:\n def solve(self,ind,flag,n,nums):\n if ind == n:\n return 0\n if flag == 0:\n a = nums[ind] + self.solve(ind+1,0,n,nums)\n b = -1*nums[ind] + self.solve(ind+1,1,n,nums)\n return max(a,b)\n else... | 0 | 0 | ['Python3'] | 0 |
maximize-total-cost-of-alternating-subarrays | My O(N) time and O(1) space most optimal dp solution | my-on-time-and-o1-space-most-optimal-dp-u6sc8 | \n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n\n# Code\n\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums | sanyaa23_ | NORMAL | 2024-07-02T17:29:54.869709+00:00 | 2024-07-02T17:29:54.869733+00:00 | 6 | false | \n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n int n = nums.size();\n long long last0 = nums[0];\n long long last1 = nums[0];\n for (int ind = 1; ind < n; ind++) {\n ... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Inclusion Exclusion Principle | DP | Java | O(N) | inclusion-exclusion-principle-dp-java-on-nkqf | Approach\n Describe your approach to solving the problem. \n1. Use Inclusion Exclusion principle which is one of form of Dynamic Programming\n\n# Video Tutorial | 21stCenturyLegend | NORMAL | 2024-07-02T15:37:12.567138+00:00 | 2024-07-02T15:39:53.795389+00:00 | 6 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n1. Use Inclusion Exclusion principle which is one of form of Dynamic Programming\n\n# Video Tutorial\n[Video Link\n](https://www.youtube.com/watch?v=Zvq458gwpwY)\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$... | 0 | 0 | ['Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | Easy Iterative Solution | easy-iterative-solution-by-kvivekcodes-rdtl | \n\n# Code\n\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n int n = nums.size();\n if(n == 1) return nums[0];\n | kvivekcodes | NORMAL | 2024-07-02T10:30:13.600868+00:00 | 2024-07-02T10:30:13.600886+00:00 | 3 | false | \n\n# Code\n```\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n int n = nums.size();\n if(n == 1) return nums[0];\n long long dp[n];\n dp[0] = nums[0];\n dp[1] = max(nums[0]+nums[1], nums[0]-nums[1]);\n for(int i = 2; i < n; i++){\n ... | 0 | 0 | ['Array', 'Dynamic Programming', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | C++ || DP | c-dp-by-riomerz-5lag | \n# Code\n\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n long long dp[nums.size()][2];\n memset(dp, 0, sizeof( | riomerz | NORMAL | 2024-07-01T20:18:14.195038+00:00 | 2024-07-01T20:18:14.195100+00:00 | 8 | false | \n# Code\n```\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n long long dp[nums.size()][2];\n memset(dp, 0, sizeof(dp));\n for(int i = 1;i<nums.size();i++){\n dp[i][0] = max(dp[i-1][0] , dp[i-1][1]) + nums[i];\n if(nums[i] >= 0){\n ... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | DP Solution || Just need to think | dp-solution-just-need-to-think-by-namang-n7qx | 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 | namangupta_05 | NORMAL | 2024-07-01T12:29:07.858886+00:00 | 2024-07-01T12:29:07.858919+00:00 | 5 | 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. $$O(n)$$ --... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | 2 Soultions | DP | DFS/Recursion -> Memoization | 2-soultions-dp-dfsrecursion-memoization-9a1j8 | Solution-1: (Recursion + Memo) -- (TLE -- 688 / 692 TCs passed -- 99.42%)\n### IDEA\n+ Simulte take & continue and take & end the subarray behaviour at each pos | shahsb | NORMAL | 2024-07-01T04:44:21.853824+00:00 | 2024-07-02T03:22:56.813545+00:00 | 8 | false | # Solution-1: (Recursion + Memo) -- (TLE -- 688 / 692 TCs passed -- 99.42%)\n### IDEA\n+ Simulte `take & continue` and `take & end the subarray` behaviour at each position.\n+ The sign would be determined using -- `pow(-1, (i-l))`\n### Complexity: \n+ **Time:** O(N^2)\n+ **Space:** O(N^2)\n\n### CODE:\n```\n# define ll... | 0 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C'] | 0 |
maximize-total-cost-of-alternating-subarrays | It's like house robber , no need [0/1] | its-like-house-robber-no-need-01-by-gues-zwm6 | \nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n int n = nums.size();\n vector<long long>dp(n + 1);\n dp[ | guesswhohas2cats | NORMAL | 2024-06-30T09:42:03.129222+00:00 | 2024-06-30T09:42:03.129255+00:00 | 6 | false | ```\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n int n = nums.size();\n vector<long long>dp(n + 1);\n dp[0] = 0;\n dp[1] = nums[0];\n for(int i = 1; i < n; i ++)\n dp[i + 1] = max(dp[i] + nums[i], dp[i - 1] + nums[i - 1] - nums[i]);\n ... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | C++ || Memoization || Easy Approach | c-memoization-easy-approach-by-gurtejsin-vzc6 | # Intuition \n\n\n# Approach\n\nSimply write all the cases that may occur to take an element and to not take it.\nRefer to this youtube video for better under | GurtejSingh84 | NORMAL | 2024-06-30T06:43:24.166412+00:00 | 2024-06-30T06:43:24.166435+00:00 | 0 | false | <!-- # Intuition -->\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nSimply write all the cases that may occur to take an element and to not take it.\nRefer to this youtube video for better understanding:\n\n[https://youtu.be/F... | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Intuitive dp solution, explained. | intuitive-dp-solution-explained-by-rahul-0ekj | Intuition\nSince k can vary a lot, and checking the answer for all possible k values is not possible. So, some way had to be thought to store answers till i ind | rahul_o15 | NORMAL | 2024-06-29T20:31:56.265711+00:00 | 2024-06-29T20:31:56.265732+00:00 | 4 | false | # Intuition\nSince ``k`` can vary a lot, and checking the answer for all possible ``k`` values is not possible. So, some way had to be thought to store answers till ``i`` index and then proceed ahead.\n\nAfter reading the problem, anyone can understand that only alterate values can be flipped from negative to positive ... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
ipo | Day 54 || C++ || Priority_Queue || Easiest Beginner Friendly Sol | day-54-c-priority_queue-easiest-beginner-m55e | Intuition of this Problem:\nThe problem asks us to maximize the total capital by selecting at most k distinct projects. We have a limited amount of initial capi | singhabhinash | NORMAL | 2023-02-23T01:02:01.039641+00:00 | 2023-04-01T10:27:57.917278+00:00 | 35,045 | false | # Intuition of this Problem:\nThe problem asks us to maximize the total capital by selecting at most k distinct projects. We have a limited amount of initial capital, and each project has a minimum capital requirement and a pure profit. We need to choose the projects in such a way that we can complete at most k distinc... | 434 | 3 | ['Sorting', 'Heap (Priority Queue)', 'C++', 'Java', 'Python3'] | 26 |
