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word-ladder-ii | 25 ms | Use less memory than 97% | BFS + DFS | Golang | | 25-ms-use-less-memory-than-97-bfs-dfs-go-o53f | Let consider each string in wordList as a vertex on a graph, and if two strings differ by at most one character, we will connect an edge with weight 1 between t | mbfibat | NORMAL | 2022-08-14T08:40:53.377936+00:00 | 2022-08-14T08:42:05.353066+00:00 | 474 | false | - Let consider each string in ```wordList``` as a vertex on a graph, and if two strings differ by at most one character, we will connect an edge with weight ```1``` between them. The problem become: "Find the shortest path, single source, single destination, on a graph where each edge\'s weight is equal to 1".\n\n- The... | 5 | 0 | ['Depth-First Search', 'Binary Search Tree', 'Go'] | 0 |
word-ladder-ii | C++ | Floyd-Warshall + BFS | O(n^3) | explanation | c-floyd-warshall-bfs-on3-explanation-by-kbcyg | \nclass Solution {\npublic:\n vector <int> isc[501];\n long dis[501][501];\n int n,tar,as;\n bool jc(string x,string y){\n int v=0;\n | SunGod1223 | NORMAL | 2022-08-14T05:05:52.716261+00:00 | 2022-08-14T11:48:54.330411+00:00 | 421 | false | ```\nclass Solution {\npublic:\n vector <int> isc[501];\n long dis[501][501];\n int n,tar,as;\n bool jc(string x,string y){\n int v=0;\n for(int i=0;i<x.size();++i)\n if(x[i]!=y[i])++v;\n return v==1;\n }\n vector<vector<string>> findLadders(string bW, string eW, vector... | 5 | 0 | ['Breadth-First Search'] | 0 |
word-ladder-ii | Java Solution using BFS and memorized DFS | java-solution-using-bfs-and-memorized-df-075w | Step1: use BFS to build a Ladder Graph start from beginWord layer and end to endWord layer\nStep2: use memorized DFS to traverse the graph and found all the sho | 1005155946 | NORMAL | 2022-07-07T10:14:54.298885+00:00 | 2022-07-07T10:14:54.298925+00:00 | 1,079 | false | Step1: use BFS to build a Ladder Graph start from beginWord layer and end to endWord layer\nStep2: use memorized DFS to traverse the graph and found all the shortest route\n``` java\nclass Solution {\n Map<String, List<List<String>>> buf;\n public List<List<String>> findLadders(String beginWord, String endWord, L... | 5 | 1 | ['Backtracking', 'Depth-First Search', 'Breadth-First Search', 'Java'] | 1 |
word-ladder-ii | Python - BFS - very simple | python-bfs-very-simple-by-edgoll-3tpo | \n\n def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:\n \n wordList.append(beginWord)\n nei | edgoll | NORMAL | 2022-01-26T05:20:10.511728+00:00 | 2022-01-26T05:20:10.511760+00:00 | 420 | false | \n\n def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:\n \n wordList.append(beginWord)\n nei = collections.defaultdict(list)\n \n ... | 5 | 0 | ['Python'] | 1 |
word-ladder-ii | Simple Python BFS solution with 20 lines of code and explanation, beats 80%! | simple-python-bfs-solution-with-20-lines-mrzz | Steps:\n1. Construct an adjecncy list from the wordList\n2. Perform normal BFS with a queue (every element is a tuple of currentWord, the list of words from beg | darshangowda0 | NORMAL | 2020-08-02T06:16:31.179273+00:00 | 2020-08-02T06:16:31.179306+00:00 | 470 | false | Steps:\n1. Construct an adjecncy list from the wordList\n2. Perform normal BFS with a queue (every element is a tuple of currentWord, the list of words from beginWord to currentWord)\n3. Everytime you find the endWord, add the list built so far to the output list!\n\n**Main STEP to help aviod TLE:**\nDo not visit the s... | 5 | 0 | [] | 3 |
word-ladder-ii | C# BFS | c-bfs-by-user638-rcn7 | \npublic class Solution\n{\n public IList<IList<string>> FindLadders(string beginWord, string endWord, IList<string> wordList)\n {\n var map = new | user638 | NORMAL | 2020-04-16T16:18:11.820378+00:00 | 2020-04-16T16:19:13.934861+00:00 | 556 | false | ```\npublic class Solution\n{\n public IList<IList<string>> FindLadders(string beginWord, string endWord, IList<string> wordList)\n {\n var map = new Dictionary<string, List<string>>();\n wordList.Add(beginWord);\n wordList = wordList.Distinct().ToList();\n for (var i = 0; i < wordList... | 5 | 0 | ['Breadth-First Search'] | 1 |
word-ladder-ii | Intuitive Javascript Solution with BFS & DFS | intuitive-javascript-solution-with-bfs-d-rfwf | \n/**\n * @param {string} beginWord\n * @param {string} endWord\n * @param {string[]} wordList\n * @return {string[][]}\n */\nvar findLadders = function(beginWo | dawchihliou | NORMAL | 2020-04-12T10:25:21.966823+00:00 | 2020-04-12T10:25:21.966862+00:00 | 937 | false | ```\n/**\n * @param {string} beginWord\n * @param {string} endWord\n * @param {string[]} wordList\n * @return {string[][]}\n */\nvar findLadders = function(beginWord, endWord, wordList) {\n if (!wordList.includes(endWord)) {\n return [];\n }\n \n /**\n * Create an adjacency list and run bfs to cr... | 5 | 0 | ['Backtracking', 'Depth-First Search', 'Breadth-First Search', 'JavaScript'] | 3 |
word-ladder-ii | C++ - solution accepted - only 1-way BFS with comments. | c-solution-accepted-only-1-way-bfs-with-cblz9 | The solution passes all tests and accepted.\nAppreciate feedback on how to optimize further.\n\n\nclass Solution {\n public:\n /*\n The idea is to | dcbusc2012 | NORMAL | 2019-11-28T07:43:15.079036+00:00 | 2019-11-28T07:43:15.079074+00:00 | 947 | false | The solution passes all tests and accepted.\nAppreciate feedback on how to optimize further.\n\n```\nclass Solution {\n public:\n /*\n The idea is to run a BFS from beginWord to endWord, while keeping track of the path.\n Once the currWord == endWord, we have found the shortest path.\n The ne... | 5 | 0 | ['C++'] | 3 |
minimum-number-of-days-to-disconnect-island | Python [you need at most 2 days] | python-you-need-at-most-2-days-by-gsan-fyec | The tricky part of this question is to notice that, you can disconnect any given island formation in at most 2 days (and you need to convince yourself that this | gsan | NORMAL | 2020-08-30T04:02:58.911391+00:00 | 2024-10-12T10:22:22.584903+00:00 | 10,868 | false | The tricky part of this question is to notice that, you can disconnect any given island formation in at most 2 days (and you need to convince yourself that this is true).\n\n**Day 0:** Check islands at day 0, return 0 if you have less than or greater than one island.\n\n**Day 1**: If not, try to add water at any given ... | 200 | 3 | ['Python3'] | 35 |
minimum-number-of-days-to-disconnect-island | DFS c++ clean code [with explanation] | dfs-c-clean-code-with-explanation-by-m05-keil | Observation\n## ans <= 2\nans is always less-equal to 2\n## why?\nfor any island we can remove the two blocks around the bottom left corner to make it disconnec | m05s | NORMAL | 2020-08-30T04:02:32.756442+00:00 | 2020-09-01T18:03:45.051353+00:00 | 10,522 | false | # Observation\n## **ans <= 2**\nans is always less-equal to 2\n## **why?**\nfor any island we can remove the two blocks around the bottom left corner to make it disconnected\n```\nx x x\nx . x\nx x x\n```\ncan be changed to\n```\nx x x\nx . .\nx . x\n```\nif you still don\'t get it read this: [comment](https://leetcode... | 91 | 3 | [] | 9 |
minimum-number-of-days-to-disconnect-island | Easy to Understand | Step by Step Detailed Explaination | Beginner Friendly | Depth-First Search | easy-to-understand-step-by-step-detailed-fzmv | Problem Statement\nWe are given an m x n binary grid grid where 1 represents land and 0 represents water. An island is defined as a maximal 4-directionally conn | tanishqsingh | NORMAL | 2024-08-11T01:12:07.143842+00:00 | 2024-08-11T01:12:07.143869+00:00 | 27,350 | false | # Problem Statement\nWe are given an `m x n` binary grid `grid` where `1` represents land and `0` represents water. An island is defined as a maximal 4-directionally connected group of `1\'s`. The grid is considered connected if there is exactly one island; otherwise, it is disconnected.\n\nOur task is to determine the... | 88 | 2 | ['Depth-First Search', 'Graph', 'C++', 'Java', 'Python3', 'JavaScript'] | 8 |
minimum-number-of-days-to-disconnect-island | ✅Beats 100% -Explained with [ Video ] -C++/Java - DFS - Explained in Detail | beats-100-explained-with-video-cjava-dfs-s21z | \n\n# YouTube Video Explanation:\n\n ### To watch the video please click on "Watch On Youtube" option available the left bottom corner of the thumbnail. \n\nIf | lancertech6 | NORMAL | 2024-08-11T00:44:06.206178+00:00 | 2024-08-11T00:44:06.206202+00:00 | 8,948 | false | \n\n# YouTube Video Explanation:\n\n<!-- ### To watch the video please click on "Watch On Youtube" option available the left bottom corner of the thumbnail. -->\n\n**If you want a vid... | 61 | 6 | ['Array', 'Depth-First Search', 'Breadth-First Search', 'Matrix', 'Strongly Connected Component', 'C++', 'Java'] | 4 |
