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minimum-number-of-taps-to-open-to-water-a-garden | Just DP. Explained Line By Line. | just-dp-explained-line-by-line-by-tr1nit-0yfc | \n// TC: O(n * m) where \n// - n is given and \n// - m is ranges[i] for inner loop (<= 100)\n// SC: O(n)\nclass Solution {\npublic:\n int minTaps(int n, vect | __wkw__ | NORMAL | 2023-08-31T04:07:23.005987+00:00 | 2023-08-31T04:08:57.834736+00:00 | 984 | false | ```\n// TC: O(n * m) where \n// - n is given and \n// - m is ranges[i] for inner loop (<= 100)\n// SC: O(n)\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n // dp[i]: min number of taps that should be open to water garden [0 .. i]\n // init with some max values\n vector<i... | 13 | 0 | ['Dynamic Programming', 'C++'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | Java 4ms | java-4ms-by-asif_hasnain-p0ev | \nclass Solution {\n public int minTaps(int n, int[] ranges) {\n int count = 0;\n int lastTap = 0;\n int min = 0;\n int max = 0;\ | asif_hasnain | NORMAL | 2020-06-13T06:12:53.526639+00:00 | 2020-06-13T06:12:53.526687+00:00 | 387 | false | ```\nclass Solution {\n public int minTaps(int n, int[] ranges) {\n int count = 0;\n int lastTap = 0;\n int min = 0;\n int max = 0;\n while(max<n){\n\t\t//Start checking from last open tap, look for the tap whose lowest range is less than or equal to \n\t\t//highest range of previo... | 13 | 0 | [] | 2 |
minimum-number-of-taps-to-open-to-water-a-garden | Video Stitching | video-stitching-by-votrubac-8f1b | Reused the solution from 1024. Video Stitching.\nCPP\nint minTaps(int n, vector<int>& rs) {\n vector<vector<int>> clips;\n for (auto i = 0; i < rs.size(); ++i | votrubac | NORMAL | 2020-01-19T10:01:28.204633+00:00 | 2020-01-19T10:03:56.676444+00:00 | 1,316 | false | Reused the solution from [1024. Video Stitching](https://leetcode.com/problems/video-stitching/discuss/269988/C%2B%2BJava-6-lines-O(n-log-n)).\n```CPP\nint minTaps(int n, vector<int>& rs) {\n vector<vector<int>> clips;\n for (auto i = 0; i < rs.size(); ++i) {\n clips.push_back({i - rs[i], i + rs[i]});\n }\n retu... | 13 | 1 | [] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | Similar to Jump Game II | similar-to-jump-game-ii-by-xzj104-qdon | I feel like this is similar to the Jump Game category question with only one change.\n\nWe first need to generate the jump gamp array. To do this, for each tap | xzj104 | NORMAL | 2020-01-19T04:01:20.827019+00:00 | 2020-01-19T04:01:51.128254+00:00 | 1,022 | false | I feel like this is similar to the Jump Game category question with only one change.\n\nWe first need to generate the jump gamp array. To do this, for each tap pos, we get the interval it can cover, and mark the left as start and the right and end. \n\nWhen checking how far one start could go, we should not stop at the... | 12 | 1 | [] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Not understanding the problem | not-understanding-the-problem-by-fultonm-nvah | Input: n = 3, ranges = [0,0,0,0]\nOutput: -1\nExplanation: Even if you activate all the four taps you cannot water the whole garden.\n\nIt seems that there is a | fultonm | NORMAL | 2021-12-15T20:30:12.900478+00:00 | 2021-12-15T20:33:44.642368+00:00 | 270 | false | Input: n = 3, ranges = [0,0,0,0]\nOutput: -1\nExplanation: Even if you activate all the four taps you cannot water the whole garden.\n\nIt seems that there is a tap for each space in the garden. Why can\'t it be watered?\n```\n[0, 1, 2, 3]\n x x x x\n```\n\n\n 1 <= n <= 104\n ranges.length == n + 1\n 0 <= r... | 11 | 0 | [] | 2 |
minimum-number-of-taps-to-open-to-water-a-garden | Logic Explanation, 100% Efficient | logic-explanation-100-efficient-by-salon-74mq | So, the question is basically asking for the minimum taps, that can cover the whole garden.\nAlgorithm: \n1. Update ranges[i] to maxReach of the respective tap. | salonix__ | NORMAL | 2021-07-10T03:31:24.551491+00:00 | 2021-07-10T03:32:15.507115+00:00 | 877 | false | So, the question is basically asking for the minimum taps, that can cover the whole garden.\nAlgorithm: \n1. Update ranges[i] to maxReach of the respective tap.\n2. Now, iterate through it in O(n)\n3. Take positon variable, update it, when the iterator becomes equal to position, also, increase your jump, i.e., number o... | 11 | 0 | ['C'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | C++ sort & Greedy vs Dijkstra's algorithm | c-sort-greedy-vs-dijkstras-algorithm-by-fm00s | Intuition\n Describe your first thoughts on how to solve this problem. \nConstruct the intervals, denoted by pair (a,b), to form an array, say I. And sort I w.r | anwendeng | NORMAL | 2023-08-31T02:48:01.760148+00:00 | 2023-08-31T15:17:48.034799+00:00 | 1,620 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nConstruct the intervals, denoted by pair (a,b), to form an array, say I. And sort I w.r.t. a.\nThen find the initial interval that covers 0 the most.\nOpen the tap with the right most range whose range has intersection with the ones alre... | 10 | 0 | ['Greedy', 'Breadth-First Search', 'Graph', 'Sorting', 'Heap (Priority Queue)', 'C++'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | O(1) Space, O(n) | o1-space-on-by-bhavesh17-lcab | Same problem as Jump II.\nWe will update i-ranges[i] index to max( i +ranges[i], ranges[i-ranges[i]] ) , so as we traverse from left to right , we have max rea | bhavesh17 | NORMAL | 2020-11-07T17:27:37.919737+00:00 | 2020-11-07T17:27:37.919785+00:00 | 1,211 | false | Same problem as Jump II.\nWe will update i-ranges[i] index to max( i +ranges[i], ranges[i-ranges[i]] ) , so as we traverse from left to right , we have max reachable index.\n\n```\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n \n for(int i=1;i<ranges.size();i++){\n ... | 10 | 0 | ['Greedy', 'C'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | Greedy O(n) solution, python | greedy-on-solution-python-by-liketheflow-se8p | This problem is an almost identical problem of an OA interview problem. See reference 1\nFor each tap, record the maximum left reach point and maximum right rea | liketheflower | NORMAL | 2020-01-21T14:52:52.521355+00:00 | 2020-01-21T15:47:34.327304+00:00 | 1,579 | false | This problem is an almost identical problem of an OA interview problem. See reference 1\nFor each tap, record the maximum left reach point and maximum right reach point.\nBased on this, we can get a list contains for each left point, what is the maximum right reaching point.\nThen using a greedy algorithm to solve this... | 10 | 0 | [] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | JAVA, O(n), use bucket sort, 2ms, well commented | java-on-use-bucket-sort-2ms-well-comment-3csb | In this given problem we need to sort all the ranges that the taps provide by their start and end. Given how the problem is nicely bounded, we can use a bucket | devincisimpson | NORMAL | 2021-08-14T19:16:02.894678+00:00 | 2021-08-14T19:16:02.894705+00:00 | 864 | false | In this given problem we need to sort all the ranges that the taps provide by their start and end. Given how the problem is nicely bounded, we can use a bucket sort to achieve an O(n) runtime and a fairly small amount of code to solve this problem. \n\n```\nclass Solution {\n public int minTaps(int n, int[] ranges) ... | 9 | 0 | ['Greedy', 'Bucket Sort', 'Java'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | Java DP Greedy | java-dp-greedy-by-hobiter-2hta | dp is simple:\n1, for each position, find the most right pos that can be coverred with the same tap.\n2, from pos 0, find a route to end. if not possible, retur | hobiter | NORMAL | 2020-01-19T04:07:18.918272+00:00 | 2020-03-28T05:21:20.673831+00:00 | 4,633 | false | dp is simple:\n1, for each position, find the most right pos that can be coverred with the same tap.\n2, from pos 0, find a route to end. if not possible, return -1;\n```\nclass Solution {\n public int minTaps(int n, int[] ranges) {\n Map<Integer, Set<Integer>> m = new HashMap<>();\n int[] dp = new int... | 9 | 0 | [] | 4 |
