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minimum-time-to-make-array-sum-at-most-x | Easy Iterative Solution | easy-iterative-solution-by-kvivekcodes-p6pc | 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 | kvivekcodes | NORMAL | 2024-09-04T12:20:28.107236+00:00 | 2024-09-04T12:20:28.107268+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:\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 | ['Array', 'Dynamic Programming', 'Sorting', 'C++'] | 0 |
minimum-time-to-make-array-sum-at-most-x | 20 Lines Simple Bottom Up DP | 20-lines-simple-bottom-up-dp-by-alex-e-x08d | Solution\nFor an optimal solution, if we have indices i1, i2... as a part of the optimal solution where we set the index to 0, then when doing the operation in | alex-e | NORMAL | 2024-08-04T21:35:54.716826+00:00 | 2024-08-04T21:37:22.251401+00:00 | 27 | false | # Solution\nFor an optimal solution, if we have indices i1, i2... as a part of the optimal solution where we set the index to 0, then when doing the operation in the problem, we should go in order of nums2[i1]<=nums2[i2]... because we want to pick the greatest value last and let the value accumulate before we set it to... | 0 | 0 | ['Dynamic Programming', 'Python3'] | 0 |
minimum-time-to-make-array-sum-at-most-x | As hinted | as-hinted-by-maxorgus-61ep | Note the dp should not be done with python builtin memoization which would cause MLE, in stead, do the tabulation yourself\n\n# Code\n\nclass Solution:\n def | MaxOrgus | NORMAL | 2024-05-14T02:52:38.130775+00:00 | 2024-05-14T02:52:38.130807+00:00 | 5 | false | Note the dp should not be done with python builtin memoization which would cause MLE, in stead, do the tabulation yourself\n\n# Code\n```\nclass Solution:\n def minimumTime(self, nums1: List[int], nums2: List[int], x: int) -> int:\n n = len(nums1)\n nums = [(n1,n2) for n1,n2 in zip(nums1,nums2)]\n ... | 0 | 0 | ['Python3'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Space Efficient Solution ... | space-efficient-solution-by-gp_777-pxto | 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 | GP_777 | NORMAL | 2024-04-23T03:55:41.499004+00:00 | 2024-04-23T03:55:41.499038+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 | ['Java'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Solution | solution-by-gp_777-d3gn | 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 | GP_777 | NORMAL | 2024-04-23T02:19:46.084231+00:00 | 2024-04-23T02:19:46.084251+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 |
minimum-time-to-make-array-sum-at-most-x | Minimum Time to Process Arrays with Constraints Beats 81.20% | minimum-time-to-process-arrays-with-cons-949c | Intuition\nThe problem seems to involve finding the minimum time required to perform certain operations on two arrays while satisfying certain conditions. \n\n# | kumarritul089 | NORMAL | 2024-03-15T07:09:40.496951+00:00 | 2024-03-15T07:09:40.496980+00:00 | 29 | false | # Intuition\nThe problem seems to involve finding the minimum time required to perform certain operations on two arrays while satisfying certain conditions. \n\n# Approach\n1. We start by calculating the total sums of both arrays, `nums1` and `nums2`, and storing the indices in a separate vector.\n2. If the sum of `num... | 0 | 0 | ['Array', 'Dynamic Programming', 'Sorting', 'Python', 'C++', 'Python3', 'JavaScript'] | 0 |
minimum-time-to-make-array-sum-at-most-x | 5 hours, mega failure | 5-hours-mega-failure-by-user3043sb-ex8v | 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 | user3043SB | NORMAL | 2024-01-21T14:06:03.398487+00:00 | 2024-01-21T14:06:03.398517+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. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Alternative to DP, a Short, Straightforward O(N^2) Greedy Solution in Python | alternative-to-dp-a-short-straightforwar-4jv3 | Approach\n Describe your approach to solving the problem. \nThe best solution with a length of l is based on the best solution with the length of l-1 and an ite | metaphysicalist | NORMAL | 2023-12-26T19:27:38.514813+00:00 | 2023-12-26T21:57:29.250663+00:00 | 17 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nThe best solution with a length of `l` is based on the best solution with the length of `l-1` and an item that is not selected yet. \nThus, my solution builds the solution from length 1 to `n`, one by one. \nBased on the solution of `l-1`, my solution... | 0 | 0 | ['Greedy', 'Python3'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Most intuitive & Easyy Top Down solution || recursion+memoizaton | most-intuitive-easyy-top-down-solution-r-i3ov | Idea & thoughts...\n\nwhat will happen after t days if we don\'t delete anything the \ns1: sum of a1\ns2: sum of a2\n\nsum will be = s1 + t*s2\nso now if we wan | demon_code | NORMAL | 2023-10-12T11:11:45.779308+00:00 | 2023-10-12T11:17:12.633698+00:00 | 26 | false | Idea & thoughts...\n\nwhat will happen after t days if we don\'t delete anything the \ns1: sum of a1\ns2: sum of a2\n\nsum will be = ```s1 + t*s2```\nso now if we want to delete some index i element at time p then \nsum will be = ```s1+t*s2 - a1[i]- p*a2[i] ```\nSo what we see here is the ```s1+ t*s2 ```remains constan... | 0 | 0 | ['Recursion', 'Memoization', 'C'] | 0 |
minimum-time-to-make-array-sum-at-most-x | O(n ^ 2) Propagation DP | on-2-propagation-dp-by-__beginner-ky8w | Intuition\n Describe your first thoughts on how to solve this problem. \nThe key observation about dp approach is that if we\'re selecting k positions to be ma | __Beginner_ | NORMAL | 2023-10-08T22:20:32.677959+00:00 | 2023-10-14T09:43:58.996026+00:00 | 21 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe key observation about dp approach is that if we\'re selecting $$ k$$ positions to be made $$0$$ at successive intervals then we should select the position with in increasing order of $$nums2[i]$$.\nLet\'s prove this below,\n# Approach... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
minimum-time-to-make-array-sum-at-most-x | java script - Minimum Time to Make Array Sum At Most x | java-script-minimum-time-to-make-array-s-10d2 | 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 | balakrishnanp | NORMAL | 2023-10-06T11:26:02.452228+00:00 | 2023-10-06T11:26:02.452246+00:00 | 6 | 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 | ['JavaScript'] | 0 |
minimum-time-to-make-array-sum-at-most-x | C++ DP solution | Intuitive | c-dp-solution-intuitive-by-aglakshya02-fivv | Approach\n Describe your approach to solving the problem. \nYou can watch Coding Mohan\'s YouTube video solution for this, it\'s really difficult to describe th | aglakshya02 | NORMAL | 2023-09-08T11:38:59.501321+00:00 | 2023-09-08T11:38:59.501344+00:00 | 16 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nYou can watch Coding Mohan\'s YouTube video solution for this, it\'s really difficult to describe the solution in a textual post for me.\nMy solution is inspired from his only.\n\n\n# Code\n```\nclass Solution {\npublic:\n\n vector<int>a,b;\n\n ... | 0 | 0 | ['C++'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Unraveling the Optimal Time Scheduler | unraveling-the-optimal-time-scheduler-by-tznv | Intuition\nImagine you have two lists of tasks with respective times required. You have a goal to optimize the sum of these tasks such that, at every step, you | janis__ | NORMAL | 2023-08-29T15:47:42.600716+00:00 | 2023-08-29T15:47:42.600733+00:00 | 11 | false | # Intuition\nImagine you have two lists of tasks with respective times required. You have a goal to optimize the sum of these tasks such that, at every step, you can either use the time from one list or swap it with a time from the other list. The problem might sound tricky, but a methodical approach can reveal pattern... | 0 | 0 | ['Array', 'Dynamic Programming', 'Sorting', 'Python'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Somehow worked, help me optimise the code! | somehow-worked-help-me-optimise-the-code-5e8x | 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 | Chhota_bheem | NORMAL | 2023-08-23T15:30:57.567267+00:00 | 2023-08-23T15:30:57.567297+00:00 | 10 | 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 |