minimum-number-of-days-to-disconnect-island | [Java] Easy to understand | java-easy-to-understand-by-mayank12559-ebub | \nclass Solution {\n public int minDays(int[][] grid) {\n if(noOfIsland(grid) != 1){\n return 0;\n }\n for(int i=0;i<grid.len | mayank12559 | NORMAL | 2020-08-30T04:08:50.626967+00:00 | 2020-08-30T04:08:50.627011+00:00 | 3,068 | false | ```\nclass Solution {\n public int minDays(int[][] grid) {\n if(noOfIsland(grid) != 1){\n return 0;\n }\n for(int i=0;i<grid.length;i++){\n for(int j=0;j<grid[0].length;j++){\n if(grid[i][j] == 1){\n grid[i][j] = 0;\n if(... | 38 | 1 | [] | 4 |
minimum-number-of-days-to-disconnect-island | ✅ Easy 3 step C++/ Java / Py / JS Solution | DFS | With Visualization🏆| | easy-3-step-c-java-py-js-solution-dfs-wi-qpj5 | Steps:\n1. Check if the Grid is Already Disconnected:\n - Use a function to count the number of islands in the grid.\n - If there is already more than one i | dev_yash_ | NORMAL | 2024-08-11T01:28:25.878442+00:00 | 2024-08-11T03:43:12.339586+00:00 | 4,050 | false | ### Steps:\n1. **Check if the Grid is Already Disconnected:**\n - Use a function to count the number of islands in the grid.\n - If there is already more than one island, the grid is disconnected, so return `0` days.\n\n2. **Try Removing One Land Cell:**\n - Iterate over each land cell in the grid.\n - For each... | 26 | 1 | ['Depth-First Search', 'Breadth-First Search', 'Graph', 'C', 'Java', 'Python3', 'JavaScript'] | 9 |
minimum-number-of-days-to-disconnect-island | [Python / Golang] There are only 3 possible answers! | python-golang-there-are-only-3-possible-ocm9j | \n## Idea\nHow to make the grid disconnected?\nWe can tell from the first official example that, the worst situation we may get into is to take 2 steps and sepa | yanrucheng | NORMAL | 2020-08-30T04:00:49.247784+00:00 | 2020-08-30T04:09:59.738439+00:00 | 2,744 | false | \n## Idea\n**How to make the grid disconnected?**\nWe can tell from the first official example that, the worst situation we may get into is to take 2 steps and separate a single island out.\nMore specifically, there are 3 situation.\n1. The number of island on the grid is not 1.\n\t- return 0\n1. The number of island o... | 25 | 2 | [] | 8 |
minimum-number-of-days-to-disconnect-island | C++ At most 2 removal; Try removing each one and see if the graph is disconnected. | c-at-most-2-removal-try-removing-each-on-q71u | See my latest update in repo LeetCode\n\n## Solution 1.\n\nYou need at most two days to separate out a 1 at the corner with all other 1s. So the answer is one o | lzl124631x | NORMAL | 2020-08-30T04:30:15.418062+00:00 | 2020-08-30T19:19:59.782215+00:00 | 2,568 | false | See my latest update in repo [LeetCode](https://github.com/lzl124631x/LeetCode)\n\n## Solution 1.\n\nYou need at most two days to separate out a `1` at the corner with all other `1`s. So the answer is one of `0`, `1`, `2`.\n\nIf the graph is already disconnected, return `0`.\n\nFor each `1`, see if removing it can disc... | 20 | 0 | [] | 3 |
minimum-number-of-days-to-disconnect-island | Most Optimal Approach (Articulation Point) || TC: O(n * m) | most-optimal-approach-articulation-point-j8hd | Intuition\n1. If there are no islands or more than 1 island, than the answer is 0, as we don\'t need to remove anything.\n2. If we can disconnct the island by r | WeirdLogic | NORMAL | 2023-11-29T19:02:52.404848+00:00 | 2023-11-29T19:05:05.741489+00:00 | 1,196 | false | # Intuition\n1. If there are no islands or more than 1 island, than the answer is 0, as we don\'t need to remove anything.\n2. If we can disconnct the island by removing exactly 1 cell, the answer is 1 (will be explaining it shortly in Case 2 of **Approach**).\n3. At last, we can always disconnect an island by removing... | 18 | 0 | ['Depth-First Search', 'Graph', 'C++'] | 2 |
minimum-number-of-days-to-disconnect-island | DFS + test vertices if any neighbor with deg<=2||7ms Beats 93.10% | dfs-test-vertices-if-any-neighbor-with-d-lydd | Intuition\n Describe your first thoughts on how to solve this problem. \nNot optimal, but much better than brute force.\n\nJust consider the vertices with degre | anwendeng | NORMAL | 2024-08-11T08:18:33.808770+00:00 | 2024-08-11T08:36:50.688940+00:00 | 1,326 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nNot optimal, but much better than brute force.\n\nJust consider the vertices with degree<=2 & its all adjacent vertices.\n\nA vertex here is a cell with 1. A vertex has at most 4 adjacent vertices.\n# Approach\n<!-- Describe your approach... | 14 | 0 | ['Array', 'Depth-First Search', 'Matrix', 'C++'] | 2 |
minimum-number-of-days-to-disconnect-island | [Java] O(MN) 2ms articulation point/bridge approach | java-omn-2ms-articulation-pointbridge-ap-vzob | All the posted solutions are using the similar basic flow: \n1. If there is no island or more than 1 island, return 0;\n2. If there is only one land, return 1;\ | caohuicn | NORMAL | 2020-08-31T04:57:43.149603+00:00 | 2020-09-01T00:14:58.620663+00:00 | 1,607 | false | All the posted solutions are using the similar basic flow: \n1. If there is no island or more than 1 island, return 0;\n2. If there is only one land, return 1;\n3. If any single cell could serve as the cut point ( divide 1 island into 2 islands), return 1;\n4. Otherwise, return 2 ( I haven\'t found a formal proof yet).... | 13 | 0 | [] | 2 |
minimum-number-of-days-to-disconnect-island | [Python] O(N*M) | Articulation Points | Beats 100% | python-onm-articulation-points-beats-100-4lbt | This question can be divided into 3 cases :\n\n1) The most number of cells to be removed in the worst case will be 2, explained later.\n2) If there is an articu | zypher27 | NORMAL | 2020-08-30T17:17:11.817957+00:00 | 2020-09-16T12:47:48.612007+00:00 | 1,269 | false | This question can be divided into 3 cases :\n\n1) The most number of cells to be removed in the worst case will be 2, explained later.\n2) If there is an articulation point, the answer will be 1.\n3) The matrix is already divided or there are no 1\'s, the answer is 0.\n\n\nSo, if you didnt understand the worst case sce... | 12 | 0 | ['Python'] | 5 |
minimum-number-of-days-to-disconnect-island | C++ easy to understand with explanation BFS | c-easy-to-understand-with-explanation-bf-qih4 | Looking at the question, it might give an impression as a very difficult question. This question actually is lot easier than it looks. On careful observation, y | mohithakhil | NORMAL | 2020-08-30T05:39:31.947800+00:00 | 2020-08-30T05:42:41.348303+00:00 | 1,115 | false | Looking at the question, it might give an impression as a very difficult question. This question actually is lot easier than it looks. On careful observation, you can notice that the answer can only lie between 0 and 2. \n\nWhen does answer can be 0? If the given matrix already contains more than one island or if the m... | 12 | 3 | ['Breadth-First Search', 'C'] | 3 |
minimum-number-of-days-to-disconnect-island | Beginner Friendly Solution with Brute Force DFS + Visualization + Step Explanation 😊😊 | beginner-friendly-solution-with-brute-fo-zs7h | Intuition\nThe problem asks us to find the minimum number of days required to disconnect a single island in a grid. The intuition is to:\n\n1. Identify if there | briancode99 | NORMAL | 2024-08-10T18:25:43.406718+00:00 | 2024-08-10T18:25:43.406748+00:00 | 1,285 | false | # Intuition\nThe problem asks us to find the minimum number of days required to disconnect a single island in a grid. The intuition is to:\n\n1. Identify if there are multiple islands: If there are multiple islands, it\'s already disconnected, and we need 0 days.\n2. Disconnect the single island: We need to find a way ... | 10 | 0 | ['Array', 'Depth-First Search', 'Matrix', 'JavaScript'] | 3 |
minimum-number-of-days-to-disconnect-island | Articulation point | articulation-point-by-shivam565-6dus | there are only 3 possible cases:\n\n if the graph already has more than 1 components\n if the graph has atleast one articulation point\n if the graph does\'t fo | shivam565 | NORMAL | 2022-08-28T09:56:30.718507+00:00 | 2022-08-28T09:59:05.068957+00:00 | 915 | false | there are only 3 possible cases:\n\n* **if the graph already has more than 1 components**\n* **if the graph has atleast one articulation point**\n* **if the graph does\'t follow those two rules**\n\n\n<br>\n<br>\n<br>\nin the first case the ans is 0 becuse we don\'t need to convert any 1 to 0 because graph is already d... | 10 | 0 | ['C'] | 1 |
minimum-number-of-days-to-disconnect-island | Simplest explanation (with diagram). Articulation Point - Tarjan Algo. Faster than 100% | simplest-explanation-with-diagram-articu-psfz | Now, first of all how many solutions can be there, I will say only 0,1 and 2, why ?\n\n### 1. for \'0\'\na. there is a initially more than one connected compone | himanshuaswal | NORMAL | 2021-06-24T10:08:23.965994+00:00 | 2021-06-24T10:08:23.966041+00:00 | 2,202 | false | # Now, first of all how many solutions can be there, I will say only 0,1 and 2, **why** ?\n\n### 1. for \'0\'\na. there is a initially more than one connected components.\nwhich is quite easy to find, just using a dfs/bfs.\nb. there is no land (all zeroes in grid).\n\n### 2. for \'2\'\n\n at max you can remove 2 connec... | 9 | 0 | ['Depth-First Search', 'Graph'] | 5 |