ipo | Very Simple (Greedy) Java Solution using two PriorityQueues | very-simple-greedy-java-solution-using-t-shbv | The idea is each time we find a project with max profit and within current capital capability.\nAlgorithm:\n1. Create (capital, profit) pairs and put them into | shawngao | NORMAL | 2017-02-04T17:18:49.802000+00:00 | 2018-10-22T16:22:07.561756+00:00 | 26,989 | false | The idea is each time we find a project with ```max``` profit and within current capital capability.\nAlgorithm:\n1. Create (capital, profit) pairs and put them into PriorityQueue ```pqCap```. This PriorityQueue sort by capital increasingly.\n2. Keep polling pairs from ```pqCap``` until the project out of current capit... | 294 | 0 | [] | 36 |
ipo | 🔥 🔥 🔥 Easy to understand | 💯 Fast | maxHeap | Sorting 🔥 🔥 🔥 | easy-to-understand-fast-maxheap-sorting-5sie7 | Check out my profile to look into solutions to more problems.\n\n# Intuition\n- The intuition behind this code is to maximize the available capital after select | bhanu_bhakta | NORMAL | 2024-06-15T00:11:16.384902+00:00 | 2024-06-15T01:39:26.985142+00:00 | 36,179 | false | Check out my [profile](https://leetcode.com/u/bhanu_bhakta/) to look into solutions to more problems.\n\n# Intuition\n- The intuition behind this code is to maximize the available capital after selecting up to k projects, by strategically choosing the projects with the highest profit that can be started within the curr... | 215 | 1 | ['Sorting', 'Heap (Priority Queue)', 'C++', 'Java', 'Go', 'Python3', 'Kotlin'] | 15 |
ipo | [Python] Priority Queue with Explanation | python-priority-queue-with-explanation-b-1rej | Explanation\nLoop k times:\nAdd all possible projects (Capital <= W) into the priority queue with the priority = -Profit.\nGet the project with the smallest pri | lee215 | NORMAL | 2017-02-04T19:12:36.162000+00:00 | 2020-01-10T03:10:31.119177+00:00 | 17,026 | false | ## Explanation\nLoop `k` times:\nAdd all possible projects (`Capital <= W`) into the priority queue with the `priority = -Profit`.\nGet the project with the smallest priority (biggest Profit).\nAdd the Profit to `W`\n<br>\n\n@tife1379: This visualisation should help understand.\n\n\n\n# YouTube Video Explanation:\n\n**If you want a video for this question please write in the comments**\n\n<!-- https://www.youtube.com/watch?v=ujU-jeO1v-k -->\nFollow me on Instag... | 86 | 6 | ['Array', 'Greedy', 'Sorting', 'Heap (Priority Queue)', 'Python', 'C++', 'Java', 'JavaScript'] | 15 |
ipo | Clean Example🔥🔥|| Full Explanation✅|| Priority Queue✅|| C++|| Java|| Python3 | clean-example-full-explanation-priority-7z0h7 | Intuition :\n- Here, We have to find maximum profit that can be achieved by selecting at most k projects to invest in, given an initial capital of W, a set of P | N7_BLACKHAT | NORMAL | 2023-02-23T02:20:15.549842+00:00 | 2023-02-23T02:53:33.596114+00:00 | 4,983 | false | # Intuition :\n- Here, We have to find maximum profit that can be achieved by selecting at most k projects to invest in, given an initial capital of W, a set of Profits and Capital requirements for each project.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach :\n- Here we are using tw... | 48 | 2 | ['Heap (Priority Queue)', 'Python', 'C++', 'Java', 'Python3'] | 5 |
ipo | ✅ JAVA Solution | java-solution-by-coding_menance-y3ig | JAVA Solution\n\nJAVA []\nclass Solution {\n class Pair implements Comparable<Pair> {\n int capital, profit;\n\n public Pair(int capital, int p | coding_menance | NORMAL | 2023-02-23T03:37:08.295694+00:00 | 2023-02-23T03:37:08.295737+00:00 | 4,292 | false | # JAVA Solution\n\n``` JAVA []\nclass Solution {\n class Pair implements Comparable<Pair> {\n int capital, profit;\n\n public Pair(int capital, int profit) {\n this.capital = capital;\n this.profit = profit;\n }\n\n public int compareTo(Pair pair) {\n retu... | 47 | 3 | ['Java'] | 1 |
ipo | C++ Priority Queue Efficient Explained Solution | c-priority-queue-efficient-explained-sol-hy3q | First we will store all the projects in projects vector as pairs {Profit(i), Capital(i)};\n2. Now we will sort all the projects according to its capital value.\ | manikgarg2000 | NORMAL | 2021-08-07T07:55:36.830093+00:00 | 2021-08-07T07:55:36.830124+00:00 | 4,354 | false | 1. First we will store all the projects in projects vector as pairs {Profit(i), Capital(i)};\n2. Now we will sort all the projects according to its capital value.\n3. Now we will fetch all the projects that we can perform for our own capital value. \n4. After fetching all these projects sotre their profit value in Max ... | 47 | 0 | ['Greedy', 'C', 'Heap (Priority Queue)'] | 9 |
ipo | ✅✅ Fast 🔥🔥 Efficient 💯💯 Simplest Explanation 🏃♂️🏃♂️Dryrun🧠 | fast-efficient-simplest-explanation-dryr-2khj | Thanks for checking out my solution. Do Upvote if this helped \uD83D\uDC4D\n#### This post has been made with \u2764 by Alok Khansali\n\n\n# \uD83C\uDFAFApproac | TheCodeAlpha | NORMAL | 2024-06-15T08:40:21.267794+00:00 | 2024-10-20T18:00:33.927818+00:00 | 2,572 | false | #### Thanks for checking out my solution. Do Upvote if this helped \uD83D\uDC4D\n#### This post has been made with \u2764 by [Alok Khansali](https://leetcode.com/u/TheCodeAlpha/)\n\n\n# \uD83C\uDFAFApproach : Priority Queue\n<!-- Describe your approach to solving the problem. -->\n\n# Intuition \uD83D\uDD2E\n<!-- Descr... | 40 | 0 | ['Greedy', 'Sorting', 'Heap (Priority Queue)', 'C++', 'Java', 'Python3'] | 3 |
ipo | 8-liner C++ 42ms beat 98% greedy algorithm (detailed explanation) | 8-liner-c-42ms-beat-98-greedy-algorithm-1oetz | Key Observation: \n1. The more capital W you have now, the more maximum capital you will eventually earn.\n2. Working on any doable project with positive P[i] > | zzg_zzm | NORMAL | 2017-02-09T20:09:28.774000+00:00 | 2017-02-09T20:09:28.774000+00:00 | 7,728 | false | **Key Observation:** \n1. The more capital `W` you have now, the more maximum capital you will eventually earn.\n2. Working on any doable project with positive `P[i] > 0` increases your capital `W`.\n3. Any project with `P[i] = 0` is useless and should be filtered away immediately (note that the problem only guarantees... | 28 | 3 | ['Greedy', 'C++'] | 8 |
ipo | 🚀Simplest Solution🚀 Beginner Friendly||🔥Priority Queue||🔥C++|| Python3🔥 | simplest-solution-beginner-friendlyprior-ypuu | Consider\uD83D\uDC4D\n\n Please Upvote If You Find It Helpful\n\n\n# Intuition\nExplanation\nThis code will first create a vector of pairs co | naman_ag | NORMAL | 2023-02-23T02:08:39.685908+00:00 | 2023-02-23T02:32:36.970746+00:00 | 2,391 | false | # Consider\uD83D\uDC4D\n```\n Please Upvote If You Find It Helpful\n```\n\n# Intuition\nExplanation\nThis code will first create a vector of pairs containing the capital and profits of each project. It will then sort this vector based on the capital required for each project. Then, it will use a prio... | 27 | 4 | ['Sorting', 'Heap (Priority Queue)', 'Python', 'C++', 'Python3'] | 2 |
ipo | Sort+Priority Queue+Binary search||75ms Beats 99.60% | sortpriority-queuebinary-search75ms-beat-u8yx | Intuition\n Describe your first thoughts on how to solve this problem. \nIPO problem is solved almost 1 year ago.\nRedo this problem with 2 approaches.\n# Appr | anwendeng | NORMAL | 2024-06-15T01:26:54.545782+00:00 | 2024-06-15T07:11:55.862101+00:00 | 5,355 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIPO problem is solved almost 1 year ago.\nRedo this problem with 2 approaches.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n[Please turn on english subtitles if necessary]\n[https://youtu.be/W4AoobL65jA?si=yj9KWW... | 25 | 1 | ['Binary Search', 'Sorting', 'Heap (Priority Queue)', 'C++', 'Python3'] | 10 |