minimum-time-to-make-array-sum-at-most-x | Get the correct DP state before thinking of solution ! | get-the-correct-dp-state-before-thinking-1uk8 | Intuition\nYou need to think of correct dp state in this question.\nIf we calculate to what extent you can reduce the sum of nums1 in each operation we can get | pshreyansh26 | NORMAL | 2023-08-21T17:40:07.426443+00:00 | 2023-08-21T17:40:33.070053+00:00 | 22 | false | # Intuition\nYou need to think of correct dp state in this question.\nIf we calculate to what extent you can reduce the sum of nums1 in each operation we can get the answer.\n\n# Approach\n* Let dp[i][j] represent the maximum total value that can be reduced if we do j operations on the first i elements\n* It can also b... | 0 | 0 | ['Dynamic Programming', 'Sorting', 'C++'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Python | DP | Concise Solution | O(n^2) | python-dp-concise-solution-on2-by-aryonb-ktre | Code\n\nclass Solution:\n def minimumTime(self, nums1: List[int], nums2: List[int], x: int) -> int:\n sum1 = sum(nums1)\n sum2 = sum(nums2)\n | aryonbe | NORMAL | 2023-08-13T11:12:14.245439+00:00 | 2023-08-13T11:12:14.245461+00:00 | 21 | false | # Code\n```\nclass Solution:\n def minimumTime(self, nums1: List[int], nums2: List[int], x: int) -> int:\n sum1 = sum(nums1)\n sum2 = sum(nums2)\n nums = sorted(list(zip(nums2, nums1)))\n n = len(nums)\n dp = [0]*(n+1)\n for i in range(n):\n for k in range(i+1, 0,... | 0 | 0 | ['Python3'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Python | DP | O(n^2) | python-dp-on2-by-aryonbe-i2ri | Code\n\nclass Solution:\n def minimumTime(self, nums1: List[int], nums2: List[int], x: int) -> int:\n sum1 = sum(nums1)\n if sum1 <= x: return | aryonbe | NORMAL | 2023-08-13T09:14:24.727368+00:00 | 2023-08-13T09:14:56.548495+00:00 | 10 | false | # Code\n```\nclass Solution:\n def minimumTime(self, nums1: List[int], nums2: List[int], x: int) -> int:\n sum1 = sum(nums1)\n if sum1 <= x: return 0\n sum2 = sum(nums2)\n nums = sorted(list(zip(nums2, nums1)))\n n = len(nums)\n dp = [0]*n\n for k in range(1,n+1):\n ... | 0 | 0 | ['Python3'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Easy to understand Recursive DP |Memorization✅ | easy-to-understand-recursive-dp-memoriza-8kvp | \n# Complexity\n- Time complexity:\n- O(N^2)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:\n- O(N^2)\n Add your space complexity here, e.g. | rahulhind | NORMAL | 2023-08-12T10:55:23.535201+00:00 | 2023-08-12T10:56:07.419527+00:00 | 41 | false | \n# Complexity\n- Time complexity:\n- O(N^2)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- O(N^2)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n\nvector<vector<long long> > dp;\nvector<pair<long long,long long> > arr;\nlong long f(int i,int k){\nif(i>=arr.siz... | 0 | 0 | ['C++'] | 1 |
minimum-time-to-make-array-sum-at-most-x | Solution in Swift, DP | solution-in-swift-dp-by-sergeyleschev-38as | Approach\nWe begin by calculating the total sum of the arrays A and B as sa and sb respectively.\n\nIf no actions are taken, at i seconds, we would have a total | sergeyleschev | NORMAL | 2023-08-12T06:16:50.832911+00:00 | 2023-08-12T06:16:50.832936+00:00 | 2 | false | # Approach\nWe begin by calculating the total sum of the arrays A and B as sa and sb respectively.\n\nIf no actions are taken, at i seconds, we would have a total of sb * i + sa.\n\nDuring the t seconds, we select t elements. When we consider these selected elements, A[i] would be removed. The sum for these selected el... | 0 | 0 | ['Swift'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Dp solution - thinking process and detailed proof | dp-solution-thinking-process-and-detaile-26kp | Intuition\n Describe your first thoughts on how to solve this problem. \nFirst thing that came to my mind afer reading the problem statement was using binary se | horizon1006 | NORMAL | 2023-08-11T08:06:25.405927+00:00 | 2023-08-11T08:10:16.699230+00:00 | 30 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirst thing that came to my mind afer reading the problem statement was using binary search but found a counter example shortly thereafter. Take a glance at the constraints, it is not hard to deduce that a $$O(n^2)$$ approach should be us... | 0 | 0 | ['Python3'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Easy dp solution in c++. | easy-dp-solution-in-c-by-isheoran-oodk | Intuition\nDP\n\n\n# Complexity\n- Time complexity:\n O(n^2)\n\n- Space complexity:\n O(n) \n\n# Code\n\nclass Solution {\npublic:\n int minimumTime(vector<i | isheoran | NORMAL | 2023-08-10T09:59:35.976926+00:00 | 2023-08-10T09:59:35.976952+00:00 | 20 | false | # Intuition\nDP\n\n\n# Complexity\n- Time complexity:\n $$O(n^2)$$\n\n- Space complexity:\n $$O(n)$$ \n\n# Code\n```\nclass Solution {\npublic:\n int minimumTime(vector<int>& nums1, vector<int>& nums2, int x) {\n int n = nums1.size();\n\n vector<pair<int,int>>pr;\n for(int i=0;i<n;i++) pr.push_b... | 0 | 0 | ['C++'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Java | java-by-tetttet-p2di | 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 | tetttet | NORMAL | 2023-08-09T13:41:48.292293+00:00 | 2023-08-09T13:41:48.292324+00:00 | 18 | 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 |
minimum-time-to-make-array-sum-at-most-x | Rust | not DP | rust-not-dp-by-tifv-yn9x | Let $n$ be the length of the given arrays, and let $a_{i}$ and $b_{i}$ represent the values of the arrays nums1 and nums2, respectively, $i = 1, \ldots, n$.\n\n | tifv | NORMAL | 2023-08-09T08:07:53.046607+00:00 | 2023-08-09T08:07:53.046629+00:00 | 16 | false | Let $n$ be the length of the given arrays, and let $a_{i}$ and $b_{i}$ represent the values of the arrays `nums1` and `nums2`, respectively, $i = 1, \\ldots, n$.\n\nThere is no point in choosing the same index twice, or delaying operations in time. Therefore, if we are to minimize the obtained sum in time $T$, we will ... | 0 | 0 | ['Rust'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Binary Search + Dp solution | binary-search-dp-solution-by-aryanfit007-mwwg | Approach\nBinary Search + Dp\n\n# Complexity\n- Time complexity: nLogn\n\n# Code\n\nclass Solution {\npublic:\n int dp[1004][1004];\n int recur(int i,int | aryanfit007 | NORMAL | 2023-08-08T20:21:17.269819+00:00 | 2023-08-08T20:21:17.269848+00:00 | 81 | false | # Approach\nBinary Search + Dp\n\n# Complexity\n- Time complexity: nLogn\n\n# Code\n```\nclass Solution {\npublic:\n int dp[1004][1004];\n int recur(int i,int k, vector<int> &v,vector<pair<int,int>> &p,int op){\n if(i==v.size() && k==0) return 0;\n if(i==v.size() && k!=0) return 100000;\n if(... | 0 | 1 | ['Binary Search', 'Dynamic Programming', 'Greedy', 'C++'] | 2 |
minimum-time-to-make-array-sum-at-most-x | Recursive DP solution with intuition | recursive-dp-solution-with-intuition-by-bf58g | Intuition\n Describe your first thoughts on how to solve this problem. \nThe actual complication in the problem is in deciding which array should we sort.\nTo h | kartikeysemwal | NORMAL | 2023-08-08T16:48:55.950351+00:00 | 2023-08-08T16:48:55.950383+00:00 | 82 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe actual complication in the problem is in deciding which array should we sort.\nTo help this simplify, understand that elements in nums1 contribute only once to every index while nums2 contribute at every second.\neg. nums1: [500, 100]... | 0 | 0 | ['Dynamic Programming', 'Recursion', 'Java'] | 0 |