minimum-number-of-days-to-disconnect-island | Java Clean - Tarjan O(mn) | java-clean-tarjan-omn-by-rexue70-4v1g | This question is similar to question 1192 https://leetcode.com/problems/critical-connections-in-a-network/description/\nwe use tarjan\'s algorithm to record (ev | rexue70 | NORMAL | 2020-09-01T05:37:42.296311+00:00 | 2020-09-01T09:27:25.746975+00:00 | 1,117 | false | This question is similar to question 1192 https://leetcode.com/problems/critical-connections-in-a-network/description/\nwe use tarjan\'s algorithm to record (every node) id and the (lowest id it can reach), if we find a critical edge, which means we can cut the island two parts by one cut.\n```\nclass Solution {\n M... | 9 | 1 | [] | 5 |
minimum-number-of-days-to-disconnect-island | Python easy | python-easy-by-valkyre_queen-a5v3 | The answer can be atmost 2 as \n--->For example number of islands are not 1 we can return 0 as there is no need to do anything\n--->If there is 1 island we can | valkyre_queen | NORMAL | 2020-08-30T08:51:32.936553+00:00 | 2020-09-02T03:49:39.690240+00:00 | 648 | false | The answer can be atmost 2 as \n--->For example number of islands are not 1 we can return 0 as there is no need to do anything\n--->If there is 1 island we can return 1 as 1 step will be sufficient to break them into\n--->If there is 1 island and we cannot break them into 2 islands in 1 day we are left with only one op... | 9 | 2 | ['Depth-First Search', 'Python'] | 1 |
minimum-number-of-days-to-disconnect-island | [Java] Remove at most two | java-remove-at-most-two-by-ycj28c-wk7t | The trick is maximum remove 2 points can get more than two islands.... - -\nThere is always one node in the cornor, for example:\n0 1 1 1 0\n0 1 1 1 0\n0 1 1 1 | ycj28c | NORMAL | 2020-08-30T04:28:26.288121+00:00 | 2020-08-30T06:13:14.474362+00:00 | 1,086 | false | The trick is maximum remove 2 points can get more than two islands.... - -\nThere is always one node in the cornor, for example:\n0 1 1 1 0\n0 1 1 1 0\n0 1 1 ***1*** 0\n0 0 0 0 0\nwe can always remove the point next to the right-bottom 1 to get more than 2 island, this case we remove [1,3],[2,2]\n0 1 1 1 0\n0 1 1 0 0\n... | 8 | 0 | [] | 2 |
minimum-number-of-days-to-disconnect-island | Easy to understand | Beginner Friendly C++ solution beats 73% | easy-to-understand-beginner-friendly-c-s-b3wr | Intuition\nThe key intuition here is to recognize that we need to find a land cell that, when removed, increases the number of islands. We can achieve this by u | _adarsh_01 | NORMAL | 2024-08-11T06:22:38.825697+00:00 | 2024-08-11T06:22:38.825732+00:00 | 84 | false | # Intuition\nThe key intuition here is to recognize that we need to find a land cell that, when removed, increases the number of islands. We can achieve this by using a depth-first search (DFS) to count the number of islands before and after removing each land cell.\n\nBy iterating through each land cell and checking i... | 7 | 0 | ['Depth-First Search', 'C++'] | 0 |
minimum-number-of-days-to-disconnect-island | Easy Java Solution 🚀 Clean Code . . . | easy-java-solution-clean-code-by-niketh_-hfj9 | Java Code\n---\n> #### Please don\'t forget to upvote if you\'ve liked my solution. \u2B06\uFE0F\n---\n\nclass Solution {\n int[] xDir = {0,0,-1,1};\n int | niketh_1234 | NORMAL | 2023-06-14T17:06:14.774840+00:00 | 2023-06-14T17:06:14.774863+00:00 | 710 | false | # Java Code\n---\n> #### *Please don\'t forget to upvote if you\'ve liked my solution.* \u2B06\uFE0F\n---\n```\nclass Solution {\n int[] xDir = {0,0,-1,1};\n int[] yDir = {-1,1,0,0};\n public boolean isSafe(int[][] grid,int i,int j,boolean[][] visited)\n {\n return(i>=0 && j>=0 && i<grid.length && j... | 7 | 1 | ['Depth-First Search', 'Recursion', 'C++', 'Java', 'Python3'] | 2 |
minimum-number-of-days-to-disconnect-island | [Java] Tarjan 4ms with explain | java-tarjan-4ms-with-explain-by-66brothe-u6jy | Explanation\n1. First let\'s construc a graph\n2. You want to make at least 2 component (if already more than one, just return 0 by doing a simple DFS/BFS check | 66brother | NORMAL | 2020-08-30T06:28:03.224314+00:00 | 2020-09-03T13:54:08.497753+00:00 | 870 | false | **Explanation**\n1. First let\'s construc a graph\n2. You want to make at least 2 component (if already more than one, just return 0 by doing a simple DFS/BFS check)\n3. You want to find a cut point (use tarjan algorithm),the cutpoint is the point where it can split the graph if you remove it and all its adjeccent edge... | 7 | 0 | [] | 5 |
minimum-number-of-days-to-disconnect-island | Beats 80% || Easy to Understand C++ Code || DFS traversal | beats-80-easy-to-understand-c-code-dfs-t-nuf1 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe code checks if the grid of 1s (land) is already disconnected. If not, it tries to d | jitenagarwal20 | NORMAL | 2024-08-11T08:45:11.630848+00:00 | 2024-08-11T08:45:11.630872+00:00 | 689 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code checks if the grid of 1s (land) is already disconnected. If not, it tries to disconnect the grid by removing one 1 at a time. If removing a single 1 disconnects the grid, it returns 1. Otherwise, if no single removal works, it re... | 6 | 0 | ['Depth-First Search', 'C++'] | 2 |
minimum-number-of-days-to-disconnect-island | 🔥 100% beats | DFS or Tarjan's Algorithm | | 100-beats-dfs-or-tarjans-algorithm-by-b_-yq3c | Solution 1 (DFS)\n# Intuition\nThe problem aims to find the minimum number of days required to disconnect a grid of islands (where each cell represents a piece | B_I_T | NORMAL | 2024-08-11T07:46:27.097331+00:00 | 2024-08-11T07:49:21.817937+00:00 | 808 | false | # Solution 1 (DFS)\n# Intuition\nThe problem aims to find the minimum number of days required to disconnect a grid of islands (where each cell represents a piece of land or water). The strategy revolves around assessing whether the grid is already disconnected, and if not, determining how quickly the disconnection can ... | 6 | 0 | ['Array', 'Depth-First Search', 'Matrix', 'Strongly Connected Component', 'Python', 'C++', 'Java', 'JavaScript'] | 1 |
minimum-number-of-days-to-disconnect-island | Java BFS | java-bfs-by-hobiter-154x | If you look carefully, the result is 0, 1, or 2;\n1, if num of islands not 1, reutrn 0;\n2, special cases, only one 1 or two 1s;\n3, try to remove any 1 that ha | hobiter | NORMAL | 2020-08-30T04:21:06.591932+00:00 | 2020-08-31T17:34:58.374344+00:00 | 574 | false | If you look carefully, the result is 0, 1, or 2;\n1, if num of islands not 1, reutrn 0;\n2, special cases, only one 1 or two 1s;\n3, try to remove any 1 that has more than 1 neighbors, if it splits the islands, return 1;\n4, otherwise return 2;\n\n```\nclass Solution {\n Queue<int[]> connects;\n public int minDay... | 6 | 3 | [] | 4 |
minimum-number-of-days-to-disconnect-island | Runtime beats 72.84%, Memory beats 89.20% [EXPLAINED] | runtime-beats-7284-memory-beats-8920-exp-qise | Intuition\nThe problem is about finding the minimum number of days needed to disconnect a single island in a grid where 1s represent land and 0s represent water | r9n | NORMAL | 2024-10-01T02:38:51.679188+00:00 | 2024-10-01T02:38:51.679228+00:00 | 11 | false | # Intuition\nThe problem is about finding the minimum number of days needed to disconnect a single island in a grid where 1s represent land and 0s represent water. An island is defined as a group of connected 1s, and we want to determine how to remove enough land cells to create multiple islands or completely empty the... | 5 | 0 | ['TypeScript'] | 0 |
minimum-number-of-days-to-disconnect-island | DFS solution with Intuition 🚀| Beginner's Approach | dfs-solution-with-intuition-beginners-ap-3dcx | Intuition\n Describe your first thoughts on how to solve this problem. \nIf we cut an island from any corner leaving smalllest possible island, we need two 0 in | hemantsingh_here | NORMAL | 2024-08-11T16:43:40.479488+00:00 | 2024-08-11T16:43:40.479520+00:00 | 209 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIf we cut an island from any corner leaving smalllest possible island, we need two 0 in diagonal setting. \n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nIf number of island is 1, then we flip all 1 to 0 by one ... | 5 | 0 | ['Depth-First Search', 'C++'] | 0 |