ipo | Python solution | python-solution-by-stefanpochmann-gxmm | Keep a max-heap of current possible profits. Insert possible profits as soon as their needed capital is reached.\n\n def findMaximizedCapital(self, k, W, Pro | stefanpochmann | NORMAL | 2017-02-04T19:35:04.213000+00:00 | 2018-09-22T14:15:31.539483+00:00 | 4,352 | false | Keep a max-heap of current possible profits. Insert possible profits as soon as their needed capital is reached.\n\n def findMaximizedCapital(self, k, W, Profits, Capital):\n current = []\n future = sorted(zip(Capital, Profits))[::-1]\n for _ in range(k):\n while future and future[-1]... | 21 | 0 | [] | 8 |
ipo | [Greedy + Proof + Tutorial] IPO | greedy-proof-tutorial-ipo-by-never_get_p-y273 | Topic : Greedy\nGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. | never_get_piped | NORMAL | 2024-06-15T02:13:38.714437+00:00 | 2024-06-21T04:51:54.192424+00:00 | 3,092 | false | **Topic** : Greedy\nGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. In these algorithms, decisions are made based on the information available at the current moment without considering the consequences of these decisions in ... | 17 | 0 | ['Greedy', 'C', 'PHP', 'Python', 'Java', 'Go', 'JavaScript'] | 3 |
ipo | [Python] Two heaps greedy solution with simple explanation | python-two-heaps-greedy-solution-with-si-8ypm | \nfrom heapq import *\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n minCapital | shub_hamburger | NORMAL | 2021-07-17T07:57:52.154596+00:00 | 2021-07-17T07:57:52.154643+00:00 | 1,810 | false | ```\nfrom heapq import *\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n minCapitalHeap = []\n maxProfitHeap = []\n\n # Insert all capitals to a min-heap\n for i in range(0, len(profits)):\n heappush(minCapit... | 17 | 0 | ['Greedy', 'Heap (Priority Queue)', 'Python', 'Python3'] | 3 |
ipo | C# Minimum Lines | c-minimum-lines-by-gregsklyanny-820j | Code\n\npublic class Solution \n{\n public int FindMaximizedCapital(int k, int w, int[] profits, int[] capital) \n {\n Array.Sort(capital, profits) | gregsklyanny | NORMAL | 2023-02-23T09:23:21.514035+00:00 | 2023-02-23T09:23:21.514081+00:00 | 512 | false | # Code\n```\npublic class Solution \n{\n public int FindMaximizedCapital(int k, int w, int[] profits, int[] capital) \n {\n Array.Sort(capital, profits);\n var pq = new PriorityQueue<int,int>();\n int index = 0;\n while(k > 0)\n {\n while (index < profits.Length && w ... | 16 | 1 | ['C#'] | 3 |
ipo | 32ms C++ beats 100% | 32ms-c-beats-100-by-f1re-crn3 | \n#define pi pair<int,int>\n#define f first\n#define s second\n\nclass Solution {\npublic:\n \n int findMaximizedCapital(int k, int W, vector<int>& p, vec | f1re | NORMAL | 2019-01-27T11:18:11.736298+00:00 | 2019-01-27T11:18:11.736367+00:00 | 1,678 | false | ```\n#define pi pair<int,int>\n#define f first\n#define s second\n\nclass Solution {\npublic:\n \n int findMaximizedCapital(int k, int W, vector<int>& p, vector<int>& c) {\n \n int n = p.size();\n vector<pair<int,int>> projects;\n for(int i=0 ; i<n ; i++) projects.push_back({p[i],c[i... | 15 | 0 | [] | 2 |
ipo | ✅ Java | Easy | Priority Queue | Greedy | With Explanation | java-easy-priority-queue-greedy-with-exp-3sg5 | The basic intuition is if you have capital then you can make profits by investing on projects and keep on adding the profits to company\'s capital and this proc | kalinga | NORMAL | 2023-02-23T05:39:47.640760+00:00 | 2023-02-23T06:09:45.288825+00:00 | 1,656 | false | **The basic intuition is if you have capital then you can make profits by investing on projects and keep on adding the profits to company\'s capital and this process will go on till the company\'s capital can be used in further project. So I used a greedy approach that if I will sort the List of Pairs according to the ... | 13 | 3 | ['Greedy', 'Sorting', 'Heap (Priority Queue)', 'Java'] | 1 |
ipo | Easy to understand C++ Solution beats 90% | easy-to-understand-c-solution-beats-90-b-8ghr | Intuition\n Describe your first thoughts on how to solve this problem. \nWhenver we see first k or last k elements based on some order, then more probably than | anuragkumar2608 | NORMAL | 2024-06-15T05:00:14.572151+00:00 | 2024-06-15T05:15:38.927310+00:00 | 1,713 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWhenver we see first k or last k elements based on some order, then more probably than not, it is a problem related to priority queues.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe first create a pair vector of... | 12 | 0 | ['C++'] | 1 |
ipo | ✅[C++] Greedy using Priority Queue | c-greedy-using-priority-queue-by-bit_leg-wuny | What? \nThe question wants us to find the maximum capital we can make by doing at most k projects. Since each project has a profit >= 0, therefore we will choos | biT_Legion | NORMAL | 2022-07-05T11:21:35.965527+00:00 | 2022-07-05T11:21:35.965573+00:00 | 1,055 | false | **What?** \nThe question wants us to find the maximum capital we can make by doing at most k projects. Since each project has a profit `>= 0`, therefore we will choose exact k projects because each project will contribute something to our answer. Greedy appraoch would be a better choice here. \n\n**Why Greedy?** \nThe ... | 12 | 0 | ['Greedy', 'C', 'Heap (Priority Queue)'] | 1 |
ipo | ✏️Without heap 🥳 || without sorting 🎁 || Beats 100% Run 🍾 ➕ 100% memory 🍾|| Proof💯 | without-heap-without-sorting-beats-100-r-fsqf | \n# \uD83C\uDF89 Screenshot \uD83D\uDCF8\n\n\n\n\n## Input \uD83D\uDCE5 \n\n Two Number Array (profits) && (capital)\n\n And k = number of max project we | Prakhar-002 | NORMAL | 2024-06-15T07:12:42.873346+00:00 | 2024-06-15T07:12:42.873367+00:00 | 1,839 | false | \n# \uD83C\uDF89 Screenshot \uD83D\uDCF8\n\n\n\n\n## Input \uD83D\uDCE5 \n\n Two Number Array (profits) && (capital)\n\n And k = number of max project we would do\n\n And w = Total wealth \uD83D\uDCB8... | 11 | 7 | ['Array', 'Greedy', 'C', 'C++', 'Java', 'Python3', 'JavaScript'] | 8 |
ipo | C++✅✅ | 2 Heaps🆗 | Self Explanatory Approach | Clean Code | | c-2-heapsok-self-explanatory-approach-cl-i3c4 | \n\n# Code\n# PLEASE DO UPVOTE!!!!!\nCONNECT WITH ME ON LINKEDIN -> https://www.linkedin.com/in/md-kamran-55b98521a/\n\n\n\nclass Solution {\npublic:\n\n int | mr_kamran | NORMAL | 2023-02-23T06:09:57.466031+00:00 | 2023-02-23T06:09:57.466384+00:00 | 1,522 | false | \n\n# Code\n# PLEASE DO UPVOTE!!!!!\n**CONNECT WITH ME ON LINKEDIN -> https://www.linkedin.com/in/md-kamran-55b98521a/**\n\n```\n\nclass Solution {\npublic:\n\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n \n priority_queue<pair<int,int>>mxh;\n priority_... | 11 | 1 | ['C++'] | 3 |