minimum-time-to-make-array-sum-at-most-x | 2d DP | O(n^2) Time, O(n) space. | 2d-dp-on2-time-on-space-by-xun6000-ozxn | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires us to find the minimum time needed to make the sum of the elements | xun6000 | NORMAL | 2023-08-08T00:32:27.655737+00:00 | 2023-08-08T00:32:27.655762+00:00 | 30 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires us to find the minimum time needed to make the sum of the elements in **nums1** less than or equal to **x**. Every second, we can increment the elements of **nums1** by corresponding values in **nums2**, and we can al... | 0 | 0 | ['Python3'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Ruby memoization & bsearch solution, explained (100%/100%) | ruby-memoization-bsearch-solution-explai-255x | Intuition\nIf you know how many numbers you\'re going to reset, you can check the total from resetting them in optimal order in O(n^2) time with memoization. U | dtkalla | NORMAL | 2023-08-07T22:53:03.058855+00:00 | 2023-08-07T22:53:03.058876+00:00 | 24 | false | # Intuition\nIf you know how many numbers you\'re going to reset, you can check the total from resetting them in optimal order in O(n^2) time with memoization. Use a binary search to figure out how many numbers to reset.\n\n# Approach\n1. Special cases: if the sum is already low enough return 0; if all numbers must be... | 0 | 0 | ['Ruby'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Easy C++ Solution || Memoization|| O(N^2)🔥🔥 | easy-c-solution-memoization-on2-by-himan-d1fe | Intuition\nBasic intution is that we will iterate a loop over number of operations and at any instant if it satisfies condition of sum of both arrays less than | Himanshu2003 | NORMAL | 2023-08-07T13:22:26.217602+00:00 | 2023-08-07T13:22:26.217630+00:00 | 42 | false | # Intuition\nBasic intution is that we will iterate a loop over number of operations and at any instant if it satisfies condition of sum of both arrays less than or equal to x the return operation else return -1 \n\n# Approach\nThe code defines a class named Solution. It uses two arrays named n1 and n2 to hold the numb... | 0 | 0 | ['C++'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Recursion to Tabulation | Java | recursion-to-tabulation-java-by-kumarnis-8h8e | Intuition\nAfter learnings all the important conclusions that :\n- At max we will do single operation on any index\n- for any two numbers a and b from nums2, if | kumarnis89 | NORMAL | 2023-08-07T11:14:19.772848+00:00 | 2023-08-07T11:14:19.772875+00:00 | 35 | false | # Intuition\nAfter learnings all the important conclusions that :\n- At max we will do single operation on any index\n- for any two numbers a and b from nums2, if b>a then b will add more value in every second to the total sum.\nso we will opearate on index i with larger nums2[i] after we operate on all other indices w... | 0 | 0 | ['Java'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Chinese DP O(n^2) | chinese-dp-on2-by-wuzhenhai-mirs | n\u4E3A\u6570\u7EC4\u957F\u5EA6.\n\n \u89C2\u5BDF\n 1: \u540C\u4E00\u4E2A\u4F4D\u7F6E\u4E0D\u5FC5\u9009\u4E2D\u4E24\u6B21, \u53EA\u9700\u4FDD\u7559\u6 | wuzhenhai | NORMAL | 2023-08-07T03:04:35.515458+00:00 | 2023-08-07T03:09:59.088908+00:00 | 32 | false | n\u4E3A\u6570\u7EC4\u957F\u5EA6.\n\n \u89C2\u5BDF\n 1: \u540C\u4E00\u4E2A\u4F4D\u7F6E\u4E0D\u5FC5\u9009\u4E2D\u4E24\u6B21, \u53EA\u9700\u4FDD\u7559\u6700\u540E\u4E00\u6B21\u9009\u4E2D\u5373\u53EF.\n 2: \u5728\u9009\u4E2D\u7684\u6570\u7EC4\u7D22\u5F15\u96C6\u5408\u4E2D, nums2[i]\u8D8A\u5927\uFF0C"i\u900... | 0 | 0 | ['Dynamic Programming'] | 0 |
minimum-time-to-make-array-sum-at-most-x | My Solution | my-solution-by-hope_ma-72xo | \n/**\n * let `n` be the length of the vector `nums1`,\n * which is the length of the vector `nums2` as well.\n *\n * some observations\n * 1. if `accumulate(nu | hope_ma | NORMAL | 2023-08-05T21:04:46.186654+00:00 | 2023-08-05T21:07:06.328804+00:00 | 21 | false | ```\n/**\n * let `n` be the length of the vector `nums1`,\n * which is the length of the vector `nums2` as well.\n *\n * some observations\n * 1. if `accumulate(nums1.begin(), nums1.end(), 0)` is less than or equal to `x`,\n * no operations are needed, so `0` should be returned.\n * 2. for any specific index `i`, 0 ... | 0 | 0 | [] | 0 |
minimum-time-to-make-array-sum-at-most-x | Sort + greedy + DP problem, beat 100% C++ | sort-greedy-dp-problem-beat-100-c-by-lon-i9qi | Intuition\n Describe your first thoughts on how to solve this problem. \n\n \n# Approach\n Describe your approach to solving the problem. \n sum = sum(num | longvatrong111 | NORMAL | 2023-08-05T19:59:09.033690+00:00 | 2023-08-05T19:59:09.033716+00:00 | 92 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n \n# Approach\n<!-- Describe your approach to solving the problem. -->\n sum = sum(nums1) + t * sums(nums2)\n Create a table:\n - Each row is values of nums1 after adding nums2 at time t\n - Element is sort by nums2\n\n``... | 0 | 0 | ['C++'] | 0 |
minimum-time-to-make-array-sum-at-most-x | python with inline explaination | python-with-inline-explaination-by-sasha-6q51 | Intuition\n Describe your first thoughts on how to solve this problem. \nWe can remove nums2 from small to big to let nums1 increase slower.\n\n# Approach\n Des | sashawanchen | NORMAL | 2023-08-05T19:50:27.466875+00:00 | 2023-08-05T19:50:27.466897+00:00 | 17 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can remove nums2 from small to big to let nums1 increase slower.\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^2)\... | 0 | 0 | ['Python'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Annotated C++ Solution | annotated-c-solution-by-gyrcpp-v74d | Bog standard c++ implementation based on the question hint.\n\n# Code\ncpp\nclass Solution {\npublic:\n int minimumTime(vector<int>& nums1, vector<int>& nums | gyrcpp | NORMAL | 2023-08-05T17:05:23.480946+00:00 | 2023-08-05T17:06:27.615230+00:00 | 58 | false | Bog standard c++ implementation based on the question hint.\n\n# Code\n```cpp\nclass Solution {\npublic:\n int minimumTime(vector<int>& nums1, vector<int>& nums2, int x) {\n \n // 1.0 Define number of nodes to be used throughout this function.\n int n = nums1.size();\n\n // 1.1 Create a m... | 0 | 0 | ['C++'] | 0 |
minimum-time-to-make-array-sum-at-most-x | C# || Solution | c-solution-by-vdmhunter-g2ev | \npublic class Solution\n{\n public int MinimumTime(IList<int> nums1, IList<int> nums2, int x)\n {\n var n = nums1.Count;\n\n var p = Enumer | vdmhunter | NORMAL | 2023-08-05T16:42:25.673668+00:00 | 2023-08-14T16:15:19.535122+00:00 | 28 | false | ```\npublic class Solution\n{\n public int MinimumTime(IList<int> nums1, IList<int> nums2, int x)\n {\n var n = nums1.Count;\n\n var p = Enumerable.Range(0, n)\n .Select(i => (a:nums1[i], b:nums2[i]))\n .OrderBy(t => t.b)\n .ToList();\n\n var dp = new int[n + ... | 0 | 0 | ['C#'] | 0 |
minimum-time-to-make-array-sum-at-most-x | Java | Best Solution | Minimum Time to Make Array Sum At Most x | java-best-solution-minimum-time-to-make-fnsjg | 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 | shrey163 | NORMAL | 2023-08-05T16:28:26.646619+00:00 | 2023-08-05T16:28:26.646647+00:00 | 104 | 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 |
online-election | [C++/Java/Python] Binary Search in Times | cjavapython-binary-search-in-times-by-le-t1he | Initialization part\nIn the order of time, we count the number of votes for each person.\nAlso, we update the current lead of votes for each time point.\nif (co | lee215 | NORMAL | 2018-09-23T03:06:50.194904+00:00 | 2021-07-25T08:27:42.440822+00:00 | 20,205 | false | ## **Initialization part**\nIn the order of time, we count the number of votes for each person.\nAlso, we update the current lead of votes for each time point.\n` if (count[person] >= count[lead]) lead = person`\n\nTime Complexity: `O(N)`\n<br>\n\n## **Query part**\nBinary search `t` in `times`,\nfind out the latest ti... | 170 | 1 | [] | 35 |
online-election | [Java/Python 3] two methods with comment- using TreeMap and binary search, respectively | javapython-3-two-methods-with-comment-us-jvek | Method 1:\nTime complexity for constructor TopVotedCandidate(int[] persons, int[] times)\tis O(nlogn), and for q(int t) is O(logn).\n\n\tprivate TreeMap tm = ne | rock | NORMAL | 2018-09-23T03:08:32.826941+00:00 | 2022-06-26T20:04:55.327025+00:00 | 5,486 | false | Method 1:\nTime complexity for constructor `TopVotedCandidate(int[] persons, int[] times)`\tis O(nlogn), and for `q(int t)` is O(logn).\n\n\tprivate TreeMap<Integer, Integer> tm = new TreeMap<>(); // time and leading candidate\n\tpublic TopVotedCandidate(int[] persons, int[] times) {\n\t int[] count = new int[person... | 43 | 0 | [] | 10 |