minimum-number-of-days-to-disconnect-island | ✅✅ BFS Solution || CPP 🚀🚀 | bfs-solution-cpp-by-baragemanish6258-jmlj | \n# Code\n\nclass Solution {\npublic:\n int dx[4] = {-1,0,1,0};\n int dy[4] = {0,1,0,-1};\n\n int gc = 0;\n\n int solve(vector<vector<int>>& grid)\n | baragemanish6258 | NORMAL | 2024-08-11T06:35:16.510545+00:00 | 2024-08-11T06:35:16.510595+00:00 | 995 | false | \n# Code\n```\nclass Solution {\npublic:\n int dx[4] = {-1,0,1,0};\n int dy[4] = {0,1,0,-1};\n\n int gc = 0;\n\n int solve(vector<vector<int>>& grid)\n {\n int n = grid.size();\n int m = grid[0].size();\n\n vector<vector<int>>vis(n, vector<int>(m, 0));\n queue<pair<int,int>>q;... | 5 | 0 | ['Array', 'Breadth-First Search', 'Matrix', 'C++'] | 3 |
minimum-number-of-days-to-disconnect-island | Python | DFS | python-dfs-by-khosiyat-72pp | see the Successfully Accepted Submission\n\n# Code\n\nclass Solution:\n def minDays(self, grid: List[List[int]]) -> int:\n def is_connected(grid):\n | Khosiyat | NORMAL | 2024-08-11T05:12:30.314191+00:00 | 2024-08-11T05:12:30.314212+00:00 | 665 | false | [see the Successfully Accepted Submission](https://leetcode.com/problems/minimum-number-of-days-to-disconnect-island/submissions/1351676215/?envType=daily-question&envId=2024-08-11)\n\n# Code\n```\nclass Solution:\n def minDays(self, grid: List[List[int]]) -> int:\n def is_connected(grid):\n visite... | 5 | 1 | ['Python3'] | 1 |
minimum-number-of-days-to-disconnect-island | Tarjan's Algorithm | tarjans-algorithm-by-anand_shukla1312-l8vb | Intuition\n Describe your first thoughts on how to solve this problem. \nAn articulation point is a cell that will split an island in two when it is changed fro | anand_shukla1312 | NORMAL | 2024-08-11T03:40:05.521981+00:00 | 2024-08-11T03:40:05.522003+00:00 | 639 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAn articulation point is a cell that will split an island in two when it is changed from land to water. If a given grid has an articulation point, we can disconnect the island in one day. Tarjan\'s algorithm efficiently finds articulation... | 5 | 0 | ['Java'] | 2 |
minimum-number-of-days-to-disconnect-island | notes Explanation .... | notes-explanation-by-ankit1478-mf2s | Intuition\nsorry for Bad Handwriting..\n\n\n\n\n\n# Complexity\n- Time complexity: O(NM) + O(NM) = O(NM)\n Add your time complexity here, e.g. O(n) \n\n- Space | ankit1478 | NORMAL | 2024-03-05T10:31:42.926434+00:00 | 2024-03-05T10:32:34.445414+00:00 | 275 | false | # Intuition\nsorry for Bad Handwriting..\n\n\n into water (`0`s). The grid is represented by a 2D array, where `1` indicates land and `0` indicates water. An island is a group of connected... | 4 | 0 | ['Depth-First Search', 'Java'] | 1 |
minimum-number-of-days-to-disconnect-island | Most --SIMPLIFIED-- using dfs stack -- EXPLAINED fully | most-simplified-using-dfs-stack-explaine-9lxf | Approach\n Describe your approach to solving the problem. \n\ncheck starting\n\n- count islands: first, count how many separate islands are in the grid. if ther | pawansingh275701 | NORMAL | 2024-08-11T01:14:25.459794+00:00 | 2024-08-11T01:14:25.459818+00:00 | 61 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n\n*check starting*\n\n- **count islands**: first, count how many separate islands are in the grid. if there are already multiple islands (or none), return 0 because no single cell removal is needed to disconnect them.\ntest each cell removal:\n\n---\n... | 4 | 0 | ['C++'] | 0 |
minimum-number-of-days-to-disconnect-island | [Python] Tarjan's Algorithm + Articulation Point detection = at most 2 days of flipping 1s to 0s | python-tarjans-algorithm-articulation-po-s7m6 | Intuition\n Describe your first thoughts on how to solve this problem. \nThis is a haaaard problem if you don\'t know that we can take at most 2 days to separat | sinclaire | NORMAL | 2023-06-03T13:50:26.138836+00:00 | 2023-06-03T13:50:26.138884+00:00 | 202 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis is a haaaard problem if you don\'t know that we can take at most 2 days to separate a given island. We solve this using SCCs concept + articulation point detection.\n\n# Approach\n<!-- Describe your approach to solving the problem. -... | 4 | 0 | ['Python3'] | 2 |
minimum-number-of-days-to-disconnect-island | [C++] 2 solutions, clean code, and detailed explanation | c-2-solutions-clean-code-and-detailed-ex-xg9f | The idea:\nFirst thing to notice is that we need atmost two days to disconnect an island. The intuition is that since diagonal does not count as a connection, g | haoran_xu | NORMAL | 2021-07-15T19:31:25.594185+00:00 | 2021-07-15T19:32:00.175049+00:00 | 315 | false | The idea:\nFirst thing to notice is that we need atmost two days to disconnect an island. The intuition is that since diagonal does not count as a connection, given a corner of the island, we can disconnect two of its connections (in fact it only has two connections) to seperate it. Notice that we can do this to any co... | 4 | 0 | [] | 0 |
minimum-number-of-days-to-disconnect-island | No Graphs Beats 100% CPP SOLUTION EASY AND INTUTIVE APPROACH | no-graphs-beats-100-cpp-solution-easy-an-lf7l | Intuition\n Describe your first thoughts on how to solve this problem. \nTo solve this problem, we need to identify the minimum number of days (or steps) requir | Iheretocode | NORMAL | 2024-08-11T05:18:00.691095+00:00 | 2024-08-11T10:41:59.543575+00:00 | 41 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo solve this problem, we need to identify the minimum number of days (or steps) required to separate the grid into two or more disconnected components or to remove all the land cells entirely.\n\n\n# Approach\n<!-- Describe your approach... | 3 | 0 | ['C++'] | 0 |
minimum-number-of-days-to-disconnect-island | C++ | Minimum Number of Days to Disconnect Island | 62ms | c-minimum-number-of-days-to-disconnect-i-7jiz | Intuition\nTo determine the minimum number of days required to disconnect a single island in a grid, we need to explore how the grid can be split into multiple | user4612MW | NORMAL | 2024-08-11T03:35:12.783380+00:00 | 2024-08-11T03:38:14.531197+00:00 | 20 | false | # Intuition\nTo determine the minimum number of days required to disconnect a single island in a grid, we need to explore how the grid can be split into multiple islands. Each day, we can change one land cell to water, and our goal is to find the minimum number of such changes needed to confirm that the grid is disconn... | 3 | 0 | ['C++'] | 0 |
minimum-number-of-days-to-disconnect-island | Easy to understand ✅ Python, C# 🔥 | Great runtime ✨ | easy-to-understand-python-c-great-runtim-e79o | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires finding the minimum number of days needed to disconnect a land mas | kotto | NORMAL | 2024-08-11T00:10:55.852950+00:00 | 2024-08-11T00:10:55.852968+00:00 | 632 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires finding the minimum number of days needed to disconnect a land mass in a grid. My initial thought is to explore whether the grid is already disconnected, and if not, determine how quickly it can be split into separate... | 3 | 0 | ['Depth-First Search', 'Matrix', 'Python3', 'C#'] | 3 |
minimum-number-of-days-to-disconnect-island | C++ DFS solution | c-dfs-solution-by-sachin_kumar_sharma-p3bl | 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 | Sachin_Kumar_Sharma | NORMAL | 2024-04-18T03:56:25.886862+00:00 | 2024-04-18T03:56:25.886885+00:00 | 9 | 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((m*n)^2)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(m*n)\n<!-- Add your space complexity here... | 3 | 0 | ['C++'] | 0 |
minimum-number-of-days-to-disconnect-island | Easy BFS Solution || C++ || Number of Island | easy-bfs-solution-c-number-of-island-by-wxvj7 | \nclass Solution {\nprivate:\n int totalLand(vector<vector<int>>& grid)\n {\n int n = grid.size() , m = grid[0].size();\n vector<vector<int> | RIOLOG | NORMAL | 2023-10-26T15:31:46.201594+00:00 | 2023-10-26T15:31:46.201626+00:00 | 712 | false | ```\nclass Solution {\nprivate:\n int totalLand(vector<vector<int>>& grid)\n {\n int n = grid.size() , m = grid[0].size();\n vector<vector<int>>vis(n,vector<int>(m,0));\n int count = 0;\n\n for (int i=0;i<n;i++)\n {\n for (int j=0;j<m;j++)\n {\n ... | 3 | 0 | ['Breadth-First Search', 'C++'] | 1 |
minimum-number-of-days-to-disconnect-island | C++| BFS | islands-problems | Easy-to-understand | c-bfs-islands-problems-easy-to-understan-638j | Intuition\n Describe your first thoughts on how to solve this problem. \nAll the posted solutions are using the similar basic flow:\n\n- If there is no island o | sko_1 | NORMAL | 2022-12-14T07:15:42.753048+00:00 | 2022-12-14T07:15:42.753093+00:00 | 741 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAll the posted solutions are using the similar basic flow:\n\n- If there is no island or more than 1 island, return 0;\n- If there is only one land, return 1;\n- If any single cell could serve as the cut point ( divide 1 island into 2 isl... | 3 | 0 | ['C++'] | 1 |