minimum-number-of-taps-to-open-to-water-a-garden | Python 3 || 11 lines, heap || T/S: 73% / 32% | python-3-11-lines-heap-ts-73-32-by-spaul-gitr | \nclass Solution:\n def minTaps(self, n: int, ranges: List[int]) -> int:\n\n pos, ans = 0, 0\n heap = [(i-r,i+r) for i, r in enumerate(ranges)] | Spaulding_ | NORMAL | 2023-08-31T16:55:59.372097+00:00 | 2024-06-04T15:49:07.666635+00:00 | 108 | false | ```\nclass Solution:\n def minTaps(self, n: int, ranges: List[int]) -> int:\n\n pos, ans = 0, 0\n heap = [(i-r,i+r) for i, r in enumerate(ranges)]\n heapify(heap)\n \n while pos < n:\n mx = 0\n while heap and heap[0][0] <= pos:\n mx = max(mx,hea... | 8 | 0 | ['Python3'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Simple and easy to understand. | simple-and-easy-to-understand-by-avanish-w9cu | \nclass Solution:\n def minTaps(self, n: int, ranges: List[int]) -> int:\n start, end = 0, 0\n taps = 0\n \n while end< n:\n | avanishtiwari | NORMAL | 2021-03-15T10:01:24.352277+00:00 | 2021-03-15T10:01:58.314507+00:00 | 1,429 | false | ```\nclass Solution:\n def minTaps(self, n: int, ranges: List[int]) -> int:\n start, end = 0, 0\n taps = 0\n \n while end< n:\n for i in range(len(ranges)):\n if i-ranges[i] <= start and i+ ranges[i]>end:\n end = i + ranges[i]\n if s... | 8 | 0 | ['Dynamic Programming', 'Python', 'Python3'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | [Java] BAD DESCRIPTION Recursion + Memoization fails on some test cases which seem wrong | java-bad-description-recursion-memoizati-gcya | 35\n[1,0,4,0,4,1,4,3,1,1,1,2,1,4,0,3,0,3,0,3,0,5,3,0,0,1,2,1,2,4,3,0,1,0,5,2]\nExpected: 6\nMy Answer: 5\n```\nclass Solution {\n public static int func(int[ | manrajsingh007 | NORMAL | 2020-01-19T04:13:05.395614+00:00 | 2020-01-19T04:17:16.309072+00:00 | 905 | false | 35\n[1,0,4,0,**4**,1,4,3,1,1,1,2,1,**4**,0,3,0,3,0,3,0,**5**,3,0,0,1,2,1,2,**4**,3,0,1,0,**5**,2]\nExpected: 6\nMy Answer: 5\n```\nclass Solution {\n public static int func(int[] ranges, int n, int i, int prev, int[][] dp) {\n if(prev >= n) return 0;\n if(i == n) return n + 1;\n if(dp[i][prev] !... | 8 | 0 | [] | 3 |
minimum-number-of-taps-to-open-to-water-a-garden | Beats 98% | With Similar Problems 🏆 | beats-98-with-similar-problems-by-hridoy-6at6 | This problem is a variation of the 1024. Video Stitching and the 45. Jump Game II\n problem.\n\nThis problem was asked Big Tech companies like Amazon, Twitter, | hridoy100 | NORMAL | 2023-08-31T09:23:22.314488+00:00 | 2023-08-31T09:28:56.320624+00:00 | 549 | false | This problem is a variation of the [1024. Video Stitching](https://leetcode.com/problems/video-stitching/) and the [45. Jump Game II\n](https://leetcode.com/problems/jump-game-ii/) problem.\n\nThis problem was asked Big Tech companies like Amazon, Twitter, Salesforce etc.\n\nBefore we dive in to the actual solution let... | 7 | 0 | ['Array', 'Greedy', 'Sorting', 'C++', 'Java'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | ✅ C++ | Recursive 🚀🚀 DP🔥 Solution | Memoization | c-recursive-dp-solution-memoization-by-p-28k6 | Code\n\nclass Solution {\npublic:\n int n, dp[10005];\n int solve(int i, vector<pair<int, int>> &v)\n {\n if(i != -1 && v[i].second >= n)\n | pkacb | NORMAL | 2023-08-31T07:52:29.295793+00:00 | 2023-08-31T07:52:29.295843+00:00 | 1,163 | false | # Code\n```\nclass Solution {\npublic:\n int n, dp[10005];\n int solve(int i, vector<pair<int, int>> &v)\n {\n if(i != -1 && v[i].second >= n)\n return 0;\n\n if(dp[i + 1] != -1)\n return dp[i + 1];\n\n int ans = 1e9;\n for(int j = i + 1; j < v.size(); j++)\n ... | 7 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | Beats 100%! With proper explanation ! Greedy || DP | beats-100-with-proper-explanation-greedy-rhg5 | \n# Approach\nThe approach we follow over here is, for every tap we find the maximum range it can go. To store this range we create an array dp to store immedia | parthdharmale008 | NORMAL | 2023-08-31T05:50:03.963164+00:00 | 2023-08-31T06:08:25.175000+00:00 | 1,235 | false | \n# Approach\nThe approach we follow over here is, for every tap we find the maximum range it can go. To store this range we create an array ```dp``` to store immediate results.\n\nNow, iterate over ```ranges``` from ```0 to n```, if we encounter 0 then continue, i.e we don\'t need to update range for this index, if it... | 7 | 1 | ['Dynamic Programming', 'Greedy', 'C++'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | Simple DP Approach 🔥 | C++ (Sorting + DP) | (Recursion+Memoization) | simple-dp-approach-c-sorting-dp-recursio-bpyg | \nclass Solution {\npublic:\n map<int,map<int,int>> dp;\n int helper(int idx,int prev,vector<vector<int>> &intervals){\n if(prev>=intervals.size()- | _maityamit | NORMAL | 2023-08-31T03:03:36.122459+00:00 | 2023-08-31T03:04:50.084682+00:00 | 746 | false | ```\nclass Solution {\npublic:\n map<int,map<int,int>> dp;\n int helper(int idx,int prev,vector<vector<int>> &intervals){\n if(prev>=intervals.size()-1) return 0;\n if(idx==intervals.size()) return 1e9;\n if(dp.count(idx) && dp[idx].count(prev)) return dp[idx][prev];\n if(intervals[idx... | 7 | 0 | ['Dynamic Programming', 'C'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | C++ | GREEDY | WITH EXPLANATION | c-greedy-with-explanation-by-tanyaaa24-b2gf | Complexity\n- Time complexity: O(N * Min taps)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. | tanyaaa24 | NORMAL | 2022-12-11T17:30:11.289342+00:00 | 2022-12-11T17:30:11.289381+00:00 | 1,990 | false | # Complexity\n- Time complexity: O(N * Min taps)\n<!-- Add your time complexity here, e.g. $$O(n)$$ --> \n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ --> \n\n# Code\n```\n\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) \n {\n int taps = 0;\n ... | 7 | 0 | ['Greedy', 'C++'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | [Python3/Java] Greedy O(n) | Explanation | python3java-greedy-on-explanation-by-mar-g7w6 | One great simplification for this is to skew the taps to the left. For example, say you have a tap at position 4, with range 2. This means this tap covers posit | mardlucca | NORMAL | 2021-11-24T03:39:08.989770+00:00 | 2021-11-29T23:03:09.869857+00:00 | 1,082 | false | One great simplification for this is to skew the taps to the left. For example, say you have a tap at position 4, with range 2. This means this tap covers positions [2, 6]. What we want to do is to transform the problem and suppose taps can only shoot water to the right. So, in order to cover the same area (i.e [2, 6])... | 7 | 0 | ['Greedy', 'Java', 'Python3'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | C++ with Explanation | O(n) Time and Space | Greedy | c-with-explanation-on-time-and-space-gre-87y3 | Approach-1 (Greedy)\nThe Greedy strategy here is to always choose the interval with the largest space that is yet to be watered. Now we could have done this by | mercuryVapor | NORMAL | 2021-06-18T13:38:31.579378+00:00 | 2021-06-18T14:05:09.033676+00:00 | 575 | false | **Approach-1** (Greedy)\nThe Greedy strategy here is to always choose the interval with the largest space that is yet to be watered. Now we could have done this by storing all the inytervals for each tap and then selecting the largest uncovered interval each time. But this choosing interval method cannot be done in O(1... | 7 | 0 | [] | 2 |
minimum-number-of-taps-to-open-to-water-a-garden | ⏰3 Minutes To Realise \\\ Greedy Approach \\\ Beats 99% | 3-minutes-to-realise-greedy-approach-bea-qhix | Make sure that the following works\nSimple algorithm O(n) space, time.\n\nclass Solution:\n def minTaps(self, n: int, ranges: List[int]) -> int:\n arr | mchlkrpch | NORMAL | 2023-08-31T10:00:39.950008+00:00 | 2023-08-31T16:11:13.843237+00:00 | 450 | false | # Make sure that the following works\n**Simple algorithm O(n) space, time.**\n```\nclass Solution:\n def minTaps(self, n: int, ranges: List[int]) -> int:\n arr = [0] * (n + 1)\n for idx, cur_radius in enumerate(ranges):\n if cur_radius == 0:\n continue\n \n ... | 6 | 0 | ['Array', 'Dynamic Programming', 'Greedy', 'Python', 'Python3'] | 3 |
minimum-number-of-taps-to-open-to-water-a-garden | BRUTE FORCE DP || C++ | brute-force-dp-c-by-ganeshkumawat8740-e4w4 | Code\n\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n int i,j;\n vector<int> dp(n+1,n+10);\n dp[0] = 0;\n | ganeshkumawat8740 | NORMAL | 2023-06-06T04:18:25.633694+00:00 | 2023-06-06T04:18:38.891900+00:00 | 1,295 | false | # Code\n```\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n int i,j;\n vector<int> dp(n+1,n+10);\n dp[0] = 0;\n for(int i = 0; i <= n; i++){\n for(j = max(0,i-ranges[i]+1); j<=min(n,i+ranges[i]); j++){\n dp[j] = min(dp[j],dp[max(i-ranges[... | 6 | 0 | ['Array', 'Dynamic Programming', 'Greedy', 'C++'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Simple Python | Explained | Comments | Greedy | simple-python-explained-comments-greedy-sjbtb | Prerequisites:\n- https://leetcode.com/problems/jump-game\n- https://leetcode.com/problems/jump-game-ii\n\nMore tips:\n- https://www.notion.so/paulonteri/Greedy | paulonteri | NORMAL | 2021-10-20T18:38:59.606859+00:00 | 2021-10-20T18:38:59.606906+00:00 | 1,271 | false | Prerequisites:\n- https://leetcode.com/problems/jump-game\n- https://leetcode.com/problems/jump-game-ii\n\nMore tips:\n- https://www.notion.so/paulonteri/Greedy-Algorithms-b9b0a6dd66c94e7db2cbbd9f2d6b50af#d7578cbb76c7423d9c819179fc749be5\n- https://leetcode.com/problems/minimum-number-of-taps-to-open-to-water-a-garden/... | 6 | 0 | ['Greedy', 'Python', 'Python3'] | 2 |