online-election | An extremely detailed explanation accompanied with Code. | an-extremely-detailed-explanation-accomp-hgvc | First lets understand the question input as I had a hard time to make sense of question input.\n\n\n ["TopVotedCandidate","q","q","q","q","q","q"], [[[0,1,1 | amrx101 | NORMAL | 2020-06-11T04:06:13.101418+00:00 | 2020-06-11T04:08:33.100407+00:00 | 1,689 | false | First lets understand the question input as I had a hard time to make sense of question input.\n\n\n ["TopVotedCandidate","q","q","q","q","q","q"], [[[0,1,1,0,0,1,0],[0,5,10,15,20,25,30]],[3],[12],[25],[15],[24],[8]]\n\t \nFirst list is the query. Now we have a list of lists. The very first list is a list of person... | 21 | 0 | ['Binary Search'] | 5 |
online-election | C++ binary search solution beats 100% | c-binary-search-solution-beats-100-by-bw-qe2p | C++\nclass TopVotedCandidate{\npublic:\n TopVotedCandidate(vector<int> persons, vector<int> times) {\n int max_count = 0, candidate = 0, len = persons | bwv988 | NORMAL | 2019-02-23T09:31:54.245316+00:00 | 2019-02-23T09:31:54.245353+00:00 | 2,059 | false | ```C++\nclass TopVotedCandidate{\npublic:\n TopVotedCandidate(vector<int> persons, vector<int> times) {\n int max_count = 0, candidate = 0, len = persons.size();\n int count[len + 1];\n memset(count, 0, sizeof count);\n // candidates.first is the time[i], candidates.second is the top vote... | 20 | 1 | [] | 2 |
online-election | Python readable short bisect solution | python-readable-short-bisect-solution-by-rbsn | There is a minor ambiguity in the question btw. If time is taken as zero, there is no chance to decide who is the winner.\n\nclass TopVotedCandidate:\n\n def | cenkay | NORMAL | 2018-09-23T03:16:16.523659+00:00 | 2018-10-23T22:01:17.915677+00:00 | 1,463 | false | There is a minor ambiguity in the question btw. If time is taken as zero, there is no chance to decide who is the winner.\n```\nclass TopVotedCandidate:\n\n def __init__(self, persons, times):\n votes = collections.defaultdict(int)\n winner = 0\n self.winners = [None] * len(times)\n self.... | 15 | 1 | [] | 4 |
online-election | JAVA TreeMap | java-treemap-by-caraxin-040c | \nclass TopVotedCandidate {\n TreeMap<Integer, Integer> map= new TreeMap<>();\n public TopVotedCandidate(int[] persons, int[] times) {\n int[] cnt= | caraxin | NORMAL | 2018-09-23T03:07:03.368967+00:00 | 2018-10-02T08:41:14.068715+00:00 | 1,098 | false | ```\nclass TopVotedCandidate {\n TreeMap<Integer, Integer> map= new TreeMap<>();\n public TopVotedCandidate(int[] persons, int[] times) {\n int[] cnt= new int[persons.length];\n int maxCnt=0;\n for (int i=0; i<persons.length; i++){\n int p= persons[i], t= times[i];\n if ... | 13 | 1 | [] | 1 |
online-election | Java || With good explaination | java-with-good-explaination-by-nobody10-7gb2 | \n# Approach\nThe TopVotedCandidate class represents the online election system. The constructor takes two arrays as input: persons, which is an array of intege | nobody10 | NORMAL | 2023-03-02T07:53:30.312722+00:00 | 2023-03-02T07:53:30.312766+00:00 | 1,424 | false | \n# Approach\nThe TopVotedCandidate class represents the online election system. The constructor takes two arrays as input: persons, which is an array of integers representing the ID of each person who voted, and times, which is an array of integers representing the time at which each vote was cast. The constructor ini... | 6 | 0 | ['Array', 'Hash Table', 'Binary Search', 'Design', 'Java'] | 0 |
online-election | Java | TreeMap | java-treemap-by-tk-x-t8cq | \nclass TopVotedCandidate {\n TreeMap<Integer, Integer> timedWinner;\n int lead;\n public TopVotedCandidate(int[] persons, int[] times) {\n int | TK-X | NORMAL | 2022-06-14T11:36:34.382475+00:00 | 2022-06-14T11:36:34.382513+00:00 | 1,038 | false | ```\nclass TopVotedCandidate {\n TreeMap<Integer, Integer> timedWinner;\n int lead;\n public TopVotedCandidate(int[] persons, int[] times) {\n int n = times.length;\n lead = -1;\n timedWinner = new TreeMap();\n Map<Integer, Integer> votes = new HashMap();\n for(int i=0; i<n; ... | 5 | 0 | ['Tree', 'Java'] | 1 |
online-election | ✅ [JS] Binary Search Approach w/ explanation 🔥 | js-binary-search-approach-w-explanation-4lpcr | \uD83D\uDC4D In the event that you found my solution to be beneficial, I would be most grateful if you would consider UPVOTING IT. This action would serve as a | ad0x99 | NORMAL | 2024-06-18T15:23:25.308986+00:00 | 2024-11-27T08:58:03.661981+00:00 | 192 | false | **\uD83D\uDC4D In the event that you found my solution to be beneficial, I would be most grateful if you would consider UPVOTING IT. This action would serve as a significant motivator for me to continue providing additional solutions in the future. \uD83D\uDE46\u200D\u2642\uFE0F**\n\n---\n\n# Binary Search Approach\n\n... | 4 | 0 | ['Binary Search', 'Rust', 'JavaScript'] | 0 |
online-election | Python | Precompute + Binary Search | python-precompute-binary-search-by-sr_vr-pcl7 | For __init__ we precompute the top voted person for each time in times in an ordered list topVoted.\n* For q we use Binary Search in topVoted (recycling code fr | sr_vrd | NORMAL | 2022-05-14T06:40:57.776676+00:00 | 2022-05-14T06:48:05.115568+00:00 | 683 | false | * For `__init__` we precompute the top voted person for each `time` in `times` in an ordered list `topVoted`.\n* For `q` we use Binary Search in `topVoted` (recycling code from <a href="https://leetcode.com/problems/search-insert-position/">35. Search Insert Position</a>) to find the top voted person at time `t`.\n\n``... | 4 | 0 | ['Binary Tree', 'Python'] | 2 |
online-election | C++, 240ms , Map + Count Vector, Very Easy to understand, Explanation given | c-240ms-map-count-vector-very-easy-to-un-7c6p | This question was really confusing after couple of errors I figured out the question and also it took some time to figure out perfect working solution that didn | phaninderg | NORMAL | 2021-12-23T05:11:40.386349+00:00 | 2021-12-23T05:11:40.386386+00:00 | 478 | false | This question was really confusing after couple of errors I figured out the question and also it took some time to figure out perfect working solution that didn\'t result in TLE\n\nHere is the code snipped which is very easy to understand\nThe vector inside constructor holds the count of votes of each person\ncurrentMa... | 4 | 0 | [] | 0 |
online-election | Anybody has a magic general formula for Binary Search? | anybody-has-a-magic-general-formula-for-seuvu | record each time, which person is leading using a hash map\n2. binary search current query time\n\nQuestion is:\nHow do you decide between\nwhile (lo <= hi) \nw | cheng_coding_attack | NORMAL | 2018-11-10T23:01:58.448807+00:00 | 2018-11-10T23:01:58.448855+00:00 | 1,354 | false | 1. record each time, which person is leading using a hash map\n2. binary search current query time\n\nQuestion is:\nHow do you decide between\n``` while (lo <= hi) ``` \n``` while (lo < hi) ```\n``` while (lo < hi - 1) ```?\nI looked at couple binary search problems, each is different when it comes to the terminal cond... | 4 | 0 | [] | 4 |
online-election | [java] Binary search | java-binary-search-by-big_news-ltm4 | ``` class TopVotedCandidate { int[] leading; int[] time; public TopVotedCandidate(int[] persons, int[] times) { time = times; leadin | big_news | NORMAL | 2018-09-23T03:12:45.188699+00:00 | 2018-10-02T14:16:55.157868+00:00 | 959 | false | ```
class TopVotedCandidate {
int[] leading;
int[] time;
public TopVotedCandidate(int[] persons, int[] times) {
time = times;
leading = new int[persons.length];
Map<Integer, Integer> map = new HashMap<>();
for (int n : persons){
map.put(n, 0);
}
in... | 4 | 0 | [] | 0 |
online-election | [C++] Easy Upper_Bound Solution | c-easy-upper_bound-solution-by-makhonya-3q28 | \nclass TopVotedCandidate {\npublic:\n unordered_map<int, int> m;\n vector<int> times;\n TopVotedCandidate(vector<int> persons, vector<int> times) {\n | makhonya | NORMAL | 2022-09-03T09:21:00.273214+00:00 | 2022-09-03T09:21:00.273253+00:00 | 814 | false | ```\nclass TopVotedCandidate {\npublic:\n unordered_map<int, int> m;\n vector<int> times;\n TopVotedCandidate(vector<int> persons, vector<int> times) {\n int n = persons.size(), lead = -1;\n this->times = times;\n unordered_map<int, int> count;\n for (int i = 0; i < n; ++i) {\n ... | 3 | 0 | ['C', 'C++'] | 0 |