minimum-number-of-days-to-disconnect-island | C++ | islands-problems | medium-hard | c-islands-problems-medium-hard-by-chikzz-3q08 | PLEASE UPVOTE IF YOU FIND MY SOLUTION HELPFUL\n\n\nclass Solution {\n int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};\n bool vis[31][31];\n int noOfIslands( | chikzz | NORMAL | 2022-03-30T05:35:34.216282+00:00 | 2022-03-30T05:35:34.216307+00:00 | 534 | false | **PLEASE UPVOTE IF YOU FIND MY SOLUTION HELPFUL**\n\n```\nclass Solution {\n int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};\n bool vis[31][31];\n int noOfIslands(vector<vector<int>>& grid,int &n,int &m)\n {\n int islands=0;\n memset(vis,0,sizeof(vis));\n for(int i=0;i<n;i++)\n {\n ... | 3 | 0 | ['Depth-First Search', 'Recursion'] | 1 |
minimum-number-of-days-to-disconnect-island | C++, Articulation Point O(m + n) | c-articulation-point-om-n-by-level7-hnwc | Guys, Comment me if you have any questions, Approach is pretty similar to AP but we need to consider some corner cases as well, \nCase 1 : there is only one isl | level7 | NORMAL | 2021-08-09T22:12:20.994331+00:00 | 2022-01-06T20:42:21.610991+00:00 | 725 | false | Guys, Comment me if you have any questions, Approach is pretty similar to AP but we need to consider some corner cases as well, \nCase 1 : there is only one island and one land return 1\nCase 2 : if no of islands are more than one -> in this case we need to return 0;\nCase 3: there is only one island and two lands retu... | 3 | 0 | ['C'] | 2 |
minimum-number-of-days-to-disconnect-island | C++ DFS 1/4 Positive | c-dfs-14-positive-by-votrubac-vrtc | The answer is:\n1. Zero if there is no islands or more than one island.\n2. One when there is a single piece of land that connects two or more semi-islands. \n3 | votrubac | NORMAL | 2020-09-01T01:48:01.181210+00:00 | 2020-09-01T01:58:59.187254+00:00 | 139 | false | The answer is:\n1. Zero if there is no islands or more than one island.\n2. One when there is a single piece of land that connects two or more semi-islands. \n3. Two otherwise.\n\nHow to tell if a land connects semi-islands? If it does not, doing DFS in one direction will explore the entire island. If it does, we need ... | 3 | 3 | [] | 0 |
minimum-number-of-days-to-disconnect-island | [JavaScript] Find the critical points and check O(MNK) (K: number of critical point) | javascript-find-the-critical-points-and-t29rk | The maximum answer could only be 2.\nEx:\n0111\n0111\n0111\n0000\n-->\n0111\n0011\n0101\n0000\n\nSince we know this rule. Now we have to do the following steps: | alanchanghsnu | NORMAL | 2020-08-30T04:02:23.381740+00:00 | 2020-08-30T04:12:54.952013+00:00 | 386 | false | The maximum answer could only be 2.\nEx:\n0111\n0111\n0111\n0000\n-->\n0111\n0011\n0101\n0000\n\nSince we know this rule. Now we have to do the following steps:\n1. Is it only one island?\n\tYes: Go next Step\n\tNo: return 0\n2. In the step 1, you can check which land connect 2 lands at most\n If connect 1 land, ret... | 3 | 0 | ['JavaScript'] | 1 |
minimum-number-of-days-to-disconnect-island | easy solution | easy-solution-by-paramdotcom-mwo1 | \n\n#include <bits/stdc++.h>\nusing namespace std;\n\nclass Solution\n{\npublic:\n void bfs(vector<vector<int>> &grid, int i, int j, int check)\n {\n | paramdotcom | NORMAL | 2024-08-11T17:04:40.627851+00:00 | 2024-08-11T17:04:40.627895+00:00 | 10 | false | \n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nclass Solution\n{\npublic:\n void bfs(vector<vector<int>> &grid, int i, int j, int check)\n {\n if (i - 1 >= 0 && grid[i - 1][j] == check)\n {\n grid[i - 1][j] = check + 1;\n bfs(grid, i - 1, j, check);\n }\n ... | 2 | 0 | ['C++'] | 0 |
minimum-number-of-days-to-disconnect-island | Easy for Beginner || Simple Approach || Full Explanation | easy-for-beginner-simple-approach-full-e-q9e7 | Approach\nHere\'s the approach:\n\nDepth-First Search (DFS): The dfs function is used to traverse the grid in a depth-first manner. It starts from a cell (i, j) | Rohan_k14 | NORMAL | 2024-08-11T13:34:19.718269+00:00 | 2024-08-11T13:34:19.718300+00:00 | 171 | false | # Approach\nHere\'s the approach:\n\nDepth-First Search (DFS): The dfs function is used to traverse the grid in a depth-first manner. It starts from a cell (i, j), and if the cell is within the grid boundaries, not visited before, and is land (grid[i][j] == 1), it marks it as visited and continues the search in all fou... | 2 | 0 | ['C++'] | 1 |
minimum-number-of-days-to-disconnect-island | Easy Solution in C | easy-solution-in-c-by-tamilarasanmurugan-kjf5 | Intuition\nThe problem is about determining the minimum number of operations (days) needed to disconnect a grid by removing cells. The goal is to break the grid | TamilarasanMurugan | NORMAL | 2024-08-11T13:05:32.203190+00:00 | 2024-08-11T13:05:32.203216+00:00 | 15 | false | # Intuition\nThe problem is about determining the minimum number of operations (days) needed to disconnect a grid by removing cells. The goal is to break the grid into multiple islands by removing as few cells as possible.\n\n# Approach\n1.Initial Island Check: First, check if the grid is already disconnected. If it is... | 2 | 0 | ['Array', 'Depth-First Search', 'Breadth-First Search', 'C', 'Matrix', 'Strongly Connected Component'] | 0 |
minimum-number-of-days-to-disconnect-island | DFS || C++ || Clean Code & Easy Solution | dfs-c-clean-code-easy-solution-by-zishan-5him | Code\n\nclass Solution {\npublic:\n int n, m;\n int vis[31][31];\n vector<pair<int, int> > dir = {{0, -1}, {0, 1}, {1, 0}, {-1, 0}};\n\n bool isVali | zishan_0point1 | NORMAL | 2024-08-11T09:09:42.989121+00:00 | 2024-08-11T09:09:42.989152+00:00 | 121 | false | # Code\n```\nclass Solution {\npublic:\n int n, m;\n int vis[31][31];\n vector<pair<int, int> > dir = {{0, -1}, {0, 1}, {1, 0}, {-1, 0}};\n\n bool isValid(int r, int c){\n return r >= 0 && r < n && c >= 0 && c < m;\n }\n void dfs(int x, int y, vector<vector<int>>& grid){\n vis[x][y] = 1;... | 2 | 0 | ['Depth-First Search', 'Matrix', 'C++'] | 0 |
minimum-number-of-days-to-disconnect-island | Easy to understand ||BFS || Beats 70% || C++ | easy-to-understand-bfs-beats-70-c-by-har-apnv | Intuition\n Describe your first thoughts on how to solve this problem. \nThe bfs or dfs function is used to mark all connected land cells starting from a given | hardik-chauhan | NORMAL | 2024-08-11T07:19:15.627841+00:00 | 2024-08-11T07:19:15.627885+00:00 | 61 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe bfs or dfs function is used to mark all connected land cells starting from a given cell as visited. The isConnected function checks if all land cells are connected by using the bfs function. The minDays function first checks if the gr... | 2 | 0 | ['C++'] | 0 |
minimum-number-of-days-to-disconnect-island | C++ || DFS | c-dfs-by-akash92-xs3s | Intuition\n Describe your first thoughts on how to solve this problem. \nNo matter what, we can always disconnect islands in 2 days at max.\nHow so?\nin a group | akash92 | NORMAL | 2024-08-11T06:52:12.827911+00:00 | 2024-08-11T06:52:12.827951+00:00 | 417 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nNo matter what, we can always disconnect islands in 2 days at max.\nHow so?\nin a group of islands, delete the two islands(right and bottom) of the top left corner island.\nthis means we need to check if it can be done in less than 2 days... | 2 | 0 | ['Depth-First Search', 'Matrix', 'C++'] | 1 |
minimum-number-of-days-to-disconnect-island | BRUTE FOCE SOLUTION || DFS || WITH EXPLANATION | brute-foce-solution-dfs-with-explanation-uzfb | Intuition\n Describe your first thoughts on how to solve this problem. \n1. Connected Components (Islands): An island is a group of connected land cells (1s). C | ManaviArora | NORMAL | 2024-08-11T06:26:01.697355+00:00 | 2024-08-11T06:26:01.697389+00:00 | 104 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1. `Connected Components (Islands):` An island is a group of connected land cells (1s). Connectivity is defined in the four cardinal directions: up, down, left, and right.\n2. `Island Removal Impact:` By removing one cell, you might eithe... | 2 | 0 | ['Array', 'Depth-First Search', 'Matrix', 'Strongly Connected Component', 'C++'] | 1 |
minimum-number-of-days-to-disconnect-island | 🌟 Mastering Island Disconnect in 2 Days with Multi-Language Solutions 🔥💯 | mastering-island-disconnect-in-2-days-wi-5ynj | \n\nExplore a collection of solutions to LeetCode problems in multiple programming languages. Each solution includes a detailed explanation and step-by-step app | withaarzoo | NORMAL | 2024-08-11T06:01:55.528465+00:00 | 2024-08-11T06:01:55.528565+00:00 | 504 | false | \n\nExplore a collection of solutions to LeetCode problems in multiple programming languages. Each solution includes a detailed explanation and step-by-step approach to solving the pr... | 2 | 0 | ['Array', 'Depth-First Search', 'Breadth-First Search', 'Matrix', 'Python', 'C++', 'Java', 'Go', 'Python3', 'JavaScript'] | 2 |