ipo | Detailed Solution With Steps | detailed-solution-with-steps-by-code_ran-3j8j | \nclass Solution {\npublic static int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int n = profits.length;\n\n // Create a list of | cOde_Ranvir25 | NORMAL | 2023-02-23T04:27:58.107463+00:00 | 2023-02-23T04:27:58.107509+00:00 | 984 | false | ```\nclass Solution {\npublic static int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int n = profits.length;\n\n // Create a list of pairs (capital, profit) for all n projects\n List<int[]> projects = new ArrayList<>();\n for (int i = 0; i < n; i++) {\n projects.add(new int[]... | 11 | 0 | ['Java'] | 4 |
ipo | JAVA Solution Explained in HINDI | java-solution-explained-in-hindi-by-the_-7byw | https://youtu.be/18FjKA210oM\n\nFor explanation, please watch the above video and do like, share and subscribe the channel. \u2764\uFE0F Also, please do upvote | The_elite | NORMAL | 2024-06-15T14:44:29.273520+00:00 | 2024-06-15T14:44:29.273545+00:00 | 908 | false | https://youtu.be/18FjKA210oM\n\nFor explanation, please watch the above video and do like, share and subscribe the channel. \u2764\uFE0F Also, please do upvote the solution if you liked it.\n\n# Subscribe:- [ReelCoding](https://www.youtube.com/@reelcoding?sub_confirmation=1)\n\nSubscribe Goal:- 500\nCurrent Subscriber:... | 10 | 0 | ['Java'] | 1 |
ipo | 🔥🔥🔥 Heap + Greedy Solution (Beats 99.69%) 🔥Python 3🔥 | heap-greedy-solution-beats-9969-python-3-g7gc | \n# Code\n\nfrom heapq import heappush, heappop, nlargest\n\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: | KingJamesH | NORMAL | 2023-02-23T00:12:06.270329+00:00 | 2023-02-27T00:17:18.702023+00:00 | 986 | false | \n# Code\n```\nfrom heapq import heappush, heappop, nlargest\n\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n if w >= max(capital):\n return w + sum(nlargest(k, profits))\n \n projects = [[capital[i],profits[i]] fo... | 10 | 0 | ['Greedy', 'Heap (Priority Queue)', 'Python3'] | 1 |
ipo | Python 3 || 7 lines, heap || T/S: 96% / 48% | python-3-7-lines-heap-ts-96-48-by-spauld-99r0 | \nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], \n capital: List[int]) -> int: | Spaulding_ | NORMAL | 2023-02-23T08:28:18.068304+00:00 | 2024-05-29T00:06:07.199505+00:00 | 526 | false | ```\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], \n capital: List[int]) -> int:\n\n heap = []\n projects = sorted(zip(capital, profits),\n key=lambda x: x[0], reverse=True)\n\n ... | 9 | 0 | ['Python3'] | 1 |
ipo | ✅✅ Priority_Queue || Beginner Friendly Sol || C++ || Java || Python3 || C# || Javascript || C | priority_queue-beginner-friendly-sol-c-j-qmva | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem asks us to maximize the total capital by selecting at most k distinct proje | sidharthjain321 | NORMAL | 2024-06-15T19:12:30.970375+00:00 | 2024-06-16T21:54:24.254126+00:00 | 638 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem asks us to maximize the total capital by selecting at most k distinct projects. We have a limited amount of initial capital, and each project has a minimum capital requirement and a pure profit. We need to choose the projects ... | 8 | 0 | ['Array', 'Greedy', 'C', 'Sorting', 'Heap (Priority Queue)', 'C++', 'Java', 'Python3', 'JavaScript', 'C#'] | 2 |
ipo | simple and easy C++ solution 😍❤️🔥 | simple-and-easy-c-solution-by-shishirrsi-n5rn | if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, ve | shishirRsiam | NORMAL | 2024-06-15T06:25:43.466999+00:00 | 2024-06-15T06:25:43.467042+00:00 | 1,109 | false | # if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) \n {\n vector<pair<int,int>>store;\n int n = capital.size();\n for(int i=0;i<n;i++)\n store.push_bac... | 7 | 0 | ['Array', 'Greedy', 'Sorting', 'Heap (Priority Queue)', 'C++'] | 4 |
ipo | PuTtA EaSY Solution C++ ✅ | Heap 🔥🔥 | | putta-easy-solution-c-heap-by-saisreeram-dfiq | \n# Code\n\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n priority_queue<pair<int | Saisreeramputta | NORMAL | 2023-02-23T09:49:32.341914+00:00 | 2023-02-23T09:49:32.341955+00:00 | 1,177 | false | \n# Code\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n priority_queue<pair<int,int>> pq;\n priority_queue<int> pq2;\n\n for(int i=0;i<profits.size();i++){\n pq.push({-1*capital[i],profits[i]}); \n }\n\... | 7 | 0 | ['Heap (Priority Queue)', 'C++'] | 3 |
ipo | JavaScript | MaxHeap | Clean and Easy | 333ms - 78.72%, 77.1MB - 72.34% | javascript-maxheap-clean-and-easy-333ms-uns9s | Credits to https://leetcode.com/explore/interview/card/leetcodes-interview-crash-course-data-structures-and-algorithms/709/greedy/4647/ for the solution explana | natalyav | NORMAL | 2023-01-20T03:44:38.848478+00:00 | 2023-01-20T03:46:26.003081+00:00 | 1,041 | false | Credits to https://leetcode.com/explore/interview/card/leetcodes-interview-crash-course-data-structures-and-algorithms/709/greedy/4647/ for the solution explanation.\n\n# Intuition\nGreedily choose the most profitable project that you can afford at each step. Use a heap to keep track of the most profitable project and ... | 7 | 0 | ['JavaScript'] | 1 |
ipo | Kotlin Heap Solution | kotlin-heap-solution-by-kotlinc-njl6 | Intuition\nWe first sort the projects by their capital.\n\nWhen we finish the previous project, we check what new projects we can now participate, then we use a | kotlinc | NORMAL | 2023-02-23T03:39:11.932176+00:00 | 2023-02-23T05:25:06.285251+00:00 | 159 | false | # Intuition\nWe first sort the projects by their capital.\n\nWhen we finish the previous project, we check what new projects we can now participate, then we use a `PriorityQueue` to track the available projects that we can participate with.\n\nThe `PriorityQueue` will give us the most lucrative project.\n\n\n# Complexi... | 6 | 0 | ['Kotlin'] | 2 |
ipo | [JavaScript] With out heap | javascript-with-out-heap-by-ky61k105-2jo2 | \nvar findMaximizedCapital = function(k, w, profits, capital) {\n if(w >= Math.max(...capital)) {\n profits.sort((a, b) => b - a);\n return profits. | ky61k105 | NORMAL | 2021-09-06T11:02:47.433112+00:00 | 2021-09-06T11:03:55.534209+00:00 | 536 | false | ```\nvar findMaximizedCapital = function(k, w, profits, capital) {\n if(w >= Math.max(...capital)) {\n profits.sort((a, b) => b - a);\n return profits.slice(0, k).reduce((acc, num) => acc + num, w);\n }\n \n for (let i = 0; i < k; i++) {\n let maxProfit = -Infinity;\n let projectIndex = ... | 6 | 0 | ['JavaScript'] | 1 |
ipo | Python | Greedy | python-greedy-by-khosiyat-va95 | see the Successfully Accepted Submission\n\n# Code\n\nimport heapq\nfrom typing import List\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: in | Khosiyat | NORMAL | 2024-06-15T05:33:33.665515+00:00 | 2024-06-15T05:33:33.665548+00:00 | 775 | false | [see the Successfully Accepted Submission](https://leetcode.com/problems/ipo/submissions/1288724380/?envType=daily-question&envId=2024-06-15)\n\n# Code\n```\nimport heapq\nfrom typing import List\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n ... | 5 | 0 | ['Python3'] | 2 |