online-election | ✅ || python || using hashmap + binary search || | python-using-hashmap-binary-search-by-ab-rdfq | \nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n \n \n self.leading =[]\n \n self.ti | Abhishen99 | NORMAL | 2021-07-16T17:18:02.716011+00:00 | 2021-07-16T19:46:20.459979+00:00 | 535 | false | ```\nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n \n \n self.leading =[]\n \n self.times = []\n \n count = defaultdict(int)\n ma = 0\n \n for i , p in enumerate(persons):\n \n count[p]+... | 3 | 1 | ['Python', 'Python3'] | 0 |
online-election | Python3 precomputed + binary search | python3-precomputed-binary-search-by-red-l1tr | \nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n self.times = times\n self.series = [0] * len(persons | redtree1112 | NORMAL | 2019-03-25T09:10:30.429480+00:00 | 2019-03-25T09:10:30.429537+00:00 | 433 | false | ```\nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n self.times = times\n self.series = [0] * len(persons)\n vote = [0] * len(persons)\n win = -1\n for i, time in enumerate(times):\n person = persons[i]\n vote[person] +... | 3 | 0 | [] | 0 |
online-election | ✅✔️Easy implementation using upper bound || C++ Solution ✈️✈️✈️✈️✈️ | easy-implementation-using-upper-bound-c-qz3mw | 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 | ajay_1134 | NORMAL | 2023-07-23T17:35:04.588794+00:00 | 2023-07-23T17:35:04.588816+00:00 | 754 | 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. $$O(n)$$ --... | 2 | 0 | ['Array', 'Hash Table', 'Binary Search', 'C++'] | 0 |
online-election | Python3 solution beats 91.11% 🚀 || quibler7 | python3-solution-beats-9111-quibler7-by-bdi7i | \n\n# Code\n\nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n self.persons = []\n self.times = []\n | quibler7 | NORMAL | 2023-05-15T06:29:19.024449+00:00 | 2023-05-15T06:29:19.024492+00:00 | 836 | false | \n\n# Code\n```\nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n self.persons = []\n self.times = []\n self.dic = collections.defaultdict(int)\n self.m = 0\n self.idx = -1\n\n for i in range(len(times)):\n self.times.append... | 2 | 0 | ['Python3'] | 1 |
online-election | Solution | solution-by-deleted_user-tx4j | C++ []\nclass TopVotedCandidate {\npublic:\n TopVotedCandidate(vector<int> persons, vector<int> times) {\n int max_count = 0, candidate = 0, len = per | deleted_user | NORMAL | 2023-05-12T04:20:48.894843+00:00 | 2023-05-12T05:21:24.489752+00:00 | 971 | false | ```C++ []\nclass TopVotedCandidate {\npublic:\n TopVotedCandidate(vector<int> persons, vector<int> times) {\n int max_count = 0, candidate = 0, len = persons.size();\n int count[len + 1];\n memset(count, 0, sizeof count);\n candidates = vector<pair<int, int>>(len);\n for(int i = 0;... | 2 | 0 | ['C++', 'Java', 'Python3'] | 1 |
online-election | Easiest Java Solution || Binary Search on Times | easiest-java-solution-binary-search-on-t-xhpw | \n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(N)\n\n# Code\n\nclass TopVotedCandidate {\n\n HashMap<Integer,Integer>time=new HashMap<> | jaiyadav | NORMAL | 2023-01-13T07:43:24.228256+00:00 | 2023-01-13T07:43:24.228315+00:00 | 1,601 | false | \n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(N)\n\n# Code\n```\nclass TopVotedCandidate {\n\n HashMap<Integer,Integer>time=new HashMap<>();\n\n int[]arr=null;\n\n public TopVotedCandidate(int[] persons, int[] times) {\n \n int maxi=Integer.MIN_VALUE;\n arr=times;\n HashM... | 2 | 0 | ['Array', 'Hash Table', 'Binary Search', 'Design', 'Java'] | 1 |
online-election | Easy JS Solution | easy-js-solution-by-dollysingh-bgqg | \n# Code\n\n/**\n * @param {number[]} persons\n * @param {number[]} times\n */\nvar TopVotedCandidate = function(persons, times) {\n let maxC = 0;\n let m | dollysingh | NORMAL | 2023-01-07T17:04:30.983340+00:00 | 2023-01-07T17:04:30.983383+00:00 | 504 | false | \n# Code\n```\n/**\n * @param {number[]} persons\n * @param {number[]} times\n */\nvar TopVotedCandidate = function(persons, times) {\n let maxC = 0;\n let maxP = null;\n\n let map = new Map();\n let tmap = new Map();\n\n for(let i = 0; i < persons.length; i++) {\n let p = persons[i];\n let... | 2 | 0 | ['Binary Search', 'JavaScript'] | 0 |
online-election | Java || Beats 90% || Complete Explanation || Binary Search | java-beats-90-complete-explanation-binar-bz7v | ```\nclass TopVotedCandidate {\n \n HashMap map = new HashMap<>();\n // declaring global arrays for the persons and times array\n int [] person;\n | kurmiamreet44 | NORMAL | 2022-12-14T07:30:53.707167+00:00 | 2022-12-14T07:30:53.707210+00:00 | 401 | false | ```\nclass TopVotedCandidate {\n \n HashMap<Integer, Integer> map = new HashMap<>();\n // declaring global arrays for the persons and times array\n int [] person;\n int [] time;\n // stores the winner at any point of time\n int [] winners;\n public TopVotedCandidate(int[] persons, int[] times) {\n... | 2 | 0 | ['Java'] | 0 |
online-election | Easily Understandable | easily-understandable-by-pritamsinghsola-gm2h | 1.Treemap is used to quickly serach on the times.\n2.mp_2 is used to store the ith person lead.\n3.lead is used just to maintain the lead for next time and assi | pritamsinghsolanki | NORMAL | 2022-07-06T06:28:09.147384+00:00 | 2023-05-05T05:40:12.587290+00:00 | 455 | false | 1.Treemap is used to quickly serach on the times.\n2.mp_2 is used to store the ith person lead.\n3.lead is used just to maintain the lead for next time and assign it at the right time as said in the question .\n \n```\nclass TopVotedCandidate {\n TreeMap<Integer,Integer> mp=new TreeMap<>(); \n HashMap<Integer,Int... | 2 | 0 | ['Java'] | 1 |
online-election | online election | online-election-by-ninjamonkey42-m6tk | \n#times: [0, 5, 10]\n#persons: [0, 1, 1]\n#leader: [0, 1, 1]\n#array item values represent the candidates\n#at a specific timestamp, the candidate | NinjaMonkey42 | NORMAL | 2022-06-16T10:39:38.316016+00:00 | 2022-07-11T16:22:25.244548+00:00 | 414 | false | ```\n#times: [0, 5, 10]\n#persons: [0, 1, 1]\n#leader: [0, 1, 1]\n#array item values represent the candidates\n#at a specific timestamp, the candidate was casted a vote by someone\n#each occurence of candidate in persons list represent the vote received by the candidate\n#https://leetcode.com/problems/onlin... | 2 | 0 | ['Binary Tree', 'Python'] | 0 |
online-election | C++ | Segment Tree | O(log N) Query | O(Nlog N) preprocessing | c-segment-tree-olog-n-query-onlog-n-prep-c1yu | \n#define fi first\n#define se second\nusing piii = pair<pair<int,int>,int>;\nstruct SegmentTree{\n vector<piii> t;\n SegmentTree(int N){\n t.assig | shadow_47 | NORMAL | 2022-06-02T12:31:14.272839+00:00 | 2022-06-02T12:31:14.272874+00:00 | 58 | false | ```\n#define fi first\n#define se second\nusing piii = pair<pair<int,int>,int>;\nstruct SegmentTree{\n vector<piii> t;\n SegmentTree(int N){\n t.assign(4 * N + 10,{{0, -1}, 0});\n Build(0,0,N-1);\n }\n piii Combine(piii &a,piii &b){\n if(a.fi.fi!=b.fi.fi)return a.fi.fi > b.fi.fi ? a : b... | 2 | 0 | ['Tree', 'Binary Tree'] | 0 |
online-election | [Java] simple solution using treetop | java-simple-solution-using-treetop-by-mw-f5h8 | \nclass TopVotedCandidate {\n TreeMap<Integer, Integer> tm = new TreeMap<>();\n\n // get lader for each moment in time one by one\n public TopVotedCand | mwacc | NORMAL | 2022-02-12T20:25:07.853928+00:00 | 2022-02-12T20:25:07.853972+00:00 | 127 | false | ```\nclass TopVotedCandidate {\n TreeMap<Integer, Integer> tm = new TreeMap<>();\n\n // get lader for each moment in time one by one\n public TopVotedCandidate(int[] persons, int[] times) {\n Map<Integer, Integer> votes = new HashMap<>(); // candidate -> number of votes\n int leader = -1;\n ... | 2 | 0 | ['Tree'] | 0 |
online-election | 100% faster than all C# submissions (at the time of writing) | 100-faster-than-all-c-submissions-at-the-6alq | \n public class TopVotedCandidate\n {\n private List<Vote> votes = new List<Vote>();\n\n public TopVotedCandidate(int[] persons, int[] times | mojo1234 | NORMAL | 2022-02-10T13:24:23.855960+00:00 | 2022-02-10T13:24:40.977355+00:00 | 248 | false | ```\n public class TopVotedCandidate\n {\n private List<Vote> votes = new List<Vote>();\n\n public TopVotedCandidate(int[] persons, int[] times) \n {\n var tally = new Dictionary<int, int>();\n int lastPerson = -1;\n\n for (int i = 0; i < persons.Length; i+... | 2 | 0 | ['C#'] | 1 |