minimum-number-of-days-to-disconnect-island | Java Solution Using GraphBFS Traversal || SCC Typically Question | java-solution-using-graphbfs-traversal-s-6tr7 | Complexity\n- Time complexity:O((n+m)*4Alpha)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:\n Add your space complexity here, e.g. O(n) \n\ | Shree_Govind_Jee | NORMAL | 2024-08-11T04:15:25.728138+00:00 | 2024-08-11T04:15:25.728163+00:00 | 277 | false | # Complexity\n- Time complexity:$$O((n+m)*4Alpha)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n\n int[] dir_X = { 1, -1, 0, 0 };\n int[] dir_Y = { 0, 0, 1, -1 };\n\n private void solveBF... | 2 | 0 | ['Array', 'Depth-First Search', 'Breadth-First Search', 'Matrix', 'Strongly Connected Component', 'Java'] | 2 |
minimum-number-of-days-to-disconnect-island | C++ || Easy Video Solution | c-easy-video-solution-by-n_i_v_a_s-o57k | The video solution for the below code is\nhttps://youtu.be/dDfY6llzsu8\n\n# Code\n\nclass Solution {\npublic:\n int minDays(vector<vector<int>>& grid) {\n | __SAI__NIVAS__ | NORMAL | 2024-08-11T04:08:34.745360+00:00 | 2024-08-11T04:08:34.745377+00:00 | 436 | false | The video solution for the below code is\nhttps://youtu.be/dDfY6llzsu8\n\n# Code\n```\nclass Solution {\npublic:\n int minDays(vector<vector<int>>& grid) {\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n // Note : Grid is said to be connected if there is excatly o... | 2 | 0 | ['C++'] | 1 |
minimum-number-of-days-to-disconnect-island | C# Solution for Minimum Number of Days to Disconnect Island Problem | c-solution-for-minimum-number-of-days-to-s9kt | Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind this approach is to treat the grid as a graph where each cell repr | Aman_Raj_Sinha | NORMAL | 2024-08-11T03:04:08.313763+00:00 | 2024-08-11T03:04:08.313789+00:00 | 122 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this approach is to treat the grid as a graph where each cell represents a node. The idea is to identify \u201Ccritical cells\u201D (similar to bridges in graph theory) that, when removed, increase the number of islan... | 2 | 0 | ['C#'] | 1 |
minimum-number-of-days-to-disconnect-island | Swift | 2 days are enough 🏄 | swift-2-days-are-enough-by-pagafan7as-0fkn | The key to solving this problem is to notice that the solution is either 0, 1 or 2 days.\n\nProof: If there is at least one square with land, go up and right as | pagafan7as | NORMAL | 2024-08-11T01:07:30.411385+00:00 | 2024-08-11T02:38:37.340870+00:00 | 48 | false | The key to solving this problem is to notice that the solution is either 0, 1 or 2 days.\n\n**Proof:** If there is at least one square with land, go up and right as long as there is land. You will end up at a land square from where you cannot proceed, with water above and to the right. This land square can be made disc... | 2 | 0 | ['Swift'] | 0 |
minimum-number-of-days-to-disconnect-island | Easy Solution | easy-solution-by-sachinonly-j8mo | Intuition\nview the hint section and for better understanding view editorial\n\n# Approach\n1. Count Islands:\n - A helper function countIslands is used to c | Sachinonly__ | NORMAL | 2024-08-11T00:44:32.717920+00:00 | 2024-08-11T12:07:16.313422+00:00 | 371 | false | # Intuition\nview the hint section and for better understanding view editorial\n\n# Approach\n1. **Count Islands:**\n - A helper function countIslands is used to count the number of islands in the grid using Depth-First Search (DFS).\n - It initializes a visited matrix to keep track of visited cells and a counter... | 2 | 0 | ['Array', 'Depth-First Search', 'Breadth-First Search', 'Matrix', 'Strongly Connected Component', 'Python', 'C++', 'Python3'] | 1 |
minimum-number-of-days-to-disconnect-island | O(m * n) | DFS | Beats 100% Java | C++ | Python | Go | Rust | JavaScript | om-n-dfs-beats-100-java-c-python-go-rust-szcp | Intuition\n\nThis problem is fundamentally a graph theory problem. The key insight is to recognize that we\'re dealing with a connected component (the island) i | kartikdevsharma_ | NORMAL | 2024-08-09T12:05:23.305727+00:00 | 2024-08-09T12:41:04.281545+00:00 | 358 | false | ### Intuition\n\nThis problem is fundamentally a graph theory problem. The key insight is to recognize that we\'re dealing with a connected component (the island) in a graph (the grid), and we need to find the most efficient way to break this component.\n\nInitially, one might think of simply removing land cells one by... | 2 | 1 | ['Array', 'Depth-First Search', 'Breadth-First Search', 'C++', 'Java', 'Go', 'Python3', 'Rust', 'JavaScript'] | 1 |
minimum-number-of-days-to-disconnect-island | Minimum Number of Days to Disconnect Island - JAVA | minimum-number-of-days-to-disconnect-isl-d2xr | Intuition\n Describe your first thoughts on how to solve this problem. \nDFS , Matrix - Flip\n\n\n# Complexity\n- Time complexity: O ( m * n )\n Add your time c | Shyam2626 | NORMAL | 2024-01-07T07:41:01.465211+00:00 | 2024-01-07T07:41:01.465242+00:00 | 247 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nDFS , Matrix - Flip\n\n\n# Complexity\n- Time complexity: O ( m * n )\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O ( m * n )\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass ... | 2 | 0 | ['Depth-First Search', 'Matrix', 'Java'] | 1 |
minimum-number-of-days-to-disconnect-island | Simple Dfs||C++||Easy | simple-dfsceasy-by-aayushi_kukreja-hxdy | \n\n# Code\n\nclass Solution {\npublic:\n void dfs(int row,int col,vector<vector<int>>& grid,vector<vector<int>>&vis)\n {\n int n = grid.size();\n | Aayushi_Kukreja | NORMAL | 2023-02-27T18:27:57.715859+00:00 | 2023-02-27T18:27:57.715899+00:00 | 182 | false | \n\n# Code\n```\nclass Solution {\npublic:\n void dfs(int row,int col,vector<vector<int>>& grid,vector<vector<int>>&vis)\n {\n int n = grid.size();\n int m = grid[0].size();\n vis[row][col]=1;\n int delrow[] = {1, -1, 0, 0};\n int delcol[] = {0, 0, 1, -1};\n\n for(int i=0... | 2 | 0 | ['C++'] | 0 |
minimum-number-of-days-to-disconnect-island | Tarjan's algorithm: O(m * n) with approach and detailed comments 💯 | tarjans-algorithm-om-n-with-approach-and-4ajh | Intuition\n Describe your first thoughts on how to solve this problem. \nThe DFS function uses Tarjan\'s algorithm to identify the articulation points in the gr | abdulahadsiddiqui11 | NORMAL | 2022-12-26T05:47:41.136869+00:00 | 2022-12-26T05:47:41.136909+00:00 | 197 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe DFS function uses Tarjan\'s algorithm to identify the articulation points in the grid. It keeps track of the insertion time and lowest time of each cell, which are used to determine whether a cell is an articulation point. The inserti... | 2 | 0 | ['Depth-First Search', 'C++'] | 0 |
minimum-number-of-days-to-disconnect-island | [C++] | The Only Observation to Solve the Problem | c-the-only-observation-to-solve-the-prob-sure | You will need at most 2 days.\n\nProof: you can always just isolate the 1 in the corner by making his 2 neighbours equal to zero.\n\nSo, first we will check the | makhonya | NORMAL | 2022-06-11T12:35:49.804473+00:00 | 2022-06-11T12:35:49.804511+00:00 | 223 | false | `You will need at most 2 days.`\n\nProof: you can always just isolate the ```1``` in the corner by making his 2 neighbours equal to zero.\n\nSo, first we will check the number of islands we have, if it is more than 1 we return 0, because islands are already disconnected. If it is equal to 1, we will replace each of 1\'... | 2 | 0 | ['Depth-First Search', 'C', 'C++'] | 1 |
minimum-number-of-days-to-disconnect-island | C++ | Similar to Finding Articulation Point | Explained | c-similar-to-finding-articulation-point-e640x | I\'ll tell you the steps:\n1) Convert the grid into a graph.\n2) If there is only one 1, then you need to remove it (because connected definition says only 1 is | harshnadar23 | NORMAL | 2021-09-15T08:46:04.470890+00:00 | 2021-09-15T08:46:04.470924+00:00 | 635 | false | I\'ll tell you the steps:\n1) Convert the grid into a graph.\n2) If there is only one 1, then you need to remove it (because connected definition says only 1 island)\n3) Find number of connected components\n4) If connected components>1 return 0\n5) Find articulation point\n6) If articulation point is present, then retu... | 2 | 1 | ['Depth-First Search', 'C'] | 0 |
minimum-number-of-days-to-disconnect-island | JAVA DFS(Connected Components) With Comments | java-dfsconnected-components-with-commen-xv6g | \nclass Solution {\n static int [][]dirs={{-1,0},{0,1},{0,-1},{1,0}};\n public void connectedComponents(int[][] grid, int i, int j, boolean[][] visited) { | abhirajsinha | NORMAL | 2021-08-16T06:18:26.060514+00:00 | 2021-08-16T15:12:02.804280+00:00 | 448 | false | ```\nclass Solution {\n static int [][]dirs={{-1,0},{0,1},{0,-1},{1,0}};\n public void connectedComponents(int[][] grid, int i, int j, boolean[][] visited) {\n \n visited[i][j] = true;\n for(int d=0;d<4;d++){\n int r=i+dirs[d][0];\n int c=j+dirs[d][1];\n \n ... | 2 | 1 | ['Depth-First Search', 'Java'] | 2 |