minimum-number-of-taps-to-open-to-water-a-garden | C++ solution (Sorting and Two Pointers) || Full explanation and appropriately commented | c-solution-sorting-and-two-pointers-full-r5u2 | Define a structure "Interval" to hold the left and right extremes of each tap (clipped at 0 and n)\n Convert the ranges array into a vector of Intervals\n Sort | Selachimorpha | NORMAL | 2021-09-21T13:11:47.474244+00:00 | 2021-09-21T13:12:29.437046+00:00 | 534 | false | * Define a structure "Interval" to hold the left and right extremes of each tap (clipped at 0 and n)\n* Convert the ranges array into a vector of Intervals\n* Sort the vector of Intervals in the increasing order of starting points. If two intervals have the same starting points give preference to the one with a higher ... | 6 | 0 | [] | 2 |
minimum-number-of-taps-to-open-to-water-a-garden | Java Solution - Greedy | java-solution-greedy-by-dixon_n-wz7x | 1024. Video Stitching\n45. Jump Game II\n1326. Minimum Number of Taps to Open to Water a Garden\n\n# Code\njava []\n// Approach -1\n\nclass Solution {\n publ | Dixon_N | NORMAL | 2024-08-30T17:40:54.531381+00:00 | 2024-08-30T17:41:33.660198+00:00 | 358 | false | [1024. Video Stitching](https://leetcode.com/problems/video-stitching/solutions/5712090/simple-solution-greedy/)\n[45. Jump Game II](https://leetcode.com/problems/jump-game-ii/solutions/5297066/all-three-solution/)\n[1326. Minimum Number of Taps to Open to Water a Garden](https://leetcode.com/problems/minimum-number-of... | 5 | 0 | ['Greedy', 'Java'] | 4 |
minimum-number-of-taps-to-open-to-water-a-garden | 🔥 Easy C++ : Minimum Jump to Reach Last Index Variation DP + Memoization🔥 | easy-c-minimum-jump-to-reach-last-index-o2gzz | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem is to find the minimum number of taps required to water a garden. Each tap | eknath_mali_002 | NORMAL | 2023-08-31T02:55:24.692960+00:00 | 2023-08-31T02:55:24.692979+00:00 | 504 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem is to find the minimum number of taps required to water a garden. Each tap has a range, and you can turn on a tap at any position within its range to water that segment of the garden. You need to cover the entire garden with t... | 5 | 0 | ['Dynamic Programming', 'Memoization', 'C++'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | GREEDY + SORTING ,WITHOUT DP | greedy-sorting-without-dp-by-comder_00-mmk3 | Okay so,\nthis is basically a question for overlapping ranges. You can find similar questions on leetcode itself.\nso what is the approach?\nwe first find the r | comder_00 | NORMAL | 2022-08-26T16:00:25.871138+00:00 | 2022-08-26T16:20:01.851736+00:00 | 587 | false | Okay so,\nthis is basically a question for overlapping ranges. You can find similar questions on leetcode itself.\nso what is the approach?\nwe first find the ranges of all the taps in the garden\nlet\'s understand with an example\nn=4;\n[2,1,1 ,2,1]\nso finding the ranges for indexes:\n0: (0-2,0+2)=(0,2)\n1:(1-1,1+1)=... | 5 | 0 | ['Greedy', 'Sorting'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | C++ simple solution using DP | c-simple-solution-using-dp-by-maitreya47-osc2 | Solution 1:\nThere are n+1 taps, so we first create our dp vector and initialise all to n+2 (impossible value as only options available == n+1).\ndp[i] will sto | maitreya47 | NORMAL | 2021-07-18T15:31:02.984691+00:00 | 2022-11-17T23:48:07.193855+00:00 | 1,176 | false | # Solution 1:\nThere are n+1 taps, so we first create our dp vector and initialise all to n+2 (impossible value as only options available == n+1).\ndp[i] will store minimum number of taps to cover range [0, i].\nStart from tap 1 and update for all the taps within its range [i-ranges[i], i+ranges[i]], we check it with I... | 5 | 1 | ['Dynamic Programming', 'C', 'C++'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Best explanation with visualisation || C++ || faster than 100% solution | best-explanation-with-visualisation-c-fa-vgut | After reading so many solutions, I got to know that whats actually happening.\nMy explanation and solution is really simple and straight forward.\n\nAlso please | arpitdhamija | NORMAL | 2021-07-18T13:17:32.323844+00:00 | 2021-07-18T13:21:07.945742+00:00 | 328 | false | After reading so many solutions, I got to know that whats actually happening.\nMy explanation and solution is really simple and straight forward.\n\nAlso please read some solutions on "most voted", you\'ll understand it more clearly then\n\nAlso do checkout jump game 2 on leetcode. Otherise I have also pasted an image ... | 5 | 1 | [] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | C++ DP | c-dp-by-wzypangpang-8w1z | \nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n int big = 1e7;\n\t\t// dp[i] stands for the minimum taps to cover the range | wzypangpang | NORMAL | 2021-03-31T20:47:45.168754+00:00 | 2021-03-31T20:51:26.063125+00:00 | 394 | false | ```\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n int big = 1e7;\n\t\t// dp[i] stands for the minimum taps to cover the range [0, i].\n\t\t// By definition, it can be deducted that dp[i] <= dp[j] for any i < j. Becasue dp[j] could always qualify dp[i].\n vector<int> dp(n+1, b... | 5 | 0 | [] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | 🏆Easy Understandble Code || 🚀Similar to Jump game-II 🚀 || O(N)🏆 | easy-understandble-code-similar-to-jump-ilf9b | \n# Complexity\n- Time complexity:\n O(N)\n\n- Space complexity:\n O(1)\n\n# Code\n\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) | rajh_3399 | NORMAL | 2023-08-31T18:39:56.267239+00:00 | 2023-08-31T18:39:56.267271+00:00 | 71 | false | \n# Complexity\n- Time complexity:\n O(N)\n\n- Space complexity:\n O(1)\n\n# Code\n```\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n vector<int> startEnd(n+1, 0);\n for(int i = 0; i<ranges.size(); i++) { \n int start = max(0, i - ranges[i]);\n int e... | 4 | 0 | ['Dynamic Programming', 'Greedy', 'C++'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | Sorting + Greedy || Easy C++ || image Explanation ✅✅ | sorting-greedy-easy-c-image-explanation-cz3y5 | Approach\n Describe your approach to solving the problem. \nTry to convert the given array into the interval format given below and sort the intervals based on | Deepak_5910 | NORMAL | 2023-08-31T14:41:26.622363+00:00 | 2023-08-31T14:41:26.622381+00:00 | 633 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nTry to convert the given array into the **interval format** given below and **sort the intervals** based on the **starting position** and now choose the intervals that give the optimal answer.\n\n {\n int[] is_reachable = new int[n+1];\n\n int inf = (int) 1e9;\n\n A | Viraj_Patil_092 | NORMAL | 2023-08-31T05:16:53.134392+00:00 | 2023-08-31T05:16:53.134412+00:00 | 301 | false | # Code\n```\nclass Solution {\n public int minTaps(int n, int[] ranges) {\n int[] is_reachable = new int[n+1];\n\n int inf = (int) 1e9;\n\n Arrays.fill(is_reachable, inf);\n is_reachable[0] = 0;\n\n for(int i = 0;i < ranges.length;i++){\n int mn = Math.max(0,i-ranges[i])... | 4 | 0 | ['Array', 'Dynamic Programming', 'Greedy', 'Java'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | [Python3][Stack] Easy Stack Solution beats 94% runtime. | python3stack-easy-stack-solution-beats-9-ewop | Approach\n Describe your approach to solving the problem. \n1. Initialize an empty stack called range_stack to keep track of the ranges being used.\n2. Initiali | near6334 | NORMAL | 2023-08-31T05:13:33.389097+00:00 | 2023-08-31T16:15:55.099857+00:00 | 552 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n1. Initialize an empty stack called `range_stack` to keep track of the ranges being used.\n2. Initialize a variable `covered` to keep track of the total coverage achieved.