ipo | C++ | 100% Faster | Priority Queue | Easy Step By Step Explanation | c-100-faster-priority-queue-easy-step-by-yi05 | \n\n# Intuition\n Describe your first thoughts on how to solve this problem. \nThe key to maximizing capital before LeetCode\'s IPO lies in strategically select | VYOM_GOYAL | NORMAL | 2024-06-15T00:17:18.252497+00:00 | 2024-06-15T00:17:18.252518+00:00 | 823 | false | \n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe key to maximizing capital before LeetCode\'s IPO lies in strategically selecting projects that offer the hig... | 5 | 0 | ['Array', 'Greedy', 'Heap (Priority Queue)', 'C++'] | 2 |
ipo | Choosing Greedily ! | choosing-greedily-by-_aman_gupta-j09u | # Intuition \n\n\n\n\n\n# Complexity\n- Time complexity: O((n + k) log (n)) = O(nlogn)\n\n\n- Space complexity: O(n)\n\n\n# Code\n\nclass Solution {\npublic:\ | _aman_gupta_ | NORMAL | 2023-02-23T06:53:08.732049+00:00 | 2023-02-23T06:54:51.686459+00:00 | 443 | 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: $$O((n + k) log (n))$$ = $$O(nlogn)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity... | 5 | 0 | ['Sorting', 'Heap (Priority Queue)', 'C++'] | 1 |
ipo | Heap (Priority Queue) | heap-priority-queue-by-alien35-xd42 | Intuition & Approach\nhttps://youtu.be/IjddNQDZxWQ\n\n# Code\n\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vect | Alien35 | NORMAL | 2023-02-23T06:34:59.076462+00:00 | 2023-02-23T06:34:59.076497+00:00 | 514 | false | # Intuition & Approach\nhttps://youtu.be/IjddNQDZxWQ\n\n# Code\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n int n = profits.size();\n vector<pair<int, int>> projects(n);\n\n for (int i = 0; i < n; ++i)\n proje... | 5 | 0 | ['Greedy', 'Sorting', 'Heap (Priority Queue)', 'C++'] | 1 |
ipo | Python short and clean. Greedy. Two heaps (PriorityQueues). | python-short-and-clean-greedy-two-heaps-p1afc | Approach\nTLDR; Same as Official solution, except another heap is used instead of sorting the projects.\n\n# Complexity\n- Time complexity: O(n * log(n))\n\n- S | darshan-as | NORMAL | 2023-02-23T04:54:29.690673+00:00 | 2023-02-23T04:54:29.690711+00:00 | 188 | false | # Approach\nTLDR; Same as [Official solution](https://leetcode.com/problems/ipo/solutions/2959870/ipo/), except another heap is used instead of sorting the projects.\n\n# Complexity\n- Time complexity: $$O(n * log(n))$$\n\n- Space complexity: $$O(n)$$\n\nwhere, `n is the number of projects`.\n\n# Code\n```python\nclass... | 5 | 0 | ['Greedy', 'Sorting', 'Heap (Priority Queue)', 'Python', 'Python3'] | 1 |
ipo | GoLang Solution with explanation | golang-solution-with-explanation-by-dr41-ec3b | Intuition\nThe problem involves selecting k projects among given n projects. Each project has an associated profit and capital. We have to start with an initial | dr41n | NORMAL | 2023-02-23T04:51:35.754755+00:00 | 2023-02-23T04:51:35.754802+00:00 | 518 | false | # Intuition\nThe problem involves selecting k projects among given n projects. Each project has an associated profit and capital. We have to start with an initial capital \'w\' and select a project whose capital is less than or equal to our current capital \'w\'. We can only select a project once. We have to maximize o... | 5 | 0 | ['Heap (Priority Queue)', 'Go'] | 3 |
ipo | 🗓️ Daily LeetCoding Challenge February, Day 23 | daily-leetcoding-challenge-february-day-xfip2 | This problem is the Daily LeetCoding Challenge for February, Day 23. Feel free to share anything related to this problem here! You can ask questions, discuss wh | leetcode | OFFICIAL | 2023-02-23T00:00:10.970903+00:00 | 2023-02-23T00:00:10.970944+00:00 | 4,507 | false | This problem is the Daily LeetCoding Challenge for February, Day 23.
Feel free to share anything related to this problem here!
You can ask questions, discuss what you've learned from this problem, or show off how many days of streak you've made!
---
If you'd like to share a detailed solution to the problem, please ... | 5 | 0 | [] | 24 |
ipo | Java + Intuition + 2 Heaps + Greedy + Explanation | java-intuition-2-heaps-greedy-explanatio-a6k8 | Intuition:\n\nThe intuition is to find the max profit for the available capital at any point of time. We add to the total profit, every time we are able to sele | learn4Fun | NORMAL | 2021-02-17T20:13:31.767304+00:00 | 2021-10-27T22:25:09.140498+00:00 | 805 | false | **Intuition:**\n\nThe intuition is to find the **max profit for the available capital** at any point of time. We add to the total profit, every time we are able to select & complete a project with maximum profit. Since our total profit is updated after each project completion, and if we haven\'t completed `k` projects ... | 5 | 0 | ['Greedy', 'Java'] | 1 |
ipo | Sort index by capital. Then use heap. C++. | sort-index-by-capital-then-use-heap-c-by-5gkm | First create a vector of index, sort by capital.\nThen go through this index vector, put profits of all qualified projects into a queue. Pick the highest profit | kwanwoo | NORMAL | 2018-11-21T03:35:22.103525+00:00 | 2018-11-21T03:35:22.103569+00:00 | 880 | false | First create a vector of index, sort by capital.\nThen go through this index vector, put profits of all qualified projects into a queue. Pick the highest profit.\nWe may not be able to finish k projects. Quit if we cannot pay the minimum capital.\n```\n\tint findMaximizedCapital(int k, int W, vector<int>& Profits, vect... | 5 | 0 | [] | 3 |
ipo | Without Heap Optimized solution beats 100% Time and Space Greedy in C++, Java, Python and Javascript | without-heap-optimized-solution-beats-10-flcw | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem is to maximize the total capital after completing at most k projects, given | gunjanbhingaradiya | NORMAL | 2024-06-15T11:18:57.657952+00:00 | 2024-06-15T11:18:57.657978+00:00 | 592 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem is to maximize the total capital after completing at most k projects, given that each project has a specific profit and requires a minimum capital to start. We need to choose projects in such a way that our capital grows as mu... | 4 | 0 | ['Array', 'Greedy', 'Sorting', 'Heap (Priority Queue)', 'Python', 'C++', 'Java', 'JavaScript'] | 1 |
ipo | Java Solution, Beats 100.00% | java-solution-beats-10000-by-mohit-005-die6 | Intuition\n\nThe goal is to maximize the capital by selecting at most k projects from the given list of projects, each characterized by its profit and required | Mohit-005 | NORMAL | 2024-06-15T09:44:01.241785+00:00 | 2024-06-15T09:44:01.241810+00:00 | 417 | false | # Intuition\n\nThe goal is to maximize the capital by selecting at most `k` projects from the given list of projects, each characterized by its profit and required capital. Initially, we have a certain amount of capital `w`. Each project can only be started if we have enough capital. Once a project is finished, its pro... | 4 | 0 | ['Java'] | 0 |