minimum-number-of-days-to-disconnect-island | Java | Easy | DFS | Using Number of Islands | java-easy-dfs-using-number-of-islands-by-065d | \nclass Solution {\n public int minDays(int[][] grid) {\n int n = grid.length ;\n int m = grid[0].length ;\n if(numIslands(grid) != 1){\ | user8540kj | NORMAL | 2021-08-05T13:01:24.194275+00:00 | 2021-08-05T13:01:24.194307+00:00 | 202 | false | ```\nclass Solution {\n public int minDays(int[][] grid) {\n int n = grid.length ;\n int m = grid[0].length ;\n if(numIslands(grid) != 1){\n return 0;\n }\n for(int i = 0; i < n;i++ ){\n for(int j = 0 ; j < m ; j++){\n if(grid[i][j] == 1){\n ... | 2 | 0 | [] | 0 |
minimum-number-of-days-to-disconnect-island | BFS C++ Approach | bfs-c-approach-by-codes_aman-nsj5 | \nclass Solution {\npublic:\n int dx[4]={-1,0,1,0};\n int dy[4]={0,1,0,-1};\n bool safe(int x,int y,int m,int n)\n {\n return (x>=0&&x<m&&y>= | codes_aman | NORMAL | 2020-11-28T13:30:36.520036+00:00 | 2020-11-28T13:30:36.520065+00:00 | 338 | false | ```\nclass Solution {\npublic:\n int dx[4]={-1,0,1,0};\n int dy[4]={0,1,0,-1};\n bool safe(int x,int y,int m,int n)\n {\n return (x>=0&&x<m&&y>=0&&y<n);\n }\n void travel(int i,int j,vector<vector<int> > &grid,vector<vector<int> > &visited)\n {\n int m=grid.size();\n int n=grid... | 2 | 0 | ['Breadth-First Search', 'C'] | 1 |
minimum-number-of-days-to-disconnect-island | Easy C++ solution, with full explanation | easy-c-solution-with-full-explanation-by-shop | Connected Components\nLook closely, the answer of the question can not be greater than 2.\n\nAlgo:\n1. First Check if the connected components are already great | hiteshgupta | NORMAL | 2020-08-30T04:13:16.635256+00:00 | 2020-08-30T04:13:58.922717+00:00 | 302 | false | # Connected Components\n**Look closely, the answer of the question can not be greater than 2.**\n\nAlgo:\n1. First Check if the connected components are already greater than 1, if yes return 0\n2. Then do the following:\n for every grid value of 1, set it up for 0, check connected components, if connect components a... | 2 | 0 | ['Breadth-First Search', 'Graph', 'C'] | 2 |
minimum-number-of-days-to-disconnect-island | C# | c-by-adchoudhary-sgjr | Code | adchoudhary | NORMAL | 2025-03-08T03:19:06.783071+00:00 | 2025-03-08T03:19:06.783071+00:00 | 2 | false | # Code
```csharp []
public class Solution {
public int MinDays(int[][] grid)
{
// Step 1: Check if the grid is already disconnected.
if (CountIslands(grid) != 1)
{
return 0;
}
// Step 2: Check for bridge-like cells.
int rows = grid.Length;
int cols = grid[0].L... | 1 | 0 | ['C#'] | 0 |
minimum-number-of-days-to-disconnect-island | Using Tarjan's Algorithm(if only one component) | using-tarjans-algorithmif-only-one-compo-oe2j | Intuition \n1. If there are no islands or more than 1 island, than the answer is 0, as we don\'t need to remove anything.\n2. If we can disconnct the island by | keshavchauhan107 | NORMAL | 2024-08-12T08:33:27.788330+00:00 | 2024-08-12T08:33:27.788357+00:00 | 5 | false | # Intuition \n1. If there are no islands or more than 1 island, than the answer is 0, as we don\'t need to remove anything.\n2. If we can disconnct the island by removing exactly 1 cell, the answer is 1 (will be finding Articulation point using Tarjan\'s Algorithm).\n3. At last, we can always disconnect an island by re... | 1 | 0 | ['Array', 'Depth-First Search', 'Breadth-First Search', 'Matrix', 'Strongly Connected Component', 'C++'] | 0 |
minimum-number-of-days-to-disconnect-island | C++,DFS | cdfs-by-daredreamer-7stk | Intuition\njust try to count diagonal one connected to one single source\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nDFS\n Desc | daredreamer | NORMAL | 2024-08-11T20:12:36.004778+00:00 | 2024-08-11T20:12:36.004797+00:00 | 23 | false | > # Intuition\njust try to count diagonal one connected to one single source\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nDFS\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- Sp... | 1 | 0 | ['Depth-First Search', 'C++'] | 0 |
minimum-number-of-days-to-disconnect-island | Simple Logic, Just observe and dry run few test cases | simple-logic-just-observe-and-dry-run-fe-kf4p | Intuition\n1. If Initial number of Islands are greater than 1 or equal to zero return 0.\n2. Now change one by one each 1 to 0, count the number of new islands | Hardy1 | NORMAL | 2024-08-11T19:31:30.676454+00:00 | 2024-08-11T19:31:30.676491+00:00 | 25 | false | # Intuition\n1. If Initial number of Islands are greater than 1 or equal to zero return 0.\n2. Now change one by one each 1 to 0, count the number of new islands and replace 0 with 1. If new islands greater than 1 or equal to 0 then return 1.\n3. In the worst case like this [[1,1,1],[1,1,1],[1,1,1]] here you can see ch... | 1 | 0 | ['Depth-First Search', 'Matrix', 'C++'] | 0 |
minimum-number-of-days-to-disconnect-island | Simple Explanation | simple-explanation-by-aayushbharti-n74h | Intuition\nTo disconnect an island, you need to break the connectivity of the land cells. This can be done by either removing a single land cell or, if the isla | AayushBharti | NORMAL | 2024-08-11T18:44:35.251315+00:00 | 2024-08-11T18:44:35.251344+00:00 | 17 | false | # Intuition\nTo disconnect an island, you need to break the connectivity of the land cells. This can be done by either removing a single land cell or, if the island is very connected, you might need to remove two or more. The minimum number of days is determined by the number of cells that need to be removed to disconn... | 1 | 0 | ['Java'] | 0 |
minimum-number-of-days-to-disconnect-island | C# Beats 100% with 113 ms | c-beats-100-with-113-ms-by-zocolini-mazy | Intuition\nFirst we can prove that the maximun number of days needed is 2. We also can prove that for islands of 1, 2 or 3 lands we already know the number o da | zocolini | NORMAL | 2024-08-11T17:25:01.572232+00:00 | 2024-08-11T17:25:01.572266+00:00 | 11 | false | # Intuition\nFirst we can prove that the maximun number of days needed is 2. We also can prove that for islands of 1, 2 or 3 lands we already know the number o days needed.\n\n# Approach\nWe first check the number os 1s to discard those cases that we already know how they end. Next step it\'s try to find one bridge poi... | 1 | 0 | ['C#'] | 0 |
minimum-number-of-days-to-disconnect-island | ✨[TYPESCRIPT]✨ 140 ms Beats 79.37% ✅ | typescript-140-ms-beats-7937-by-rain84-nxpl | Intuition\n\nThere is a lot of LOC\'s, but it is not a most complicated solution.\nRealy. \n\n# Approach\n\n1. Use dfs and use it to count the number of "island | rain84 | NORMAL | 2024-08-11T17:18:29.327404+00:00 | 2024-08-13T08:04:22.301155+00:00 | 30 | false | # Intuition\n\nThere is a lot of LOC\'s, but it is not a most complicated solution.\nRealy. \n\n# Approach\n\n1. Use dfs and use it to count the number of "islands"\n2. If there are 0 islands or more than the 1st, it means that nothing in the matrix needs to be changed and we return a zero\n3. Go through the "island" c... | 1 | 0 | ['Array', 'Depth-First Search', 'Breadth-First Search', 'Matrix', 'TypeScript', 'JavaScript'] | 0 |
minimum-number-of-days-to-disconnect-island | Easy to understand C++ solution with comments. | easy-to-understand-c-solution-with-comme-tivb | 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 | BhavyaGautam899 | NORMAL | 2024-08-11T17:10:54.143429+00:00 | 2024-08-11T17:10:54.143461+00:00 | 15 | 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. $$O(n)$$ --... | 1 | 0 | ['Depth-First Search', 'C++'] | 0 |
minimum-number-of-days-to-disconnect-island | solution -python | solution-python-by-shrutisingh64-unmb | Intuition\nThe grid consists of 1s (land) and 0s (water)\nThe goal is to find the minimum number of days required to disconnect the island, meaning that no two | shrutisingh64 | NORMAL | 2024-08-11T16:17:54.587525+00:00 | 2024-08-11T16:17:54.587558+00:00 | 48 | false | # Intuition\nThe grid consists of 1s (land) and 0s (water)\nThe goal is to find the minimum number of days required to disconnect the island, meaning that no two 1s are connected.\nIn one day, you can remove one 1 (land cell) from the grid.\nAfter removing a 1, the grid is updated, and the remaining 1s may become disco... | 1 | 0 | ['Python3'] | 0 |
minimum-number-of-days-to-disconnect-island | C++ Approach || Time complexity O(m*n)^2 & space complexity O(m&n) || Simple Dfs Approach | c-approach-time-complexity-omn2-space-co-a3j6 | Complexity\n- Time complexity:O(mn)^2\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(mn)\n Add your space complexity here, e.g. O(n) \n\n# | sneha029 | NORMAL | 2024-08-11T16:15:52.227596+00:00 | 2024-08-11T16:15:52.227625+00:00 | 35 | false | # Complexity\n- Time complexity:O(m*n)^2\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(m*n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int m, n;\n void dfs(vector<vector<int>>& grid, int i, int j, vector<vector<bool>>& vis){... | 1 | 0 | ['C++'] | 0 |