\n3. Iterate through the given ranges. For each range, check if the current rang... | 4 | 0 | ['Python3'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | BEST C++ SOLUTION U CAN GET | best-c-solution-u-can-get-by-2005115-vz5y | PLEASE UPVOTE MY SOLUTION AND COMMENT FOR ANY DOUBT\n# Doubt shall be cleared within 3 hr )\nConnect with me here\n- https://www.linkedin.com/in/pratay-nandy-9b | 2005115 | NORMAL | 2023-08-31T04:05:43.958282+00:00 | 2023-08-31T04:05:43.958302+00:00 | 509 | false | # **PLEASE UPVOTE MY SOLUTION AND COMMENT FOR ANY DOUBT**\n# Doubt shall be cleared within 3 hr )\n**Connect with me here**\n- [https://www.linkedin.com/in/pratay-nandy-9ba57b229/\n]()\n- [https://www.instagram.com/pratay_nandy/\n]()\n\n# Approach\nThe provided code is implementing a solution to the "Minimum Number of ... | 4 | 0 | ['Array', 'Greedy', 'C++'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Java | Progression from recursion to DP solution | java-progression-from-recursion-to-dp-so-79t0 | \n# Intuition\nFirst try simple recursion, get TLE, then try to return an output like integer, then try memoization using DP\n\n# Code (TLE, only recursion)\n\n | pacificlion | NORMAL | 2023-08-31T01:30:12.082938+00:00 | 2023-08-31T01:34:14.948870+00:00 | 622 | false | \n# Intuition\nFirst try simple recursion, get TLE, then try to return an output like integer, then try memoization using DP\n\n# Code (TLE, only recursion)\n```\nclass Solution {\n\n\n int res = Integer.MAX_VALUE;\n public int minTaps(int n, int[] ranges) {\n helper(0,0,0,ranges,0);\n\n return res=... | 4 | 0 | ['Java'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | C++,Greedy,Interval based | cgreedyinterval-based-by-nideesh45-aiwx | \nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n vector<pair<int,int>> v;\n for(int i=0;i<ranges.size();i++){\n | nideesh45 | NORMAL | 2022-07-08T03:49:12.244641+00:00 | 2022-07-08T03:49:12.244684+00:00 | 648 | false | ```\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n vector<pair<int,int>> v;\n for(int i=0;i<ranges.size();i++){\n // making ranges\n v.push_back({i-ranges[i],i+ranges[i]});\n }\n // sorting the intervals\n sort(v.begin(),v.end());\n ... | 4 | 0 | ['Greedy', 'C'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | ✔️ EXPLANATION || PYTHON ✔️ | explanation-python-by-karan_8082-tkfy | First fid the ranges of all taps.\nHere ranges are stored in same array a.\na[ i ][ 0 ] = lower range of tap i\na[ i ][ 1 ] = upper range of tap i\n\nSort the r | karan_8082 | NORMAL | 2022-06-05T10:35:32.806431+00:00 | 2022-06-05T10:35:32.806456+00:00 | 359 | false | First fid the ranges of all taps.\nHere ranges are stored in same array a.\n**a[ i ][ 0 ] = lower range of tap i**\n**a[ i ][ 1 ] = upper range of tap i**\n\nSort the ranges according to first value.\nThen start from garden index to be watered. Choose the pair that has maximum upper range and has garden index between l... | 4 | 0 | ['Array', 'Sorting'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Simple Javascript solution 1-dimensional DP Bottom Up | simple-javascript-solution-1-dimensional-5tr6 | Use bottom up approach dp to record the minimum number of taps needed. Iterate through all the ranges, and at location k, take the minimum of between the memoiz | liangpeter | NORMAL | 2022-04-23T03:13:39.355875+00:00 | 2022-04-23T03:13:39.355909+00:00 | 534 | false | Use bottom up approach dp to record the minimum number of taps needed. Iterate through all the ranges, and at location k, take the minimum of between the memoized result and 1 + (taps need to cover the rest of the ranges k - 1).\n```\nvar minTaps = function(n, ranges) {\n const dp = new Array(n + 1).fill(Infinity);\... | 4 | 0 | ['Dynamic Programming', 'JavaScript'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | C++ || GREEDY || SHORT AND CLEAN CODE || NO-SORTING | c-greedy-short-and-clean-code-no-sorting-z9s8 | ```\nclass Solution {\npublic:\n int minTaps(int n, vector& ranges) {\n int mn = 0;\n int reach = 0;\n int taps = 0;\n \n | Easy_coder | NORMAL | 2022-01-12T09:58:50.506878+00:00 | 2022-01-12T09:58:50.506912+00:00 | 621 | false | ```\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n int mn = 0;\n int reach = 0;\n int taps = 0;\n \n while(reach < n){\n for(int i=0; i<ranges.size(); i++){\n if(i-ranges[i] <= mn && i+ranges[i] > reach)\n ... | 4 | 0 | ['Greedy', 'C'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Java - O(N) - Jump Game II + I Concept | java-on-jump-game-ii-i-concept-by-protya-txpp | One of the classic Hard problem.\n\n\nSo, we need to build intervals of [left,right] for each of the given tap.\nTo convert this into a Jump Game II array, we n | protyay | NORMAL | 2021-07-04T18:34:07.924510+00:00 | 2021-07-04T18:34:07.924554+00:00 | 354 | false | One of the classic Hard problem.\n\n\nSo, we need to build intervals of _[left,right]_ for each of the given tap.\nTo convert this into a Jump Game II array, we need to store the max jump for each index(tap).\n\n*We re-iterate through the jumps array and check the maxReach at each index*\nOne concept of jump game I is ... | 4 | 0 | [] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | C++, O(n) with a stack, only a few lines of code. | c-on-with-a-stack-only-a-few-lines-of-co-1925 | The idea is to just iterate though taps one by one, and see which previous taps the current tap can replace, and the put current tap onto the stack if it can ex | deeperblue | NORMAL | 2021-05-24T00:52:23.319447+00:00 | 2021-05-24T00:52:23.319488+00:00 | 428 | false | The idea is to just iterate though taps one by one, and see which previous taps the current tap can replace, and the put current tap onto the stack if it can expand the coverage.\n\n```\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n stack<pair<int, int>> tapsInfo;\n tapsInfo.p... | 4 | 0 | [] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | [Java] clean O(n) Greedy Solution || with detailed comments | java-clean-on-greedy-solution-with-detai-nmh7 | This problem is basically Leetcode 45. Jump Game II but construct the array (furthestIdx in the solution here) by ourselves. Before you solving this problem, I | xieyun95 | NORMAL | 2021-03-20T00:12:07.034957+00:00 | 2021-03-20T00:56:21.250993+00:00 | 1,116 | false | This problem is basically Leetcode 45. Jump Game II but construct the array (furthestIdx in the solution here) by ourselves. Before you solving this problem, I strongly recommand to get familiar with these "prototype" questions or revisit the solutions. \n* [55. Jump Game](https://leetcode.com/problems/jump-game/discus... | 4 | 0 | ['Greedy', 'Java'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | JavaScript Interval Merging | javascript-interval-merging-by-daleighan-opwe | \nfunction minTaps(n, ranges) {\n let intervals = [];\n for (let i = 0; i < ranges.length; i++) {\n intervals.push([i - ranges[i], i + ranges[i]]);\n }\n | daleighan | NORMAL | 2020-01-21T02:42:09.324485+00:00 | 2020-01-21T02:42:09.324576+00:00 | 296 | false | ```\nfunction minTaps(n, ranges) {\n let intervals = [];\n for (let i = 0; i < ranges.length; i++) {\n intervals.push([i - ranges[i], i + ranges[i]]);\n }\n intervals.sort((a, b) => a[0] - b[0]);\n let currentL = 0, currentR = 0, taps = 0, i = 0;\n while (i < intervals.length) {\n while (i < intervals.lengt... | 4 | 0 | ['JavaScript'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | O(n^2) DP, O(nLog(n)) greedy, O(n) greedy solution | on2-dp-onlogn-greedy-on-greedy-solution-msxly | O(n^2) DP\n\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& r) {\n int dp[n+1];\n for(int i =0;i<n+1;i++) dp[i] = n+3;\n dp | objectobject | NORMAL | 2020-01-20T18:41:11.592283+00:00 | 2020-01-20T20:00:58.033407+00:00 | 465 | false | **O(n^2) DP**\n```\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& r) {\n int dp[n+1];\n for(int i =0;i<n+1;i++) dp[i] = n+3;\n dp[0] = 0;\n for(int i = 0;i<=n;i++)\n {\n for(int j = max(0,i-r[i]);j<=min(i+r[i],n);j++)\n {\n dp[j] =... | 4 | 0 | [] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | JavaScript DP with detailed comments | javascript-dp-with-detailed-comments-by-7x3ce | js\nvar minTaps = function(n, ranges) {\n // dp[i] means the smallest number of taps required to water under i;\n const dp = Array(n + 1).fill(n + 2);\n dp[0 | paradite | NORMAL | 2020-01-19T06:08:30.847363+00:00 | 2020-01-19T06:08:30.847399+00:00 | 901 | false | ```js\nvar minTaps = function(n, ranges) {\n // dp[i] means the smallest number of taps required to water under i;\n const dp = Array(n + 1).fill(n + 2);\n dp[0] = 0;\n // iterate all taps\n for (let i = 0; i <= n; i++) {\n const start = i - ranges[i] > 0 ? i - ranges[i] : 0;\n const end = ranges[i] + i > n ... | 4 | 0 | ['Dynamic Programming', 'JavaScript'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | is this a wrong test case? | is-this-a-wrong-test-case-by-ltt198612-qiyk | ```\n35\n[1,0,4,0,4,1,4,3,1,1,1,2,1,4,0,3,0,3,0,3,0,5,3,0,0,1,2,1,2,4,3,0,1,0,5,2]\n 4 4 5 4 5\n\t\t | ltt198612 | NORMAL | 2020-01-19T04:08:12.946936+00:00 | 2020-01-19T04:08:12.946972+00:00 | 371 | false | ```\n35\n[1,0,4,0,4,1,4,3,1,1,1,2,1,4,0,3,0,3,0,3,0,5,3,0,0,1,2,1,2,4,3,0,1,0,5,2]\n 4 4 5 4 5\n\t\t \nThe result should be 5,\nHowever the test case is expecting 6?\n | 4 | 1 | [] | 3 |