online-election | Easy to understand - TreeMap Solution || O(N + Q*log(N)) | easy-to-understand-treemap-solution-on-q-xli6 | \nclass TopVotedCandidate \n{\n //Time Complexity: O(N + Q*logN), where N is the number of votes, and Q is the number of queries.\n \n TreeMap<Integer, | dhake_baba_met17 | NORMAL | 2021-07-31T21:30:36.694483+00:00 | 2021-08-27T11:41:41.538994+00:00 | 216 | false | ```\nclass TopVotedCandidate \n{\n //Time Complexity: O(N + Q*logN), where N is the number of votes, and Q is the number of queries.\n \n TreeMap<Integer, Integer> map; // Our main Red-Black tree for searching\n HashMap<Integer, Integer> store;\n int person, maxVotes; // Stores the person who could be th... | 2 | 0 | ['Tree', 'Binary Search Tree', 'Java'] | 2 |
online-election | Python3 Binary Search with short explanation | python3-binary-search-with-short-explana-99tq | You\'re told that times is sorted, so you can already begin thinking of binary search. Now all you need to know is who is leading at what time which is why you | jsn667 | NORMAL | 2021-05-24T14:49:16.979088+00:00 | 2021-05-24T14:50:49.813214+00:00 | 168 | false | You\'re told that times is sorted, so you can already begin thinking of binary search. Now all you need to know is who is leading at what time which is why you keep count of who is leading at each index of the "times" array. Then, when you are asked to find a leader for a certain time, you binary search for the time in... | 2 | 0 | [] | 1 |
online-election | [Python3] binary search O(logN) for query | python3-binary-search-ologn-for-query-by-er1g | Algo\nWhile running initializer, we compute the frequency table of each candidates votes and decide winner at each time stamp. When running query, we binary sea | ye15 | NORMAL | 2020-12-01T03:41:52.658126+00:00 | 2020-12-01T03:41:52.658166+00:00 | 295 | false | **Algo**\nWhile running initializer, we compute the frequency table of each candidates votes and decide winner at each time stamp. When running query, we binary search the proper moment to decide who the winner is at that time. \n\n**Implementation**\n```\nclass TopVotedCandidate:\n\n def __init__(self, persons: Lis... | 2 | 0 | ['Python3'] | 0 |
online-election | C++ | Unordered_map | Upper bound | c-unordered_map-upper-bound-by-wh0ami-idpo | \nclass TopVotedCandidate {\n unordered_map<int, int>m;\n vector<int>time;\npublic:\n TopVotedCandidate(vector<int>& persons, vector<int>& times) {\n | wh0ami | NORMAL | 2020-06-17T09:29:01.999455+00:00 | 2020-06-17T09:29:01.999484+00:00 | 110 | false | ```\nclass TopVotedCandidate {\n unordered_map<int, int>m;\n vector<int>time;\npublic:\n TopVotedCandidate(vector<int>& persons, vector<int>& times) {\n unordered_map<int, int>count;\n int lead = -1;\n time = times;\n for (int i = 0; i < persons.size(); i++) {\n count[per... | 2 | 0 | [] | 0 |
online-election | C# Binary Search based implementation | c-binary-search-based-implementation-by-be90y | (Side note :By the time I submitted this solution it beated the existed solution 100%\nQuote: "Runtime: 696 ms, faster than 100.00% of C# online submissions for | captain_u | NORMAL | 2019-01-20T00:29:09.799270+00:00 | 2019-01-20T00:29:09.799310+00:00 | 283 | false | (Side note :By the time I submitted this solution it beated the existed solution 100%\nQuote: "Runtime: 696 ms, faster than 100.00% of C# online submissions for Online Election.")\n\n*Time Complexity:* \nFor Method ```TopVotedCandidate```: O(N) where N is the size of the times array. \nFor Method ```Q```: O(log(N)) for... | 2 | 0 | [] | 0 |
online-election | [Java] TreeMap: O(NlogN) + O(logN) and Binary Search: O(N) + O(logN) | java-treemap-onlogn-ologn-and-binary-sea-ad63 | Below its standard TreeMap solution, as map.put takes O(logN) so the constructor takes O(NlogN).\n\nclass TopVotedCandidate {\n\n TreeMap<Integer, Integer> m | goku_wukong | NORMAL | 2018-10-01T06:37:11.336912+00:00 | 2018-10-01T06:42:05.852718+00:00 | 553 | false | Below its standard TreeMap solution, as map.put takes O(logN) so the constructor takes O(NlogN).\n```\nclass TopVotedCandidate {\n\n TreeMap<Integer, Integer> map = new TreeMap<>();\n public TopVotedCandidate(int[] persons, int[] times) {\n \n HashMap<Integer, Integer> count = new HashMap<>();\n ... | 2 | 0 | [] | 1 |
online-election | JavaScript solution | javascript-solution-by-dva2-ug6z | https://github.com/dva22/leetcode/blob/master/problems/911.%20Online%20Election/index.js\n\nvar TopVotedCandidate = module.exports = function(persons, times) { | dva2 | NORMAL | 2018-09-26T00:31:48.320237+00:00 | 2018-09-26T00:31:48.320286+00:00 | 299 | false | https://github.com/dva22/leetcode/blob/master/problems/911.%20Online%20Election/index.js\n```\nvar TopVotedCandidate = module.exports = function(persons, times) {\n this.winningTimes = {};\n this.times = times;\n var votesCount = [];\n \n var currentWinningPerson = -1;\n var winningVotes = 0;\n for (var i = 0; ... | 2 | 0 | [] | 0 |
online-election | Java solution using PriorityQueue and BinarySearch (O(n*log(n)) | java-solution-using-priorityqueue-and-bi-94gd | Code | swapit | NORMAL | 2025-03-21T12:04:49.507274+00:00 | 2025-03-21T12:04:49.507274+00:00 | 50 | false | # Code
```java []
class TopVotedCandidate {
public class Candidate {
int id;
int votes;
int latestVoteIndex;
public Candidate(int id, int votes, int latestVoteIndex) {
this.id = id;
this.votes = votes;
this.latestVoteIndex = latestVoteIndex;
}
}
HashMap<Integer, Candidate> db;
PriorityQ... | 1 | 0 | ['Java'] | 0 |
online-election | ✅ 🎯 📌 Simple Solution || Dictionary || Prefix Sum || Binary Search || Fastest Solution ✅ 🎯 📌 | simple-solution-dictionary-prefix-sum-bi-mn6h | Code\npython3 []\nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n self.persons=[]\n self.times=[]\n | vvnpais | NORMAL | 2024-09-14T19:34:09.371597+00:00 | 2024-09-14T19:34:09.371632+00:00 | 339 | false | # Code\n```python3 []\nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n self.persons=[]\n self.times=[]\n self.d=defaultdict(int)\n self.mx=0\n for person,time in zip(persons,times):\n self.times.append(time)\n self.d[pe... | 1 | 0 | ['Array', 'Hash Table', 'Binary Search', 'Design', 'Python3'] | 0 |
online-election | Simple python solution beats 85% of users (HashMap-BinarySearch) | simple-python-solution-beats-85-of-users-o915 | Code\n\nfrom collections import defaultdict\nimport bisect\n\nclass TopVotedCandidate:\n\n def __init__(self, persons, times):\n self.leadings = []\n | youssefkhalil320 | NORMAL | 2024-05-17T18:28:11.331704+00:00 | 2024-05-17T18:28:11.331731+00:00 | 197 | false | # Code\n```\nfrom collections import defaultdict\nimport bisect\n\nclass TopVotedCandidate:\n\n def __init__(self, persons, times):\n self.leadings = []\n self.times = times\n self.vote_counts = defaultdict(int)\n leading = -1\n \n for i in range(len(persons)):\n ... | 1 | 0 | ['Hash Table', 'Binary Search', 'Python3'] | 0 |
online-election | python3 solution beats 97% 🚀 || quibler7 | python3-solution-beats-97-quibler7-by-qu-uek6 | Code\n\nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n self.persons = []\n self.times = []\n s | quibler7 | NORMAL | 2023-05-07T03:46:48.412533+00:00 | 2023-05-07T03:46:48.412562+00:00 | 518 | false | # Code\n```\nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n self.persons = []\n self.times = []\n self.dic = collections.defaultdict(int)\n self.m = 0\n self.idx = -1\n\n for i in range(len(times)):\n self.times.append(tim... | 1 | 0 | ['Python3'] | 1 |
online-election | ✅ beats 100% Java code | beats-100-java-code-by-abstractconnoisse-eeid | Java Code\n\nclass TopVotedCandidate {\n private final int[] time;\n private final int[] leader;\n public TopVotedCandidate(int[] persons, int[] times) | abstractConnoisseurs | NORMAL | 2023-02-13T07:34:24.921300+00:00 | 2023-02-13T07:34:24.921334+00:00 | 224 | false | # Java Code\n```\nclass TopVotedCandidate {\n private final int[] time;\n private final int[] leader;\n public TopVotedCandidate(int[] persons, int[] times) {\n LeaderChange head = new LeaderChange();\n LeaderChange tail = head;\n int n = persons.length;\n int[] votes = new int[n];\... | 1 | 1 | ['Array', 'Hash Table', 'Binary Search', 'Design', 'Java'] | 0 |