ipo | C# Solution for IPO Problem | c-solution-for-ipo-problem-by-aman_raj_s-g410 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind this solution is to efficiently manage the selection of projects u | Aman_Raj_Sinha | NORMAL | 2024-06-15T08:57:20.156794+00:00 | 2024-06-15T08:57:20.156833+00:00 | 240 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this solution is to efficiently manage the selection of projects using two heaps (priority queues) to always have access to the most profitable project that can be started given the current available capital. By separ... | 4 | 0 | ['C#'] | 0 |
ipo | Best solution | very optimized | with vid explanation to make concept super easy | best-solution-very-optimized-with-vid-ex-yq8v | detailed problem statement explanation + approach + optimization\nwatch in 2x to save your time\n\nhttps://youtu.be/yGket_WqTqA\n\n# Intuition\n Describe your f | Atharav_s | NORMAL | 2024-06-15T02:47:56.287153+00:00 | 2024-06-15T02:47:56.287176+00:00 | 487 | false | detailed problem statement explanation + approach + optimization\nwatch in 2x to save your time\n\nhttps://youtu.be/yGket_WqTqA\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe approach involves sorting the projects based on their capital requirements and using a max-heap (priori... | 4 | 0 | ['C++'] | 2 |
ipo | Daily Challenge is Good | daily-challenge-is-good-by-raviparihar-gn2e | Solution Approach\nCombine and Sort Projects:\n\nCreate a list of projects, where each project is represented by its required capital and profit.\nSort this lis | raviparihar_ | NORMAL | 2024-06-15T01:53:42.163167+00:00 | 2024-06-15T01:53:42.163183+00:00 | 942 | false | Solution Approach\nCombine and Sort Projects:\n\nCreate a list of projects, where each project is represented by its required capital and profit.\nSort this list based on the capital required to start the projects.\nUse a Max-Heap for Profits:\n\nUse a max-heap to keep track of the profits of the projects that can be s... | 4 | 0 | ['Greedy', 'C', 'Heap (Priority Queue)', 'Python', 'Java'] | 2 |
ipo | Typescript/Priority-Queue (Heap) Clean Intuitive Solution | typescriptpriority-queue-heap-clean-intu-3bz8 | Intuition\n Describe your first thoughts on how to solve this problem. \nFor each iteration, we basically need to get a list of project that can be done with th | Adetomiwa | NORMAL | 2023-02-23T15:33:42.568181+00:00 | 2023-02-23T15:35:48.299924+00:00 | 246 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFor each iteration, we basically need to get a list of project that can be done with the current capital at hand.\n\nThen we need to get the most profitable out of that list and add to our current capital at hand.\n\n# Approach\n<!-- Desc... | 4 | 0 | ['Sorting', 'Heap (Priority Queue)', 'TypeScript'] | 0 |
ipo | Easy JAVA solution | O(n log n) in best case | Greedy Approach | easy-java-solution-on-log-n-in-best-case-ovi5 | \n\n\n# Intuition\n Describe your first thoughts on how to solve this problem. \nUsing greedy algorithm to solve the problem. The algorithm can be described as | Yaduttam_Pareek | NORMAL | 2023-02-23T03:02:11.147460+00:00 | 2023-02-23T03:03:13.817796+00:00 | 459 | false | \n\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUsing greedy algorithm to solve the problem. The algorithm can be described as follows:\n\n- Find the maximum capital ... | 4 | 0 | ['Java'] | 1 |
ipo | Sqrt Decomposition | C++ Both 100% | 108ms/72.8MB | explanation | sqrt-decomposition-c-both-100-108ms728mb-6fbq | \nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& p, vector<int>& c) {\n int n=p.size(),ans=0,it=0,nn=0,ma=0,mb=0,sz=0 | SunGod1223 | NORMAL | 2022-08-04T09:28:21.038253+00:00 | 2023-02-23T01:09:18.800315+00:00 | 535 | false | ```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& p, vector<int>& c) {\n int n=p.size(),ans=0,it=0,nn=0,ma=0,mb=0,sz=0,sq[101][101]={};\n for(int i=0;i<n;++i)\n if(c[i]<=w){\n sq[p[i]/100][p[i]%100]++;\n sq[p[i]/100][100]++;\n ... | 4 | 0 | ['C++'] | 1 |
ipo | [JavaScript] 16 lines heap solution with explanation | javascript-16-lines-heap-solution-with-e-qg3b | Runtime: 552 ms, faster than 61.11% of JavaScript online submissions for IPO.\nMemory Usage: 87.7 MB, less than 27.78% of JavaScript online submissions for IPO. | 0533806 | NORMAL | 2021-07-16T02:54:58.483850+00:00 | 2021-07-16T02:57:33.673649+00:00 | 638 | false | Runtime: 552 ms, faster than 61.11% of JavaScript online submissions for IPO.\nMemory Usage: 87.7 MB, less than 27.78% of JavaScript online submissions for IPO.\n\nidea: two heaps, respectively max(priority: profits) and min(priority: capital)\nstep1. put every group in maxheap\nstep2. dequeue item from maxheap, if its... | 4 | 0 | ['JavaScript'] | 1 |
ipo | Detailed Explanation: Java Single Heap Solution. 86% time, 100% memory | detailed-explanation-java-single-heap-so-yhj3 | The approach in this problem is very similar to the LeetCode 857 problem, but easier.\nGiven:\n w -> initial working capital\n k -> maximum number of projects t | snc120 | NORMAL | 2020-05-24T10:28:00.321236+00:00 | 2020-05-24T10:33:21.993112+00:00 | 337 | false | The approach in this problem is very similar to the LeetCode 857 problem, but easier.\nGiven:\n* w -> initial working capital\n* k -> maximum number of projects that may be done to earn profit\n* profits -> Array containing "pure" profits of the projects. (Note: this array contains pure profit and not revenue. So the v... | 4 | 0 | ['Heap (Priority Queue)', 'Java'] | 1 |
ipo | Simple || Easy to Understand | simple-easy-to-understand-by-kdhakal-ysv5 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | kdhakal | NORMAL | 2025-04-08T16:47:53.971836+00:00 | 2025-04-08T16:47:53.971836+00:00 | 36 | 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
`... | 3 | 0 | ['Java'] | 0 |
ipo | beginner-friendly solution ✅|| simple Priority Queue technique 📘 | beginner-friendly-solution-simple-priori-17qh | Intuition\n Describe your first thoughts on how to solve this problem. \nTo maximize profit at any given capital w, we need to select a task that offers the hig | yousufmunna143 | NORMAL | 2024-06-15T13:06:06.471883+00:00 | 2024-06-15T13:06:06.471908+00:00 | 135 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo maximize profit at any given capital `w`, we need to select a task that offers the highest profit and has a capital requirement less than or equal to `w`. We can use a **max heap (priority queue)** to keep track of the profits of tasks... | 3 | 0 | ['Sorting', 'Heap (Priority Queue)', 'Java'] | 0 |
ipo | LC Hard made Easy | Well Explained⭐💯 | lc-hard-made-easy-well-explained-by-_ris-xtca | Problem Description\nYou are given k projects, each with a capital requirement and a profit. You have an initial capital w. Your goal is to find the maximum cap | _Rishabh_96 | NORMAL | 2024-06-15T10:36:34.670185+00:00 | 2024-06-15T10:36:34.670215+00:00 | 78 | false | ## Problem Description\nYou are given `k` projects, each with a capital requirement and a profit. You have an initial capital `w`. Your goal is to find the maximum capital you can achieve after completing at most `k` projects. You can only start a project if you have the required capital.\n\n## Detailed Approach\n\n###... | 3 | 0 | ['Array', 'Greedy', 'Sorting', 'Heap (Priority Queue)', 'C++'] | 0 |