minimum-number-of-taps-to-open-to-water-a-garden | Greedy solution using stack. Detailed explanation | greedy-solution-using-stack-detailed-exp-22xj | Python code\nInline explanation in the code\nNotes at the end\n\n\nclass Solution(object):\n def minTaps(self, n, ranges):\n ranges = sorted([(i - r, | kaiwensun | NORMAL | 2020-01-19T04:02:13.833889+00:00 | 2020-01-19T06:47:43.640273+00:00 | 790 | false | Python code\nInline explanation in the code\nNotes at the end\n\n```\nclass Solution(object):\n def minTaps(self, n, ranges):\n ranges = sorted([(i - r, min(i + r, n)) for i, r in enumerate(ranges)]) # sort by ranges\' left point. the min is needed. see notes below\n stack = [(0, 0), (0, 0)] ... | 4 | 1 | [] | 2 |
minimum-number-of-taps-to-open-to-water-a-garden | Minimum Number of Taps to Open to Water a Garden | minimum-number-of-taps-to-open-to-water-vqpyb | 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 | rissabh361 | NORMAL | 2023-08-31T15:33:07.611548+00:00 | 2023-08-31T15:33:07.611568+00:00 | 910 | 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 |
minimum-number-of-taps-to-open-to-water-a-garden | O(NlogN + N) Approach || Beginner Friendly | onlogn-n-approach-beginner-friendly-by-l-5yjo | Approach\n- Creating new array consisting of start and end ranges of each tap.\n- sorting the array according to their starting position.\n- traversing array ti | Lil_ToeTurtle | NORMAL | 2023-08-31T03:39:43.867169+00:00 | 2023-08-31T03:39:43.867194+00:00 | 149 | false | # Approach\n- Creating new array consisting of start and end ranges of each tap.\n- sorting the array according to their starting position.\n- traversing array till the furthest point that can be reached, if the new range is encountered, then loops continues and `res++` because that is a valid range\n- if no new range ... | 3 | 0 | ['Greedy', 'Sorting', 'Java'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | Beats 92.85 in runtime 85.93 in memory Greedy approach | beats-9285-in-runtime-8593-in-memory-gre-h0mk | \n# Code\n\nclass Solution:\n def minTaps(self, n: int, ranges: List[int]) -> int:\n arr = [0] * (n + 1)\n for i, r in enumerate(ranges):\n | ayan_101 | NORMAL | 2023-08-31T02:51:58.574312+00:00 | 2023-08-31T02:53:21.560815+00:00 | 27 | false | \n# Code\n```\nclass Solution:\n def minTaps(self, n: int, ranges: List[int]) -> int:\n arr = [0] * (n + 1)\n for i, r in enumerate(ranges):\n if r == 0:continue\n left = max(0, i - r);arr[left] = max(arr[left], i + r)\n end, far_can_reach, cnt = 0, 0, 0\n for i, rea... | 3 | 0 | ['Array', 'Divide and Conquer', 'Dynamic Programming', 'Greedy', 'Brainteaser', 'Sliding Window', 'Suffix Array', 'Sorting', 'Counting', 'Python3'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Python3 Solution | python3-solution-by-motaharozzaman1996-vwkh | \n\nclass Solution:\n def minTaps(self, n: int, ranges: List[int]) -> int:\n dp=[0]+[n+2]*n\n for i,x in enumerate(ranges):\n for j | Motaharozzaman1996 | NORMAL | 2023-08-31T02:31:12.087729+00:00 | 2023-08-31T02:31:12.087750+00:00 | 337 | false | \n```\nclass Solution:\n def minTaps(self, n: int, ranges: List[int]) -> int:\n dp=[0]+[n+2]*n\n for i,x in enumerate(ranges):\n for j in range(max(i-x+1,0),min(i+x,n)+1):\n dp[j]=min(dp[j],dp[max(0,i-x)]+1)\n\n return dp[n] if dp[n]<n+2 else -1 \n``` | 3 | 0 | ['Python', 'Python3'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | JAVA Easily Understandable Solution with Explanation and Comments | java-easily-understandable-solution-with-2hxb | Approach: We just have to find the tap with largest rightMax range && which should start atleast from our current minimum ranged reached. Now the current minimu | abhi71k | NORMAL | 2022-08-09T08:26:48.853961+00:00 | 2022-08-09T17:41:13.607162+00:00 | 642 | false | **Approach:** We just have to find the tap with largest rightMax range && which should start **atleast** from our current minimum ranged reached. Now the current minimum point will be the rightMax range of previously selected tap.\nHence if no tap is selected, previous minimum would be 0, so next tap should be selected... | 3 | 0 | ['Greedy', 'Java'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | Great Explained Greedy Solution in Python, Very Easy to Understand!!! | great-explained-greedy-solution-in-pytho-c267 | Use Greedy method!\n\nTwo parts:\n\nPart 1:\nGenerate a jump list, which keeps the rightmost index that every index could reach in one move.\nFor Example:\n[3 4 | GeorgePot | NORMAL | 2022-05-08T18:33:08.739776+00:00 | 2022-05-21T19:17:09.430165+00:00 | 352 | false | Use **Greedy** method!\n\n**Two parts:**\n\n**Part 1:**\nGenerate a **jump list**, which keeps the rightmost index that every index could reach in one move.\n**For Example:**\n[3 4 1 1 0 0]\nindex 0: range [0 3], jump list[0] = 3 (**Note:** if left range is smaller than 0, left range would be 0)\nindex 1: range [0 5], ... | 3 | 0 | ['Greedy', 'Python'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Similair to 871. Minimum Number of Refueling Stops | similair-to-871-minimum-number-of-refuel-j1hc | I am writing this post as I was not able to find for some time that this question is almost similar to 871. Minimum Number of Refueling Stops. \nOther posts do | ahen445 | NORMAL | 2022-04-18T18:23:49.859360+00:00 | 2022-04-18T18:27:43.015334+00:00 | 457 | false | I am writing this post as I was not able to find for some time that this question is almost similar to 871. Minimum Number of Refueling Stops. \nOther posts do give many other problems which are almost similar to this, maybe we can add it to the same list. \nSharing code for both:\n## Minimum number of refueling stops\... | 3 | 0 | ['Greedy', 'Heap (Priority Queue)', 'C++'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | C++ | Recursive DP | c-recursive-dp-by-codingsuju-61nn | ```\nLet dp[pos] be the minimum number of taps required to cover everything under [0,pos]. \nUsing top-down approach as we all are familiar with.\nsolve(pos) {\ | codingsuju | NORMAL | 2022-01-26T11:21:30.758751+00:00 | 2022-01-26T11:47:00.728050+00:00 | 886 | false | ```\nLet dp[pos] be the minimum number of taps required to cover everything under [0,pos]. \nUsing top-down approach as we all are familiar with.\nsolve(pos) {\nThen for each i from pos-100 to pos-1 {\n if(pos>=i+r[i]) //reachable to pos from i tap so that we can use i tap for pos\n dp[pos]=min(dp[pos],so... | 3 | 0 | ['Dynamic Programming', 'Memoization'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | Jump Game || Logic, TC: O(N), SC:O(N) | jump-game-logic-tc-on-scon-by-najimali-rcdv | \nclass Solution {\n static int minTaps(int n, int[] ranges) {\n\t\tint dp[] = new int[ranges.length];\n\t\t\n\t\tfor (int i = 0; i < dp.length; i++) {\n\t\t | najimali | NORMAL | 2022-01-18T05:54:44.057737+00:00 | 2022-01-18T05:54:44.057814+00:00 | 284 | false | ```\nclass Solution {\n static int minTaps(int n, int[] ranges) {\n\t\tint dp[] = new int[ranges.length];\n\t\t\n\t\tfor (int i = 0; i < dp.length; i++) {\n\t\t\tif (ranges[i] == 0) {\n\t\t\t\tcontinue;\n\t\t\t}\n\t\t\tint left = Math.max(i - ranges[i], 0);\n\t\t\tint right = Math.min(i + ranges[i], n);\n\t\t\tdp[le... | 3 | 0 | [] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Totally Equal to Jump Game 2 -- Java | totally-equal-to-jump-game-2-java-by-fre-v01q | I tried Jump Game 2 first with the traditional BFS method and also a concise version of BFS (top-most voted).\n\nI found this question is totally the same as Ju | freya250817 | NORMAL | 2021-11-27T01:00:21.645629+00:00 | 2022-03-16T03:46:12.539650+00:00 | 397 | false | I tried Jump Game 2 first with the traditional BFS method and also a concise version of BFS `(top-most voted)`.\n\nI found this question is totally the same as Jump Game 2, where 2 steps are needed.\n1) construct an array `reached[]`, `reached[i]` means the furthest point can be reached at current position `i`\n - no... | 3 | 0 | ['Breadth-First Search'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | C++ Solution | Greedy Approach | c-solution-greedy-approach-by-ajainuary-4kz1 | Solution Approach\n\nWe can solve this problem by using a Greedy Approach. Till we reach the end, determine the largest fountain that covers this position and a | ajainuary | NORMAL | 2021-11-11T07:40:50.996716+00:00 | 2021-11-11T07:40:50.996750+00:00 | 529 | false | ## Solution Approach\n\nWe can solve this problem by using a Greedy Approach. Till we reach the end, determine the largest fountain that covers this position and activate that. Implemented in a naive manner, this will be O(n^2) but with some simple preprocessing we can do this in O(n). We maintain a <code>jumps</code> ... | 3 | 0 | [] | 2 |