online-election | C++✅ || Easy || Solution | c-easy-solution-by-prinzeop-4sfl | \nclass TopVotedCandidate {\npublic:\n map<int, int> m2;\n unordered_map<int, int> m;\n\tTopVotedCandidate(vector<int>& persons, vector<int>& times) {\n\t | casperZz | NORMAL | 2022-08-16T11:01:42.134987+00:00 | 2022-08-16T11:04:17.510195+00:00 | 664 | false | ```\nclass TopVotedCandidate {\npublic:\n map<int, int> m2;\n unordered_map<int, int> m;\n\tTopVotedCandidate(vector<int>& persons, vector<int>& times) {\n\t\tint lead = -1;\n\t\tm[-1] = 0;\n\t\tint n = persons.size();\n\t\tfor (int i = 0; i < n; ++i) {\n\t\t\tm[persons[i]]++;\n\t\t\tif (m[persons[i]] >= m[lead])... | 1 | 0 | ['C', 'Binary Tree'] | 0 |
online-election | Easy Solution using a HashMap and a TreeMap | easy-solution-using-a-hashmap-and-a-tree-61bn | \nclass TopVotedCandidate {\n // <key: voting candidate, value: count of votes a candidate has received>\n HashMap<Integer, Integer> voteCount;\n // <k | oori3280 | NORMAL | 2022-08-08T06:48:25.653230+00:00 | 2022-08-08T06:48:25.653263+00:00 | 264 | false | ```\nclass TopVotedCandidate {\n // <key: voting candidate, value: count of votes a candidate has received>\n HashMap<Integer, Integer> voteCount;\n // <key: timestamp, value: who is the leader with max vote>\n TreeMap<Integer, Integer> timeToLeader;\n int currentLeader;\n public TopVotedCandidate(int... | 1 | 0 | [] | 0 |
online-election | [Python] Clean, simple with explanation | python-clean-simple-with-explanation-by-widei | \nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n counter = defaultdict(int)\n\n mostVotePersons = [0] | npq | NORMAL | 2022-07-22T15:34:11.075381+00:00 | 2022-07-22T15:34:11.075418+00:00 | 322 | false | ```\nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n counter = defaultdict(int)\n\n mostVotePersons = [0] * len(persons) # mostVotePersons[i] is the most vote person at times[i]\n largestVote = -1 # keep largest vote person index\n for i in range(l... | 1 | 0 | ['Binary Tree', 'Python'] | 0 |
online-election | [Python 3] Binary Search | python-3-binary-search-by-gabhay-hlk7 | We first Precompute all winners at a given time if the winner is not changed we don\'t consider to add it to our list of winners at a given time. after that we | gabhay | NORMAL | 2022-07-07T13:56:15.512248+00:00 | 2022-07-07T13:56:15.512287+00:00 | 135 | false | We first Precompute all winners at a given time if the winner is not changed we don\'t consider to add it to our list of winners at a given time. after that we will find the lower bound of the time for which we are querying and just provide the result from our precomputed list.\nO(N)-- to precompute\nO(log(N)--> to que... | 1 | 0 | ['Binary Search'] | 0 |
online-election | C++ SOLUTION | c-solution-by-kunal_patil-dctc | \n unordered_map mp;\n map chk;\n vector times1;\n int maxi=0;\n int ele;\n \n int binser(int k)\n {\n int i = 0;\n int j | kunal_patil | NORMAL | 2022-05-31T08:53:42.411838+00:00 | 2022-05-31T08:53:42.411877+00:00 | 104 | false | \n unordered_map<int,int> mp;\n map<int,int> chk;\n vector<int> times1;\n int maxi=0;\n int ele;\n \n int binser(int k)\n {\n int i = 0;\n int j = times1.size()-1;\n \n while(i<=j)\n {\n int mid = (i+j)/2;\n if(times1[mid]==k)\n ... | 1 | 0 | ['Binary Tree'] | 0 |
online-election | (C++) 911. Online Election | c-911-online-election-by-qeetcode-ezgu | \n\nclass TopVotedCandidate {\n vector<int> times, lead; \npublic:\n TopVotedCandidate(vector<int>& persons, vector<int>& times) : times(times) {\n | qeetcode | NORMAL | 2021-12-10T20:30:10.545470+00:00 | 2021-12-10T20:30:10.545520+00:00 | 300 | false | \n```\nclass TopVotedCandidate {\n vector<int> times, lead; \npublic:\n TopVotedCandidate(vector<int>& persons, vector<int>& times) : times(times) {\n unordered_map<int, int> freq; \n int pp = 0; \n for (auto& p : persons) {\n ++freq[p]; \n if (freq[p] >= freq[pp]) pp = ... | 1 | 0 | ['C'] | 1 |
online-election | Precompute + Binary Search [Java] | precompute-binary-search-java-by-itkan-p3cl | \nimport java.util.HashMap;\n\npublic class TopVotedCandidate {\n int[] winners;\n int[] times;\n\n public TopVotedCandidate(int[] persons, int[] times | itkan | NORMAL | 2021-11-20T09:52:31.374601+00:00 | 2021-11-20T09:52:31.374638+00:00 | 187 | false | ```\nimport java.util.HashMap;\n\npublic class TopVotedCandidate {\n int[] winners;\n int[] times;\n\n public TopVotedCandidate(int[] persons, int[] times) {\n this.times = times;\n winners = new int[times.length];\n\n HashMap<Integer, Integer> voteMap = new HashMap<>();\n int maxVo... | 1 | 0 | ['Binary Tree', 'Java'] | 0 |
online-election | 📌📌 Easy and Well Explained || 98% faster || Simple Approach 🐍 | easy-and-well-explained-98-faster-simple-z78h | IDEA :\nIn the order of time, we count the number of votes for each person.\nAlso, we update the current lead of votes for each time point.\nif (count[person] | abhi9Rai | NORMAL | 2021-11-06T14:36:21.442661+00:00 | 2021-11-06T14:36:21.442690+00:00 | 118 | false | ## IDEA :\nIn the order of time, we count the number of votes for each person.\nAlso, we update the current lead of votes for each time point.\n`if (count[person] >= count[lead]) lead = person`\n\nTime Complexity: O(N)\n\n**Implementation :**\n\n\'\'\'\n\n\tclass TopVotedCandidate:\n def __init__(self, persons: Lis... | 1 | 0 | [] | 0 |
online-election | Pre-computation + Binary Search [C++] | pre-computation-binary-search-c-by-siddh-7wmq | \n unordered_map<int, int> mp;\n vector<int> pref, times;\n int best = -1;\n \n TopVotedCandidate(vector<int>& persons, vector<int>& times) {\n | siddhantchimankar | NORMAL | 2021-11-04T13:52:36.978078+00:00 | 2021-11-04T13:52:36.978107+00:00 | 86 | false | ```\n unordered_map<int, int> mp;\n vector<int> pref, times;\n int best = -1;\n \n TopVotedCandidate(vector<int>& persons, vector<int>& times) {\n \n pref = vector<int>(persons.begin(), persons.end());\n this->times = times;\n \n for(int i = 0 ; i < times.size() ; i++) ... | 1 | 0 | [] | 0 |
online-election | ✔️ C# - Simple to understand With Comments. Map & Binary Search | c-simple-to-understand-with-comments-map-t682 | Thumbs up if you find this helpful \uD83D\uDC4D\n\nThe main idea in this solution is to create a result array called leaders that keeps track of who the current | keperkjr | NORMAL | 2021-09-30T07:19:50.645429+00:00 | 2021-09-30T07:40:47.120428+00:00 | 99 | false | **Thumbs up if you find this helpful** \uD83D\uDC4D\n\nThe main idea in this solution is to create a result array called ```leaders``` that keeps track of who the current leader is up to that point, for each ```time[i]```.\n\nFor example:\n```\nleaders[i] = The current leading vote getter up to the point of time[i]\n``... | 1 | 0 | ['Binary Tree'] | 0 |
online-election | C++ || Map || binary search | c-map-binary-search-by-priyanka1230-xqks | \n\npublic:\n mapmp;\n vector>v;\n TopVotedCandidate(vector& p, vector& times) {\n int maxx=0,curr=0;\n for(int i=0;i=maxx)\n | priyanka1230 | NORMAL | 2021-09-29T12:48:24.580399+00:00 | 2021-09-29T12:48:24.580425+00:00 | 82 | false | ```\n\n```public:\n map<int,int>mp;\n vector<pair<int,int>>v;\n TopVotedCandidate(vector<int>& p, vector<int>& times) {\n int maxx=0,curr=0;\n for(int i=0;i<p.size();i++)\n {\n mp[p[i]]++;\n if(mp[p[i]]>=maxx)\n {\n maxx=mp[p[i]];\n ... | 1 | 0 | [] | 0 |
online-election | C# BinarySearch | c-binarysearch-by-nick_l-4dho | \npublic class TopVotedCandidate\n{\n private static int[] _top;\n private static int[] _times;\n \n public TopVotedCandidate(int[] persons, int[] t | nick_l | NORMAL | 2021-07-19T15:42:49.437905+00:00 | 2021-07-19T15:42:49.437935+00:00 | 77 | false | ```\npublic class TopVotedCandidate\n{\n private static int[] _top;\n private static int[] _times;\n \n public TopVotedCandidate(int[] persons, int[] times)\n {\n _times = times;\n _top = new int[persons.Length];\n \n var map = new Dictionary<int, int>();\n var winner =... | 1 | 0 | [] | 0 |
online-election | C++, Probably the easiest to understand, Binary Search for querying | c-probably-the-easiest-to-understand-bin-2uxo | Precomputing the maximum voted person at each possible timestep, and performing binary search on time to find the correct time. \n\nclass TopVotedCandidate {\np | aditya_trips | NORMAL | 2021-05-30T12:20:11.324841+00:00 | 2021-05-30T12:20:11.324884+00:00 | 237 | false | Precomputing the maximum voted person at each possible timestep, and performing binary search on time to find the correct time. \n```\nclass TopVotedCandidate {\npublic:\n vector<int> persons, times,topVoted;\n TopVotedCandidate(vector<int>& p, vector<int>& t) {\n persons = p; \n times = t; \n ... | 1 | 0 | ['C'] | 0 |