ipo | Mere jaise dimag wale logo ke liye easy hindi solution!! | mere-jaise-dimag-wale-logo-ke-liye-easy-v2iai | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nInput Processing:\nPehl | bakree | NORMAL | 2024-06-15T08:20:13.686723+00:00 | 2024-06-15T08:20:13.686742+00:00 | 107 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nInput Processing:\nPehle, profits aur capital ke basis pe ek 2D array arr banate hain jahan har sub-array mein capital aur profit ka pair hota hai.\n\nSorting:\nUske b... | 3 | 0 | ['Heap (Priority Queue)', 'Java'] | 0 |
ipo | Easy python with explanation | easy-python-with-explanation-by-ekambare-ervi | Intuition
first sort based on capital (tagging profit) , as we need to choose min capital
at every iteration we need to choose max profit for the respective 'w' | ekambareswar1729 | NORMAL | 2024-06-15T04:22:12.653292+00:00 | 2025-03-21T10:43:56.346980+00:00 | 80 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
- first sort based on capital (tagging profit) , as we need to choose min capital
- at every iteration we need to choose max profit for the respective 'w'
- so use max heap to obtain max profit
# Complexity
- Time complexity:O(n*log(n))... | 3 | 0 | ['Python3'] | 0 |
ipo | Easy Java Solution | Priority Queue | Greedy | Beats 100💯✅ | easy-java-solution-priority-queue-greedy-9n3u | Find Maximized Capital\n\n## Problem Statement\n\nThe function findMaximizedCapital is designed to maximize the capital by selecting up to k projects from a lis | shobhitkushwaha1406 | NORMAL | 2024-06-15T04:00:41.903163+00:00 | 2024-06-15T04:00:41.903190+00:00 | 1,063 | false | # Find Maximized Capital\n\n## Problem Statement\n\nThe function `findMaximizedCapital` is designed to maximize the capital by selecting up to `k` projects from a list of available projects, where each project has an associated profit and capital requirement. Given initial capital `w`, the function returns the maximum ... | 3 | 0 | ['Array', 'Math', 'Greedy', 'Heap (Priority Queue)', 'Java'] | 2 |
ipo | Priority Queue || Optimized || Beats 100% || Explained line by line || C++, Python, Java | priority-queue-optimized-beats-100-expla-udcf | Intuition\n Describe your first thoughts on how to solve this problem. \n * The intuition behind this code is to maximize the available capital after selecting | avinash_singh_13 | NORMAL | 2024-06-15T03:39:45.688437+00:00 | 2024-06-15T03:39:45.688466+00:00 | 20 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n * The intuition behind this code is to maximize the available capital after selecting up to k projects, by strategically choosing the projects with the highest profit that can be started within the current capital constraints. It does th... | 3 | 0 | ['C++', 'Java', 'Python3'] | 0 |
ipo | Easiest ✅|| Beats 100% 🔥🔥|| Java✅ || 0ms | easiest-beats-100-java-0ms-by-wankhedeay-fe77 | Intuition\nThe intuition behind this code is to maximize the available capital after selecting up to k projects, by strategically choosing the projects with the | wankhedeayush90 | NORMAL | 2024-06-15T03:08:36.924859+00:00 | 2024-06-15T03:08:36.924888+00:00 | 304 | false | # Intuition\nThe intuition behind this code is to maximize the available capital after selecting up to k projects, by strategically choosing the projects with the highest profit that can be started within the current capital constraints. It does this by sorting projects by their capital requirements to quickly find the... | 3 | 1 | ['Array', 'Sorting', 'Heap (Priority Queue)', 'C++', 'Java'] | 1 |
ipo | Python3 Solution | python3-solution-by-motaharozzaman1996-f49g | \n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n max_profit=[]\n min_capi | Motaharozzaman1996 | NORMAL | 2024-06-15T02:18:47.829794+00:00 | 2024-06-15T02:18:47.829837+00:00 | 826 | false | \n```\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n max_profit=[]\n min_capital=[(c,p) for c,p in zip(capital,profits)]\n heapq.heapify(min_capital) \n for i in range(k):\n while min_capital and min_capital[0... | 3 | 0 | ['Python', 'Python3'] | 1 |
ipo | Maximizing Capital for LeetCode's IPO Through Optimal Project Selection | maximizing-capital-for-leetcodes-ipo-thr-p6q1 | To solve the problem of maximizing LeetCode\'s capital before its IPO by choosing at most \( k \) distinct projects, we can employ a greedy approach. This metho | sleekmind | NORMAL | 2024-06-15T01:43:55.467179+00:00 | 2024-06-15T01:49:17.172219+00:00 | 477 | false | To solve the problem of maximizing LeetCode\'s capital before its IPO by choosing at most \\( k \\) distinct projects, we can employ a greedy approach. This method ensures that at each step, we choose the project that offers the maximum profit while meeting the current capital requirement.\n\n### Approach Breakdown\n\n... | 3 | 0 | ['Greedy', 'Heap (Priority Queue)', 'C++'] | 4 |
ipo | Beats 97% users... Trust this code guys | beats-97-users-trust-this-code-guys-by-a-anud | \n\n# Code\n\nclass T {\n public int pro;\n public int cap;\n public T(int pro, int cap) {\n this.pro = pro;\n this.cap = cap;\n }\n}\n\nclass Solutio | Aim_High_212 | NORMAL | 2024-06-15T01:42:17.240273+00:00 | 2024-06-15T01:42:17.240292+00:00 | 39 | false | \n\n# Code\n```\nclass T {\n public int pro;\n public int cap;\n public T(int pro, int cap) {\n this.pro = pro;\n this.cap = cap;\n }\n}\n\nclass Solution {\n public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) {\n Queue<T> minHeap = new PriorityQueue<>((a, b) -> a.cap - b.cap);\n ... | 3 | 0 | ['Python', 'C++', 'Java', 'Python3'] | 0 |
ipo | 🔥 🔥 🔥 Simple Approch 🔥 🔥 🔥 | simple-approch-by-vivek_kumr-k6c9 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal is to maximize the available capital w after k investment rounds. Each project | Vivek_kumr | NORMAL | 2024-06-15T01:27:00.137056+00:00 | 2024-06-15T01:27:00.137085+00:00 | 435 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to maximize the available capital w after k investment rounds. Each project has a specific profit and capital requirement. We need to strategically choose projects that can be funded with the current capital and yield the high... | 3 | 0 | ['C++'] | 1 |
ipo | Two- Heap Pattern ✅✔️💯 | two-heap-pattern-by-dixon_n-d9n8 | This is a new Pattern (Two- Heap Pattern.Min heap and Max Heap Pattern)\n\n215. Kth Largest Element in an Array same pattern\n451. Sort Characters By Frequency\ | Dixon_N | NORMAL | 2024-05-30T22:25:20.512438+00:00 | 2024-05-30T22:25:20.512474+00:00 | 144 | false | This is a new Pattern (Two- Heap Pattern.Min heap and Max Heap Pattern)\n\n[215. Kth Largest Element in an Array](https://leetcode.com/problems/kth-largest-element-in-an-array/solutions/5195848/k-pattern-heaps-peiority/) same pattern\n451. Sort Characters By Frequency\n[703. Kth Largest Element in a Stream](https://lee... | 3 | 0 | ['Greedy', 'Heap (Priority Queue)', 'Java'] | 4 |