minimum-number-of-taps-to-open-to-water-a-garden | Greedy | 8 line solution explained | greedy-8-line-solution-explained-by-hars-sy8p | The idea is we check the maximum ground (mx) we can water after opening the \'open\'th tap.\nwe check that in for loop.\nAfter each openth step, we set mn=mx, b | harshnadar23 | NORMAL | 2021-08-31T13:10:28.247322+00:00 | 2021-08-31T13:10:28.247365+00:00 | 376 | false | The idea is we check the maximum ground (mx) we can water after opening the \'open\'th tap.\nwe check that in for loop.\nAfter each openth step, we set mn=mx, because we need to find a tap which can water the ground from somewhere less than mn to any distance greater than mn.\n\n```\nclass Solution {\npublic:\n int ... | 3 | 0 | ['Greedy'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | C++ || O(n) Solution || Well Commented | c-on-solution-well-commented-by-anshumaa-332d | \nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n\t\n\t\t// dp array shows how far from current index you can give water, if hose is | anshumaaa | NORMAL | 2021-07-29T08:27:36.892788+00:00 | 2021-07-29T08:28:46.041772+00:00 | 450 | false | ```\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n\t\n\t\t// dp array shows how far from current index you can give water, if hose is somewhere ahead\n vector<int>dp(n+1,-1);\n for(int i=0;i<=n;i++)\n {\n int left=max(0,i-ranges[i]);\n int right=mi... | 3 | 2 | [] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | CPP | Very easy | like jump game | cpp-very-easy-like-jump-game-by-jammera5-b7g0 | \n// 1326. Minimum Number of Taps to Open to Water a Garden\n// similar to the jump game\n\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& range | jammera5 | NORMAL | 2021-07-14T06:21:11.920692+00:00 | 2021-07-14T06:21:11.920742+00:00 | 304 | false | ```\n// 1326. Minimum Number of Taps to Open to Water a Garden\n// similar to the jump game\n\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n vector<int> dp(n+1,0);\n for(int i=0; i<ranges.size(); i++){\n int piche = max(0,i-ranges[i]);\n int aage = min(n,... | 3 | 4 | ['C', 'C++'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | O(N) Time, O(1) space, C++ | on-time-o1-space-c-by-ajaysaiva-0rao | \nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n\tfor(int i=0;i<=n;i++) {\n\t\tint start = max(i-ranges[i],0);\n\t\tranges[start] = | ajaysaiva | NORMAL | 2021-05-14T15:55:19.869861+00:00 | 2021-05-14T15:55:19.869908+00:00 | 344 | false | ```\nclass Solution {\npublic:\n int minTaps(int n, vector<int>& ranges) {\n\tfor(int i=0;i<=n;i++) {\n\t\tint start = max(i-ranges[i],0);\n\t\tranges[start] = max(ranges[start],i+ranges[i]);\n\t}\n\tint curEnd = ranges[0],maxEnd = 0,ans = 1;\n\tfor(int i=0;i<=n;i++) {\n\t\tif(i>curEnd) {\n\t\t\tif(curEnd<maxEnd) {\... | 3 | 0 | [] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | GoLang DP Solution with Explanation | golang-dp-solution-with-explanation-by-n-bnq0 | Inspired By https://leetcode.com/problems/minimum-number-of-taps-to-open-to-water-a-garden/discuss/736944/O(n)-time-complexity-greedy-implementation-in-C%2B%2B- | najis | NORMAL | 2020-12-06T06:00:17.602482+00:00 | 2020-12-06T17:05:59.911424+00:00 | 232 | false | Inspired By https://leetcode.com/problems/minimum-number-of-taps-to-open-to-water-a-garden/discuss/736944/O(n)-time-complexity-greedy-implementation-in-C%2B%2B-with-comments\n\n```\nfunc minTaps(n int, ranges []int) int {\n // Find the reach of each position, NOT each tap. We do this by finding the left\n // and ... | 3 | 0 | ['Dynamic Programming', 'Greedy', 'Go'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | JAVA | O(n) time O(1) space | Simple Solution | java-on-time-o1-space-simple-solution-by-70lu | \'\'\'class Solution {\n public int minTaps(int n, int[] ranges) {\n \n \n int left,right;\n for(int i=0;i<n+1;i++)\n {\n | smasher | NORMAL | 2020-09-21T09:44:00.470974+00:00 | 2020-09-21T09:44:00.471005+00:00 | 412 | false | \'\'\'class Solution {\n public int minTaps(int n, int[] ranges) {\n \n \n int left,right;\n for(int i=0;i<n+1;i++)\n {\n if(ranges[i] != 0)\n {\n left = i-ranges[i];\n right = i+ranges[i];\n \n if(left <... | 3 | 0 | [] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Easy Sorting | Python | Top 92% Speed | easy-sorting-python-top-92-speed-by-arag-so4k | Easy Sorting | Python | Top 92% Speed\n\nThe code below takes a very simple approach to this problem, yet it manages to achieve a Top 92% Speed rating.\n\nThe a | aragorn_ | NORMAL | 2020-07-21T21:50:59.818039+00:00 | 2020-07-21T21:51:49.957089+00:00 | 535 | false | **Easy Sorting | Python | Top 92% Speed**\n\nThe code below takes a very simple approach to this problem, yet it manages to achieve a Top 92% Speed rating.\n\nThe algorithm is the following:\n\n1. We first create explicit watering ranges for each Tap, and sort them. This solves 99% of the problem :)\n\n2. By looking at... | 3 | 0 | ['Python', 'Python3'] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | Java, merge intervals, explained, clean code | java-merge-intervals-explained-clean-cod-ehs0 | This problem can be viewed as a variation of interval merge problem. Thus the approach can be same - form intervals, sort by starting coordinate, then traverse. | gthor10 | NORMAL | 2020-01-20T16:52:50.117942+00:00 | 2020-01-20T16:52:50.117982+00:00 | 301 | false | This problem can be viewed as a variation of interval merge problem. Thus the approach can be same - form intervals, sort by starting coordinate, then traverse.\n\nWe can exclude intervals with 0 distance - it doesn\'t add anything to solution. \n\nComplexity is O(nlgn) - while we iterate over each interval == range on... | 3 | 0 | ['Java'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Java greedy sol by using sort | java-greedy-sol-by-using-sort-by-cuny-66-s4e1 | Explanation: First eliminate all zero since they are useless. Then sort the array base on the first index,then the second index. After that, we get the first ta | CUNY-66brother | NORMAL | 2020-01-19T05:10:54.839212+00:00 | 2020-01-19T05:29:11.083546+00:00 | 147 | false | Explanation: First eliminate all zero since they are useless. Then sort the array base on the first index,then the second index. After that, we get the first tap by used the 1st loop. With the first tap, we try to find the next tap with the fartest end point. The next tap should overlap with the pre tap. After we find ... | 3 | 1 | [] | 1 |
minimum-number-of-taps-to-open-to-water-a-garden | Easy Java nlogn Solution with PriorityQueue | easy-java-nlogn-solution-with-priorityqu-zes9 | java\nclass Solution {\n public int minTaps(int n, int[] ranges) {\n if (ranges == null || ranges.length == 0 || n <= 0 || ranges.length != n + 1) ret | dreamyjpl | NORMAL | 2020-01-19T04:04:56.950501+00:00 | 2020-01-19T04:15:14.444996+00:00 | 648 | false | ```java\nclass Solution {\n public int minTaps(int n, int[] ranges) {\n if (ranges == null || ranges.length == 0 || n <= 0 || ranges.length != n + 1) return -1;\n Queue<int[]> pq = new PriorityQueue<>((e1, e2) -> Integer.compare(e1[0], e2[0]));\n for (int i = 0; i < ranges.length; i++) {\n ... | 3 | 0 | ['Greedy', 'Heap (Priority Queue)', 'Java'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Python heap based solution | python-heap-based-solution-by-cubicon-absj | \ndef minTaps(self, n: int, ranges: List[int]) -> int:\n opens, closes, closed = [[] for _ in range(n)], [[] for _ in range(n)], set()\n for i in | Cubicon | NORMAL | 2020-01-19T04:02:15.285445+00:00 | 2020-01-19T04:02:15.285485+00:00 | 238 | false | ```\ndef minTaps(self, n: int, ranges: List[int]) -> int:\n opens, closes, closed = [[] for _ in range(n)], [[] for _ in range(n)], set()\n for i in range(len(ranges)):\n idx = 0\n if i - ranges[i] > 0:\n idx = i - ranges[i] \n if idx < len(opens):\n ... | 3 | 0 | [] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Best C++ Solution || Dynamic Programming || Bottom Up | best-c-solution-dynamic-programming-bott-rxuf | 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 | ravikumar50 | NORMAL | 2023-10-28T07:42:49.643359+00:00 | 2023-10-28T07:42:49.643384+00:00 | 149 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['C++'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | C++ || BFS || Jump Game -> Shortest distance from 0 to N | c-bfs-jump-game-shortest-distance-from-0-cflj | 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 | Anmol_Singh098 | NORMAL | 2023-09-01T16:43:38.984943+00:00 | 2023-09-01T16:44:01.811619+00:00 | 137 | 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(100*n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(100*n)\n<!-- Add your space complexity here, ... | 2 | 0 | ['Breadth-First Search', 'Graph', 'C++'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | Java Easy Solution with comments! | java-easy-solution-with-comments-by-kris-fftg | 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 | KrishnaaV | NORMAL | 2023-08-31T19:48:19.107040+00:00 | 2023-08-31T19:48:19.107061+00:00 | 59 | 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 | ['Java'] | 0 |