minimum-number-of-days-to-disconnect-island | 🧠✅ PYTHON SIMPLE SOLUTION EXPLAINED IN DETAILS ✔️🔥🧠 | python-simple-solution-explained-in-deta-wgln | \n## Intuition \uD83E\uDD14\nThe key to solving this problem is to understand that we need to find the minimum number of days required to disconnect the grid, w | sagarsaini_ | NORMAL | 2024-08-11T14:08:07.972986+00:00 | 2024-08-11T14:08:07.973020+00:00 | 12 | false | \n## Intuition \uD83E\uDD14\nThe key to solving this problem is to understand that we need to find the minimum number of days required to disconnect the grid, which means we need to find the minimum number of cells that, when changed from land to water, will result in the grid having more than one island or no islands ... | 1 | 0 | ['Array', 'Depth-First Search', 'Matrix', 'Strongly Connected Component', 'Python3'] | 0 |
minimum-number-of-days-to-disconnect-island | Easy solution | Step by Step Detailed Full Solution | C++ Solution | easy-solution-step-by-step-detailed-full-j8xl | For the full code, check at bottom. \n### Problem Understanding:\nYou have a grid where each cell is either land (1) or water (0). The goal is to determine the | aizagazyani16 | NORMAL | 2024-08-11T13:38:59.921533+00:00 | 2024-08-11T13:38:59.921561+00:00 | 17 | false | > # For the full code, check at bottom. \n### Problem Understanding:\nYou have a grid where each cell is either land (1) or water (0). The goal is to determine the minimum number of days needed to disconnect an island in the grid. An island is defined as a group of connected land cells (horizontally or vertically). \n\... | 1 | 0 | ['C++'] | 0 |
minimum-number-of-days-to-disconnect-island | ✅EASY JAVA SOLUTION BEATS 100% | easy-java-solution-beats-100-by-swayam28-btvi | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | swayam28 | NORMAL | 2024-08-11T13:38:46.957869+00:00 | 2024-08-11T13:38:46.957898+00:00 | 53 | 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. $$O(n)$$ --... | 1 | 0 | ['Java'] | 0 |
minimum-number-of-days-to-disconnect-island | DFS Brute-Force Beats 78% | dfs-brute-force-beats-78-by-robin_rathor-9wuj | \n# Code\n\nclass Solution {\n private:\n void dfs(vector<vector<int>>& temp, int i, int j){\n if(i < 0 || j < 0 || i >= temp.size() || j >= temp[i | Robin_Rathore | NORMAL | 2024-08-11T13:24:21.814288+00:00 | 2024-08-11T13:24:21.814314+00:00 | 24 | false | \n# Code\n```\nclass Solution {\n private:\n void dfs(vector<vector<int>>& temp, int i, int j){\n if(i < 0 || j < 0 || i >= temp.size() || j >= temp[i].size() || temp[i][j] == 0) return;\n\n if(temp[i][j])\n temp[i][j] = 0;\n dfs(temp, i+1, j);\n dfs(temp, i, j+1);\n dfs(... | 1 | 0 | ['C++'] | 0 |
minimum-number-of-days-to-disconnect-island | Minimum Number of Days to Disconnect Island: Efficient DFS-Based Solution | minimum-number-of-days-to-disconnect-isl-dm0y | Intuition\nIn this problem, the goal is to determine the minimum number of days required to disconnect an island, where an island is defined as a connected grou | ruturaj_dm | NORMAL | 2024-08-11T12:52:02.575443+00:00 | 2024-08-11T12:53:35.154759+00:00 | 31 | false | # Intuition\nIn this problem, the goal is to determine the minimum number of days required to disconnect an island, where an island is defined as a connected group of land cells (1s) in a grid.\n\n# Approach:\n1. Initial Check: We first count the number of islands in the grid. If the island count is not exactly one, th... | 1 | 0 | ['Java'] | 0 |
minimum-number-of-days-to-disconnect-island | EASY CODE EXPLAINED || Disconnecting a Grid: Minimum Days to Disconnect Land Cells in a Matrix | easy-code-explained-disconnecting-a-grid-85dv | Problem Statement\nThe function minDays determines the minimum number of days required to disconnect a grid (represented as a 2D list) into two or more disconne | Person_UK | NORMAL | 2024-08-11T11:52:36.424717+00:00 | 2024-08-11T11:52:36.424740+00:00 | 39 | false | # Problem Statement\nThe function minDays determines the minimum number of days required to disconnect a grid (represented as a 2D list) into two or more disconnected components by removing land cells. The grid consists of land cells (represented by 1s) and water cells (represented by 0s).\n\n# Key Concepts\n**Grid Tra... | 1 | 0 | ['Array', 'Depth-First Search', 'Python3'] | 0 |
minimum-number-of-days-to-disconnect-island | Ruby || DFS | ruby-dfs-by-alecn2002-w4en | \n# Code\nruby\nclass Grid\n include Enumerable\n\n attr_reader :grid, :h, :w, :hr, :wr\n\n def initialize(grid)\n @grid, @h, @w = grid, grid.si | alecn2002 | NORMAL | 2024-08-11T11:26:14.342906+00:00 | 2024-08-11T11:26:14.342931+00:00 | 10 | false | \n# Code\n```ruby\nclass Grid\n include Enumerable\n\n attr_reader :grid, :h, :w, :hr, :wr\n\n def initialize(grid)\n @grid, @h, @w = grid, grid.size, grid.first.size\n @hr, @wr = (0...h), (0...w)\n end\n\n def each\n grid.each_with_index {|row, ridx|\n row.each_with_index... | 1 | 0 | ['Ruby'] | 0 |
minimum-number-of-days-to-disconnect-island | easy||c++||only BFS||graph||beginner friendly||striver|| | easyconly-bfsgraphbeginner-friendlystriv-j7u8 | Intuition\n Describe your first thoughts on how to solve this problem. \nyou need to get no of island problem before doing it;\nthat qsn is the prerequisite of | sidvi | NORMAL | 2024-08-11T10:46:17.670567+00:00 | 2024-08-11T10:46:17.670597+00:00 | 23 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nyou need to get no of island problem before doing it;\nthat qsn is the prerequisite of this problem\n\nlink:- https://leetcode.com/problems/number-of-islands/description/\nsolution to prerequisite problem:- https://leetcode.com/problems/n... | 1 | 0 | ['Array', 'Breadth-First Search', 'Matrix', 'Strongly Connected Component', 'C++'] | 0 |
minimum-number-of-days-to-disconnect-island | C++ | Max 2 Removal | c-max-2-removal-by-sarvesh225-odq7 | Approach\n Describe your approach to solving the problem. \nIf there are more than 1 disconnected islands return 0.\nIf all the elements of grid are 0 return 0. | sarvesh225 | NORMAL | 2024-08-11T10:32:59.475509+00:00 | 2024-08-11T10:35:14.494387+00:00 | 3 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nIf there are more than 1 disconnected islands return 0.\nIf all the elements of grid are 0 return 0.\nOne by one change the elements of the islands to 0, one at a time and check if any conversion disconnects the island. If it does return 1, else retur... | 1 | 0 | ['C++'] | 0 |
minimum-number-of-days-to-disconnect-island | Simple Approach Beats 78.74% in time and 97.99% in space🔥🔥 | simple-approach-beats-7874-in-time-and-9-4n0r | Intuition\nSimple bombing approach grid to get the number of islands\n Describe your first thoughts on how to solve this problem. \n\n\n# Approach\n1)If number | pranjalkishor5 | NORMAL | 2024-08-11T09:15:15.717545+00:00 | 2024-08-11T09:15:15.717580+00:00 | 6 | false | # Intuition\nSimple bombing approach grid to get the number of islands\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n# Approach\n1)If number of islands is not one return 0.\n2)Try... | 1 | 0 | ['C++'] | 0 |
minimum-number-of-days-to-disconnect-island | Simple BFS ✏️✏️✏️ | Python3 🔥🔥🔥 | simple-bfs-python3-by-lightning_mc_queen-54qv | Complexity\n- Time complexity: O(MN)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(MN)\n Add your space complexity here, e.g. O(n) \n\n# | Lightning_Mc_Queen | NORMAL | 2024-08-11T08:30:40.890867+00:00 | 2024-08-11T08:30:40.890907+00:00 | 10 | false | # Complexity\n- Time complexity: $$O(M*N)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(M*N)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def minDays(self, grid: List[List[int]]) -> int:\n m, n = len(grid), len(grid[0])\n ... | 1 | 0 | ['Breadth-First Search', 'Matrix', 'Python3'] | 0 |
minimum-number-of-days-to-disconnect-island | Interview Solution || Easy to Understand || Most Optimized | interview-solution-easy-to-understand-mo-81s4 | \n\n# Complexity\n- Time complexity: O (n * m)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O (n * m)\n Add your space complexity here, e. | muntajir | NORMAL | 2024-08-11T08:07:08.202912+00:00 | 2024-08-11T08:07:08.202937+00:00 | 109 | false | \n\n# Complexity\n- Time complexity: O (n * m)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O (n * m)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n private final int[] delRow= {1,-1,0,0};\n private final int[] delCol= {0,0,-1,1};\n pr... | 1 | 0 | ['Depth-First Search', 'Matrix', 'Strongly Connected Component', 'Java'] | 2 |
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