online-election | Easy and simple solution by using map | easy-and-simple-solution-by-using-map-by-m6hp | class TopVotedCandidate {\n vectorp;\n vectort;\n vectorres;\n int sum=0;\npublic:\n TopVotedCandidate(vector& persons, vector& times) {\n | bhoomi12356789 | NORMAL | 2021-05-05T20:24:40.485542+00:00 | 2021-05-05T20:24:40.485582+00:00 | 88 | false | class TopVotedCandidate {\n vector<int>p;\n vector<int>t;\n vector<int>res;\n int sum=0;\npublic:\n TopVotedCandidate(vector<int>& persons, vector<int>& times) {\n p=persons;\n t=times;\n unordered_map<int,int>mp;\n int mx=0,mval=-1;;\n for(int i=0;i<p.size();i++)\n ... | 1 | 2 | ['C'] | 0 |
online-election | JavaScript Solution - Binary Search Approach | javascript-solution-binary-search-approa-flei | \nvar TopVotedCandidate = function(persons, times) {\n this.times = times;\n this.len = times.length;\n this.votes = new Array(this.len).fill(0);\n | Deadication | NORMAL | 2021-03-22T17:16:46.188029+00:00 | 2021-03-22T17:19:07.280598+00:00 | 220 | false | ```\nvar TopVotedCandidate = function(persons, times) {\n this.times = times;\n this.len = times.length;\n this.votes = new Array(this.len).fill(0);\n \n let max = 0; // max votes received by any single candidate so far.\n let leader = -1l;\n \n this.leaders = persons.map((person, i) => {\n ... | 1 | 0 | ['Binary Tree', 'JavaScript'] | 0 |
online-election | Java Using HashMap And TreeMap | java-using-hashmap-and-treemap-by-tara-9-s0rs | The question is quite simple.We just have to find which person is leading at a given time\n\nclass TopVotedCandidate {\n Map<Integer,Integer> voteCount = new | tara-97 | NORMAL | 2021-01-12T06:06:33.132294+00:00 | 2021-01-12T06:06:33.132329+00:00 | 101 | false | The question is quite simple.We just have to find which person is leading at a given time\n```\nclass TopVotedCandidate {\n Map<Integer,Integer> voteCount = new HashMap<>();\n TreeMap<Integer,Integer> lead = new TreeMap<>();\n int maxVotes = 0;\n public TopVotedCandidate(int[] persons, int[] times) {\n ... | 1 | 0 | [] | 0 |
online-election | C++ simple precompute and then query | c-simple-precompute-and-then-query-by-yo-zonk | \nclass TopVotedCandidate {\npublic:\n array<int,5001> votes = {};\n map<int,int> time2winner;\n \n TopVotedCandidate(vector<int>& persons, vector<in | youngkobe | NORMAL | 2021-01-02T21:16:15.409216+00:00 | 2021-01-02T21:16:15.409260+00:00 | 98 | false | ```\nclass TopVotedCandidate {\npublic:\n array<int,5001> votes = {};\n map<int,int> time2winner;\n \n TopVotedCandidate(vector<int>& persons, vector<int>& times) {\n if(persons.empty())\n return;\n int currentWinner = persons[0];\n time2winner[times[0]] = currentWinner;\n ... | 1 | 0 | [] | 0 |
online-election | Simple Java Solution using TreeMap | simple-java-solution-using-treemap-by-mi-7719 | \nclass TopVotedCandidate {\n // Space is O(N)\n TreeMap<Integer, Integer> timeMaxVoteMap = new TreeMap<>();\n // 5000 is based on the limits provided | milind9179367800 | NORMAL | 2020-12-26T08:31:03.838628+00:00 | 2020-12-26T08:31:03.838678+00:00 | 203 | false | ```\nclass TopVotedCandidate {\n // Space is O(N)\n TreeMap<Integer, Integer> timeMaxVoteMap = new TreeMap<>();\n // 5000 is based on the limits provided in the question.\n int[] votes = new int[5000];\n int maxVotePerson = -1;\n int maxVote = 0;\n public TopVotedCandidate(int[] persons, int[] time... | 1 | 0 | ['Tree', 'Java'] | 1 |
online-election | Java easy to understand binary search | java-easy-to-understand-binary-search-by-9ugz | ```\nclass TopVotedCandidate {\n int[] persons;\n int[] times;\n\n public TopVotedCandidate(int[] persons, int[] times) {\n int max = persons[0] | zmz999 | NORMAL | 2020-10-24T07:16:32.143825+00:00 | 2020-10-24T07:16:32.143860+00:00 | 112 | false | ```\nclass TopVotedCandidate {\n int[] persons;\n int[] times;\n\n public TopVotedCandidate(int[] persons, int[] times) {\n int max = persons[0];\n Map<Integer, Integer> map = new HashMap<>();\n for (int i = 0; i < times.length; i++) {\n int count = map.getOrDefault(persons[i], ... | 1 | 0 | [] | 0 |
online-election | Python 3 | Binary Search + Hash Table | Explanations | python-3-binary-search-hash-table-explan-18qt | Explanation\n- For binary search problem we often need hash table for help\n- Given persion and times, we can form a time-series data, which store the informati | idontknoooo | NORMAL | 2020-09-11T00:57:58.859791+00:00 | 2020-09-11T00:57:58.859833+00:00 | 403 | false | ### Explanation\n- For binary search problem we often need hash table for help\n- Given `persion` and `times`, we can form a time-series data, which store the information indicates who\'s winning at time `t`\n\t- `times` is already an increasing series, but we need to know who\'s winning at these time points\n\t- `vote... | 1 | 0 | ['Hash Table', 'Binary Search', 'Python', 'Python3'] | 0 |
online-election | C++ DP+Hash solution | c-dphash-solution-by-rom111-0lzq | \nclass TopVotedCandidate {\npublic:\n unordered_map<int,int>freq;\n unordered_map<int,int>votes;\n vector<int>ans;\n int maxi=INT_MIN;\n int a;\ | rom111 | NORMAL | 2020-08-09T17:35:26.712896+00:00 | 2020-08-09T17:35:26.712931+00:00 | 74 | false | ```\nclass TopVotedCandidate {\npublic:\n unordered_map<int,int>freq;\n unordered_map<int,int>votes;\n vector<int>ans;\n int maxi=INT_MIN;\n int a;\n TopVotedCandidate(vector<int>& persons, vector<int>& times) {\n a=0;\n for(int i=0;i<times.size();i++){\n a=max(a,times[i]);\n ... | 1 | 0 | [] | 0 |
online-election | [C++] 100% faster | c-100-faster-by-kangchunhung-4ork | \nclass TopVotedCandidate {\npublic:\n int *votes, *timestamp, last_time;\n TopVotedCandidate(vector<int>& persons, vector<int>& times) {\n ios_bas | kangchunhung | NORMAL | 2020-07-16T08:33:23.541398+00:00 | 2020-07-16T08:33:23.541440+00:00 | 119 | false | ```\nclass TopVotedCandidate {\npublic:\n int *votes, *timestamp, last_time;\n TopVotedCandidate(vector<int>& persons, vector<int>& times) {\n ios_base::sync_with_stdio(0);\n cin.tie(0); cout.tie(0);\n last_time = times.back() + 1;\n int L = persons.size();\n votes = new int[L];... | 1 | 0 | [] | 0 |
online-election | Python Clean Solution | python-clean-solution-by-olivia_xie_93-a1bm | python\nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n self.nums = []\n count = collections.Counter() | olivia_xie_93 | NORMAL | 2020-07-05T03:32:44.325040+00:00 | 2020-07-05T03:32:44.325076+00:00 | 106 | false | ```python\nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n self.nums = []\n count = collections.Counter()\n mx = 0\n major = None\n \n for people, time in zip(persons, times):\n count[people] += 1\n if count[peop... | 1 | 0 | [] | 0 |
online-election | Python3 dict look up or array binary search - Online Election | python3-dict-look-up-or-array-binary-sea-1kfq | \nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n self.times = times\n self.winner = {times[0]: person | r0bertz | NORMAL | 2020-07-01T03:13:47.024067+00:00 | 2020-07-01T03:14:06.294853+00:00 | 259 | false | ```\nclass TopVotedCandidate:\n\n def __init__(self, persons: List[int], times: List[int]):\n self.times = times\n self.winner = {times[0]: persons[0]}\n c = Counter({persons[0]: 1})\n maxCount = 1\n for i in range(1, len(times)):\n p = persons[i]\n c[p] += 1\... | 1 | 0 | ['Binary Search', 'Python', 'Python3'] | 0 |
online-election | O(N) + O(QLogN) | on-oqlogn-by-shivam529-qjih | \nclass TopVotedCandidate {\n int []leaders;\n int []times;\n public TopVotedCandidate(int[] persons, int[] times) {\n leaders=new int[persons.l | shivam529 | NORMAL | 2020-05-24T20:45:14.902053+00:00 | 2020-05-24T20:45:14.902107+00:00 | 108 | false | ```\nclass TopVotedCandidate {\n int []leaders;\n int []times;\n public TopVotedCandidate(int[] persons, int[] times) {\n leaders=new int[persons.length];\n this.times=times;\n Map<Integer,Integer> map=new HashMap<>();\n int leader=-1;\n int max=-1;\n for(int i=0;i<per... | 1 | 0 | ['Java'] | 0 |
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