minimum-number-of-taps-to-open-to-water-a-garden | using Intervals | using-intervals-by-tanya-1109-6g2i | Intuition\nThis Problem can be solved using intervals and sorting\n\n# Approach\n\n1. An intervals list is created to store the intervals that each tap can cove | Tanya-1109 | NORMAL | 2023-08-31T19:48:16.017353+00:00 | 2023-08-31T19:48:16.017378+00:00 | 10 | false | # Intuition\nThis Problem can be solved using intervals and sorting\n\n# Approach\n\n1. An `intervals` list is created to store the intervals that each tap can cover. The interval for each tap is calculated as `[max(0, i - ranges[i]), min(n, i + ranges[i])]`, ensuring that it doesn\'t exceed the boundaries of the garde... | 2 | 0 | ['Array', 'Python3'] | 0 |
first-letter-to-appear-twice | Java 5-line solution👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍 | java-5-line-solution-by-ouyang2021-onwc | Upvote please, if it helps. \nThank you!\n\npublic char repeatedCharacter(String s) {\n Set<Character> set = new HashSet();\n for (char c : s.toC | Ouyang2021 | NORMAL | 2022-07-27T20:00:44.912997+00:00 | 2022-07-27T21:05:05.521662+00:00 | 4,626 | false | Upvote please, if it helps. \nThank you!\n```\npublic char repeatedCharacter(String s) {\n Set<Character> set = new HashSet();\n for (char c : s.toCharArray()) \n if (!set.add(c)) \n return c;\n\t\t\t\t\n return \'a\';// can\'t reach to this line, because there must be a... | 42 | 1 | ['Java'] | 10 |
first-letter-to-appear-twice | CPP | Easy | O(1) Time | O(26) Space | cpp-easy-o1-time-o26-space-by-amirkpatna-v74e | \nclass Solution {\npublic:\n char repeatedCharacter(string s) {\n vector<int> v(26);\n for(char c:s){\n v[c-\'a\']++;\n | amirkpatna | NORMAL | 2022-07-24T04:04:22.887551+00:00 | 2022-08-17T01:55:37.289533+00:00 | 5,161 | false | ```\nclass Solution {\npublic:\n char repeatedCharacter(string s) {\n vector<int> v(26);\n for(char c:s){\n v[c-\'a\']++;\n if(v[c-\'a\']>1)return c;\n }\n return \'a\';\n }\n};\n```\n\n**Update** :\n```\nclass Solution {\npublic:\n char repeatedCharacter(strin... | 28 | 0 | ['C++'] | 11 |
first-letter-to-appear-twice | ✅Python || Map || Easy Approach | python-map-easy-approach-by-chuhonghao01-kopd | Algorithm:\n(1) use setS to store the letter have ever seen;\n(2) one pass the input string:\n(3) if x is already in setS:\n\t\tx is what we need (it is the fir | chuhonghao01 | NORMAL | 2022-07-24T04:02:41.850260+00:00 | 2022-08-23T02:22:29.027617+00:00 | 3,544 | false | Algorithm:\n(1) use setS to store the letter have ever seen;\n(2) one pass the input string:\n(3) if x is already in setS:\n\t\tx is what we need (it is the first occurrence twice).\n(4) else (x is not in setS):\n\t\twe put the x into the setS to store it.\n```\nclass Solution:\n def repeatedCharacter(self, s: str) ... | 21 | 0 | ['Python', 'Python3'] | 4 |
first-letter-to-appear-twice | ✅C++ | ✅Simple Solution | ✅Use Hashmap | c-simple-solution-use-hashmap-by-yash2ar-75qa | \nclass Solution \n{\npublic:\n char repeatedCharacter(string s) \n {\n unordered_map<char, int> mp; //for storing occurrences of char\n \n | Yash2arma | NORMAL | 2022-07-24T04:03:58.767415+00:00 | 2022-07-24T04:03:58.767507+00:00 | 3,101 | false | ```\nclass Solution \n{\npublic:\n char repeatedCharacter(string s) \n {\n unordered_map<char, int> mp; //for storing occurrences of char\n \n char ans;\n for(auto it:s)\n {\n if(mp.find(it) != mp.end()) //any char which comes twice first will be the ans; \n ... | 18 | 1 | ['String', 'C', 'C++'] | 2 |
first-letter-to-appear-twice | Return When First seen | return-when-first-seen-by-xxvvpp-vp0z | C++\n char repeatedCharacter(string s) {\n int cnt[26]{};\n for(char ch:s) if(++cnt[ch-\'a\']==2) return ch;\n return 0;\n }\n \n# | xxvvpp | NORMAL | 2022-07-24T04:01:44.160667+00:00 | 2022-07-24T04:01:44.160696+00:00 | 2,581 | false | # C++\n char repeatedCharacter(string s) {\n int cnt[26]{};\n for(char ch:s) if(++cnt[ch-\'a\']==2) return ch;\n return 0;\n }\n \n# Java\n public char repeatedCharacter(String s) {\n int cnt[]= new int[26];\n for(char ch:s.toCharArray()) if(++cnt[ch-\'a\']==2) return ch;\... | 14 | 0 | ['C', 'Java'] | 6 |
first-letter-to-appear-twice | easy solution in c++ using unordered_map | easy-solution-in-c-using-unordered_map-b-ztuc | \nclass Solution {\npublic:\n char repeatedCharacter(string s) {\n int n=s.size();\n int i;\n unordered_map<char,int>map;\n for(i | vinaykumar333j | NORMAL | 2024-10-22T07:49:45.898924+00:00 | 2024-10-22T07:50:07.041234+00:00 | 341 | false | ```\nclass Solution {\npublic:\n char repeatedCharacter(string s) {\n int n=s.size();\n int i;\n unordered_map<char,int>map;\n for(i=0;i<n;i++){\n if(map.find(s[i])!=map.end()){\n return s[i];\n }\n map[s[i]];\n }\n return \'0\... | 12 | 0 | ['C'] | 0 |
first-letter-to-appear-twice | Count Chars | count-chars-by-votrubac-m9f7 | C++\ncpp\nchar repeatedCharacter(string s) {\n int seen[128] = {}, i = 0;\n for (i = 0; i < s.size() && ++seen[s[i]] < 2; ++i);\n return s[i];\n}\n | votrubac | NORMAL | 2022-07-24T04:01:57.919638+00:00 | 2022-07-24T04:01:57.919668+00:00 | 1,573 | false | **C++**\n```cpp\nchar repeatedCharacter(string s) {\n int seen[128] = {}, i = 0;\n for (i = 0; i < s.size() && ++seen[s[i]] < 2; ++i);\n return s[i];\n}\n``` | 12 | 0 | [] | 5 |
first-letter-to-appear-twice | C++ solutions | c-solutions-by-infox_92-5gu2 | \nclass Solution {\npublic:\n char repeatedCharacter(string s) {\n vector<bool> v(26);\n for(char c:s){\n if(v[c-\'a\'])return c;\n\ | Infox_92 | NORMAL | 2022-11-12T06:13:54.907395+00:00 | 2022-11-12T06:13:54.907440+00:00 | 1,159 | false | ```\nclass Solution {\npublic:\n char repeatedCharacter(string s) {\n vector<bool> v(26);\n for(char c:s){\n if(v[c-\'a\'])return c;\n\t\t\tv[c-\'a\']=true;\n }\n return \'a\';\n }\n};\n``` | 10 | 0 | ['C', 'C++'] | 0 |
first-letter-to-appear-twice | JavaScript Set less and faster than 85% | javascript-set-less-and-faster-than-85-b-x3rj | \nconst repeatedCharacter = s => {\n const set = new Set();\n \n for (let i = 0; i < s.length; i += 1) {\n const currentLetter = s[i]\n i | seredulichka | NORMAL | 2022-08-24T21:59:55.663403+00:00 | 2022-08-24T21:59:55.663435+00:00 | 1,311 | false | ```\nconst repeatedCharacter = s => {\n const set = new Set();\n \n for (let i = 0; i < s.length; i += 1) {\n const currentLetter = s[i]\n if(set.has(currentLetter)) {\n return currentLetter \n }\n set.add(currentLetter);\n }\n};\n``` | 10 | 0 | ['JavaScript'] | 0 |
first-letter-to-appear-twice | Python Simple Code (Hash Table , Set, List) | python-simple-code-hash-table-set-list-b-xy2y | If you got help from this,... Plz Upvote .. it encourage me\n\n# Code\n> # HashTable\n\nclass Solution:\n def repeatedCharacter(self, s: str) -> str:\n | vvivekyadav | NORMAL | 2023-10-05T17:50:37.199659+00:00 | 2023-10-05T17:52:38.067488+00:00 | 292 | false | **If you got help from this,... Plz Upvote .. it encourage me**\n\n# Code\n> # HashTable\n```\nclass Solution:\n def repeatedCharacter(self, s: str) -> str:\n d = {}\n for char in s:\n if char in d:\n return char\n else:\n d[char] = 1\n\n ... | 8 | 0 | ['Hash Table', 'String', 'Counting', 'Python', 'Python3'] | 0 |
first-letter-to-appear-twice | Here Simple✔ and Easy ✌Solution With Clear Explanation|| 100% beats✨ | here-simple-and-easy-solution-with-clear-0vvd | Intuition\nHere im Initializing an array count of 26 and set the values 0\'s for all index and traverse the string and adding the char in the array by index and | sanjuzzz28 | NORMAL | 2024-10-14T17:06:11.948370+00:00 | 2024-10-14T17:09:55.591107+00:00 | 341 | false | # Intuition\nHere im Initializing an array count of 26 and set the values 0\'s for all index and traverse the string and adding the char in the array by index and if the index is equal to two return the character of the first letter appear twice string.\n\n# Approach\n##### ==>Character Count Array Initialization:\nc[2... | 7 | 0 | ['String', 'C', 'C++'] | 0 |