ipo | Simple || Concise || Beginner Friendly || Greedy ✅✅ | simple-concise-beginner-friendly-greedy-jzvwn | Complexity\n- Time complexity: O(n*log n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(n)\n Add your space complexity here, e.g. O(n) \n | Lil_ToeTurtle | NORMAL | 2023-09-09T07:53:52.005369+00:00 | 2023-09-09T07:53:52.005388+00:00 | 165 | false | # Complexity\n- Time complexity: $$O(n*log n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n```\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int n=capita... | 3 | 0 | ['Greedy', 'Sorting', 'Heap (Priority Queue)', 'Java'] | 1 |
ipo | 502: Solution with step by step explanation | 502-solution-with-step-by-step-explanati-1fds | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n1. Create a list of tuples containing (capital, profit) for each project. | Marlen09 | NORMAL | 2023-03-12T04:21:24.910303+00:00 | 2023-03-12T04:21:24.910332+00:00 | 886 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Create a list of tuples containing (capital, profit) for each project. We can use a list comprehension to achieve this.\n\n2. Sort the list of projects by increasing capital required to start them. We can use the built-in... | 3 | 0 | ['Array', 'Greedy', 'Sorting', 'Python', 'Python3'] | 1 |
ipo | Sort + Priority Queue | C++ | sort-priority-queue-c-by-tusharbhart-kp2s | \nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n int n = profits.size(), p = 0;\n | TusharBhart | NORMAL | 2023-02-25T10:34:09.931508+00:00 | 2023-02-25T10:34:09.931553+00:00 | 274 | false | ```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n int n = profits.size(), p = 0;\n vector<pair<int, int>> v;\n for(int i=0; i<n; i++) v.push_back({capital[i], profits[i]});\n\n sort(v.begin(), v.end());\n priorit... | 3 | 0 | ['Greedy', 'Sorting', 'Heap (Priority Queue)', 'C++'] | 1 |
ipo | Java | Priority Queue | 10 lines | O(n log n) time | java-priority-queue-10-lines-on-log-n-ti-4gi8 | 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 | judgementdey | NORMAL | 2023-02-23T21:32:42.855804+00:00 | 2023-02-23T21:40:40.370298+00:00 | 153 | 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: $$O(n*log(n))$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexi... | 3 | 0 | ['Heap (Priority Queue)', 'Java'] | 0 |
ipo | 📌📌Python3 || ⚡783 ms, faster than 98.25% of Python3 | python3-783-ms-faster-than-9825-of-pytho-muzq | \n\ndef findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n if w >= max(capital):\n return w + sum(nl | harshithdshetty | NORMAL | 2023-02-23T12:57:08.979107+00:00 | 2023-02-23T12:57:08.979148+00:00 | 418 | false | \n```\ndef findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n if w >= max(capital):\n return w + sum(nlargest(k, profits)) \n projects =... | 3 | 0 | ['Heap (Priority Queue)', 'Python', 'Python3'] | 1 |
ipo | require two heaps, max and min | require-two-heaps-max-and-min-by-mr_star-rkw8 | \nclass Solution {\npublic:\n struct compare {\n bool operator()(pair<int ,int> p, pair<int ,int> q) {\n return p.first>q.first;\n } | mr_stark | NORMAL | 2023-02-23T12:20:42.195238+00:00 | 2023-02-23T12:20:42.195282+00:00 | 251 | false | ```\nclass Solution {\npublic:\n struct compare {\n bool operator()(pair<int ,int> p, pair<int ,int> q) {\n return p.first>q.first;\n }\n };\n int findMaximizedCapital(int k, int w, vector<int>& p, vector<int>& c) {\n priority_queue<pair<int,int>, vector<pair<int,int>>, compare>... | 3 | 0 | ['C'] | 1 |
ipo | Solution in C++ | solution-in-c-by-ashish_madhup-0zdc | 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 | ashish_madhup | NORMAL | 2023-02-23T11:47:57.523539+00:00 | 2023-02-23T11:47:57.523566+00:00 | 1,001 | 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. $$O(n)$$ --... | 3 | 0 | ['C++'] | 1 |
ipo | ✅ beats 99% Java code | beats-99-java-code-by-abstractconnoisseu-y95k | Java Code\n\nclass Solution {\n public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) {\n int maxCapital = 0;\n for (int | abstractConnoisseurs | NORMAL | 2023-02-23T09:27:54.936411+00:00 | 2023-02-23T09:27:54.936452+00:00 | 451 | false | # Java Code\n```\nclass Solution {\n public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) {\n int maxCapital = 0;\n for (int i = 0; i < Capital.length; i++) {\n maxCapital = Math.max(Capital[i], maxCapital);\n }\n\n if (W >= maxCapital) {\n Pri... | 3 | 0 | ['Array', 'Greedy', 'Sorting', 'Heap (Priority Queue)', 'Java'] | 1 |
ipo | C++ Solution | | Greedy approach | c-solution-greedy-approach-by-vaibhav_sa-4jr8 | 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 | Vaibhav_saroj | NORMAL | 2023-02-23T08:22:03.811395+00:00 | 2023-02-23T08:22:03.811421+00:00 | 676 | 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. $$O(n)$$ --... | 3 | 0 | ['C++'] | 0 |
ipo | ✅ Java | Brute Force Approach | java-brute-force-approach-by-prashantkac-uwx7 | \n// Approach 1: Brute Force Approach - TLE\n\n// Time complexity: O(n^2)\n// Space complexity: O(n)\n\npublic int findMaximizedCapital(int k, int w, int[] prof | prashantkachare | NORMAL | 2023-02-23T05:33:47.750117+00:00 | 2023-02-23T05:33:47.750156+00:00 | 216 | false | ```\n// Approach 1: Brute Force Approach - TLE\n\n// Time complexity: O(n^2)\n// Space complexity: O(n)\n\npublic int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n\tint n = profits.length;\n\tboolean[] finished = new boolean[n];\n\n\tfor (int i = 0; i < k; i++) {\n\t\tint prjIndex = -1;\n\n\t\tfo... | 3 | 0 | ['Greedy', 'Java'] | 2 |
ipo | [Python] greey with min heap queue, Explained | python-greey-with-min-heap-queue-explain-n6ky | Step 1, sort the profits based on the captial we need to gain the profit\nStep 2, based on the capital we have, add the profit into a min heap (add the negative | wangw1025 | NORMAL | 2023-02-23T04:20:27.101161+00:00 | 2023-02-23T04:20:27.101219+00:00 | 355 | false | Step 1, sort the profits based on the captial we need to gain the profit\nStep 2, based on the capital we have, add the profit into a min heap (add the negative profit)\nStep 3, every loop, we pick the head from the heap, which is the largest profit we can gain\nStep 4, check the project list and add more projects base... | 3 | 0 | ['Heap (Priority Queue)', 'Python3'] | 1 |
ipo | JS priority queue +explanation | js-priority-queue-explanation-by-crankyi-we9o | Performance\nRuntime 447ms Beats 66.27% Memory 101.1MB Beats 10.84%\n\n### Intuition\nWe need to find the k projects which yield the greatest overall profit. S | crankyinmv | NORMAL | 2023-02-23T01:25:42.620481+00:00 | 2023-02-23T01:25:42.620518+00:00 | 387 | false | ### Performance\nRuntime 447ms Beats 66.27% Memory 101.1MB Beats 10.84%\n\n### Intuition\nWe need to find the `k` projects which yield the greatest overall profit. Since *the operating capital never goes down* we don\'t have a house robbber situation where choosing one project has an opportunity cost which would keep ... | 3 | 0 | ['JavaScript'] | 0 |
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