first-letter-to-appear-twice | ✅C++ || ✅Simple || ✅Commented || ✅Easy to Understand | c-simple-commented-easy-to-understand-by-8apr | \nclass Solution {\npublic:\n char repeatedCharacter(string s) \n {\n map<char,int> mp; // for storing the frequency of char\n | mayanksamadhiya12345 | NORMAL | 2022-07-24T04:00:50.338457+00:00 | 2022-07-24T04:17:17.858992+00:00 | 896 | false | ```\nclass Solution {\npublic:\n char repeatedCharacter(string s) \n {\n map<char,int> mp; // for storing the frequency of char\n \n for(auto i: s)\n {\n // if we found that this char is already present menas it is our ans return it\n if(mp[i]==1)\n ... | 7 | 0 | [] | 0 |
first-letter-to-appear-twice | Finding the First Repeated Character in a String Using Set | finding-the-first-repeated-character-in-aklmt | IntuitionApproachTo solve the problem of finding the first repeated character in a string, the idea is to keep track of characters that have already been seen. | lalith_chandra | NORMAL | 2025-02-23T03:58:55.454095+00:00 | 2025-02-23T03:58:55.454095+00:00 | 201 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
To solve the problem of finding the first repeated character in a string, the idea is to keep track of characters that have already been seen. As soon as a character is enc... | 6 | 0 | ['C++'] | 1 |
first-letter-to-appear-twice | C++ ✅ Easy to Understand O(n) Solution 🔥 | c-easy-to-understand-on-solution-by-_sxr-fbln | Intuition\nThe given code aims to find the first character in the string that appears twice. It uses an unordered map to keep track of the frequency of each cha | _sxrthakk | NORMAL | 2024-07-24T13:15:20.301985+00:00 | 2024-07-24T13:15:20.302019+00:00 | 469 | false | # Intuition\nThe given code aims to find the first character in the string that appears twice. It uses an unordered map to keep track of the frequency of each character as it iterates through the string. As soon as it detects a character that has appeared twice, it immediately returns that character.\n\n# Approach\n1. ... | 6 | 0 | ['C++'] | 0 |
first-letter-to-appear-twice | bit manipulation | O(n) time | O(1) space | no Map, hash, array | bit-manipulation-on-time-o1-space-no-map-fml0 | Instead of allocating 26 boolean array, use 26 bits as flag. This reduces many spaces from O(26) to O(1).\n\n### Python\npython\nclass Solution:\n def repeat | wf9a5m75 | NORMAL | 2022-08-16T01:20:48.378578+00:00 | 2022-08-16T01:25:26.790742+00:00 | 331 | false | Instead of allocating 26 boolean array, use `26 bits` as flag. This reduces many spaces from `O(26)` to `O(1)`.\n\n### Python\n```python\nclass Solution:\n def repeatedCharacter(self, s: str) -> str:\n appear = 0\n \n ordA = ord("a")\n for char in s:\n mask = 1 << (ord(char) - ... | 6 | 0 | ['Bit Manipulation', 'Python', 'Java'] | 1 |
first-letter-to-appear-twice | ✅💯Beats 100.00% of users || Best solution in C++/C#/C || Beginner friendly👍🤝 | beats-10000-of-users-best-solution-in-cc-egbn | PLEASE DO UPVOTE\n\n\n# Intuition\n Describe your first thoughts on how to solve this problem. \nThe code finds the first repeated character in a given string b | anand18402 | NORMAL | 2024-01-20T09:29:31.289168+00:00 | 2024-01-20T09:29:31.289196+00:00 | 389 | false | # PLEASE DO UPVOTE\n\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code finds the first repeated character in a given string by maintaining ... | 5 | 0 | ['String', 'Ordered Map', 'C', 'String Matching', 'C++', 'C#'] | 1 |
first-letter-to-appear-twice | O(n) || C++||Beats 100% ||Easy 4 line code ||counting || Simple Solution ||With Explanation | on-cbeats-100-easy-4-line-code-counting-tk09w | Intuition\n Describe your first thoughts on how to solve this problem. \n\n\n# Approach\n Describe your approach to solving the problem. \ncount the frequency o | nikhil_2002_singh | NORMAL | 2023-05-16T12:20:17.287079+00:00 | 2023-05-16T16:32:46.531913+00:00 | 65 | 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. -->\ncount the frequency of ech element \nwhen the frequeny becomes equals to 2 return the character.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity h... | 5 | 0 | ['Hash Table', 'String', 'Counting', 'C++'] | 0 |
first-letter-to-appear-twice | Easy Java Solution using HashSet | easy-java-solution-using-hashset-by-abhi-l06y | \nclass Solution {\n public char repeatedCharacter(String s) {\n HashSet<Character> set = new HashSet<>();//Create a set of characters\n for(in | Abhinav_0561 | NORMAL | 2022-10-10T03:43:12.567367+00:00 | 2022-10-10T03:43:12.567395+00:00 | 2,525 | false | ```\nclass Solution {\n public char repeatedCharacter(String s) {\n HashSet<Character> set = new HashSet<>();//Create a set of characters\n for(int i = 0 ; i < s.length() ; i++){\n if(set.contains(s.charAt(i))) return s.charAt(i);//If the set already contains the current character, then it i... | 5 | 0 | ['Java'] | 1 |
first-letter-to-appear-twice | C++ || Simple Solution || O(26) Space || Easy | c-simple-solution-o26-space-easy-by-sura-dseo | C++ Code:\n\nclass Solution {\npublic:\n //using O(26) space \n char repeatedCharacter(string s) {\n vector<int> cnt(26,0);\n \n char | suraj_jha989 | NORMAL | 2022-07-25T02:47:08.956331+00:00 | 2022-07-25T02:48:28.902837+00:00 | 218 | false | # C++ Code:\n```\nclass Solution {\npublic:\n //using O(26) space \n char repeatedCharacter(string s) {\n vector<int> cnt(26,0);\n \n char ans;\n for(char ch: s){\n if(cnt[ch-\'a\'] == 1){\n ans = ch;\n break;\n }\n cnt[ch-... | 5 | 0 | ['Array', 'C', 'C++'] | 0 |
first-letter-to-appear-twice | Count and check | Easy and simple | count-and-check-easy-and-simple-by-nadar-g49j | Code is self explanatory\n\n\nclass Solution:\n def repeatedCharacter(self, s: str) -> str:\n occurences = defaultdict(int)\n for char in s:\n | nadaralp | NORMAL | 2022-07-24T04:01:46.976658+00:00 | 2022-07-24T04:01:46.976688+00:00 | 921 | false | Code is self explanatory\n\n```\nclass Solution:\n def repeatedCharacter(self, s: str) -> str:\n occurences = defaultdict(int)\n for char in s:\n occurences[char] += 1\n if occurences[char] == 2:\n return char\n``` | 5 | 0 | ['Python'] | 0 |
first-letter-to-appear-twice | Simple Java Code ☠️ | simple-java-code-by-abhinandannaik1717-6vw8 | \n# Code\n\nclass Solution {\n public char repeatedCharacter(String s) {\n int n = s.length();\n HashSet<Character> set = new HashSet<>();\n | abhinandannaik1717 | NORMAL | 2024-07-18T18:50:55.893254+00:00 | 2024-07-18T18:50:55.893285+00:00 | 423 | false | \n# Code\n```\nclass Solution {\n public char repeatedCharacter(String s) {\n int n = s.length();\n HashSet<Character> set = new HashSet<>();\n for(int i=0;i<n;i++){\n if(set.contains(s.charAt(i))){\n return s.charAt(i);\n }\n set.add(s.charAt(i));... | 4 | 0 | ['Hash Table', 'String', 'Java'] | 1 |
first-letter-to-appear-twice | Easy Solution Using HashMap [ JAVA ] [ 100% ] [ 0ms ] | easy-solution-using-hashmap-java-100-0ms-fy35 | Approach\n1. Initialize a HashMap st to store characters and their counts.\n2. Iterate through each character in the input string s.\n3. Check if the character | RajarshiMitra | NORMAL | 2024-04-20T10:56:51.376680+00:00 | 2024-06-16T08:32:09.217304+00:00 | 71 | false | # Approach\n1. Initialize a HashMap `st` to store characters and their counts.\n2. Iterate through each character in the input string `s`.\n3. Check if the character is already present in the HashMap:\n - If it is present, return the character as it\'s the first repeated character.\n - If it\'s not present, add it ... | 4 | 0 | ['Java'] | 0 |
first-letter-to-appear-twice | 100% Beginners Friendly | 100-beginners-friendly-by-bhaskarkumar07-74b5 | \n# Approach\n Describe your approach to solving the problem. \n as the hashmap value of the key went to 2 return the char\n\n# Complexity\n- Time comp | bhaskarkumar07 | NORMAL | 2023-04-18T18:55:48.574796+00:00 | 2023-04-18T18:55:48.574851+00:00 | 655 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\n as the hashmap value of the key went to 2 return the char\n\n# Complexity\n- Time complexity:O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n... | 4 | 0 | ['Java'] | 1 |
first-letter-to-appear-twice | Solution in C++ | solution-in-c-by-ashish_madhup-vdf0 | 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 | 2023-04-18T17:48:29.482312+00:00 | 2023-04-18T17:48:29.482363+00:00 | 412 | 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. $$O(n)$$ --... | 4 | 0 | ['C++'] | 0 |
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