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detonate-the-maximum-bombs | JavaScript - DFS Solution | javascript-dfs-solution-by-harsh07bharva-fwbu | \n/**\n * @param {number[][]} bombs\n * @return {number}\n */\nvar maximumDetonation = function(bombs) {\n if(bombs.length <= 1) return bombs.length;\n \n | harsh07bharvada | NORMAL | 2021-12-21T04:25:29.956323+00:00 | 2021-12-21T04:25:29.956359+00:00 | 1,161 | false | ```\n/**\n * @param {number[][]} bombs\n * @return {number}\n */\nvar maximumDetonation = function(bombs) {\n if(bombs.length <= 1) return bombs.length;\n \n let adj = {}, maxSize = 0;\n const checkIfInsideRange = (x, y, center_x, center_y, radius) =>{\n return ( (x-center_x)**2 + (y-center_y)**2 <= ... | 6 | 0 | ['Depth-First Search', 'JavaScript'] | 0 |
detonate-the-maximum-bombs | ✅ [c++] || BFS | c-bfs-by-xor09-m5ui | \n#define ll long long\nclass Solution {\npublic:\n bool isInside(ll circle_x, ll circle_y, ll rad, ll x, ll y){\n if ((x - circle_x) * (x - circle_x) | xor09 | NORMAL | 2021-12-17T14:31:16.141895+00:00 | 2021-12-17T14:31:16.141922+00:00 | 541 | false | ```\n#define ll long long\nclass Solution {\npublic:\n bool isInside(ll circle_x, ll circle_y, ll rad, ll x, ll y){\n if ((x - circle_x) * (x - circle_x) + (y - circle_y) * (y - circle_y) <= rad * rad) return true;\n else return false;\n }\n \n int bfs(int i, unordered_map<int,vector<int>> &ma... | 6 | 0 | ['Breadth-First Search', 'Graph', 'C', 'C++'] | 0 |
detonate-the-maximum-bombs | Easy Math and Graph Solution || C++ | easy-math-and-graph-solution-c-by-abhi_p-dvob | 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 | abhi_pandit_18 | NORMAL | 2023-06-02T16:36:21.876764+00:00 | 2023-06-02T16:36:21.876801+00:00 | 50 | 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. $$O(n)$$ --... | 5 | 0 | ['C++'] | 0 |
detonate-the-maximum-bombs | Detonate Maximum Bombs, C++ Explained Solution | detonate-maximum-bombs-c-explained-solut-8uw3 | The problem here can be solved in 2 ways. Using BFS or DFS. We have discussed BFS here because it can reduce down the code needed in DFS to form the graph first | ShuklaAmit1311 | NORMAL | 2022-07-15T09:07:23.021802+00:00 | 2022-07-15T09:07:23.021844+00:00 | 659 | false | The problem here can be solved in 2 ways. Using **BFS** or **DFS**. We have discussed BFS here because it can reduce down the code needed in DFS to form the graph first. Basically if you see, we have been given a **DIRECTED GRAPH**. Note : **The graph might not be a DAG (Directed Acyclic Graph) cause it might be possib... | 5 | 0 | ['Depth-First Search', 'Breadth-First Search', 'Graph', 'C'] | 2 |
detonate-the-maximum-bombs | Video solution | Intuition explained in detail | C++ | BFS | Hindi | video-solution-intuition-explained-in-de-0amh | Video\nHey everyone i have created a video solution for this problem (its in hindi), it involves intuitive explanation with code, this video is part of my playl | _code_concepts_ | NORMAL | 2024-10-30T09:51:04.944331+00:00 | 2024-10-30T09:59:21.627138+00:00 | 339 | false | # Video\nHey everyone i have created a video solution for this problem (its in hindi), it involves intuitive explanation with code, this video is part of my playlist "Master Graphs"\nVideo link : https://youtu.be/hkD8JUuWbkA\nPlaylist link: : https://www.youtube.com/playlist?list=PLICVjZ3X1AcZ5c2oXYABLHlswC_1LhelY\n\n\... | 4 | 0 | ['C++'] | 0 |
detonate-the-maximum-bombs | EASY 5 POINTER APPROACH || BFS || GRAPH | easy-5-pointer-approach-bfs-graph-by-abh-dh6d | Intuition\n Describe your first thoughts on how to solve this problem. \nThe algorithm aims to find the maximum number of bombs that can be detonated by choosin | Abhishekkant135 | NORMAL | 2023-12-13T01:06:34.481714+00:00 | 2023-12-13T01:06:34.481735+00:00 | 214 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe algorithm aims to find the maximum number of bombs that can be detonated by choosing only one bomb and triggering its chain reaction.\nBFS is an option to solve this problem as in this case we gotta travel from ont to another bomb and... | 4 | 0 | ['Breadth-First Search', 'Graph', 'Java'] | 0 |
detonate-the-maximum-bombs | C# | BFS | Detailed Explanation + Dry Run | Runtime Beats 87.47% | c-bfs-detailed-explanation-dry-run-runti-nld6 | Intuition\nWhile reading the problem we get to understand that this is a sort of problem where a relationship needs to be established between bombs so that when | anaken | NORMAL | 2023-07-15T07:43:33.122914+00:00 | 2023-07-15T08:58:28.354843+00:00 | 114 | false | # Intuition\nWhile reading the problem we get to understand that this is a sort of problem where a relationship needs to be established between bombs so that when one explodes the other one does too. This becomes more clearer by the fact that the range is involved while deciding that. So with this information it become... | 4 | 0 | ['Breadth-First Search', 'Graph', 'C#'] | 1 |
detonate-the-maximum-bombs | short and sweet JS solution using BFS | short-and-sweet-js-solution-using-bfs-by-8k9m | Intuition\nA new bomb explodes if its distance to the current bomb is less than the radius of the current bomb. \nWe can calculate this with a^2 + b^2 = c^2\nWe | Mister_CK | NORMAL | 2023-06-02T20:54:38.173274+00:00 | 2024-04-07T21:00:42.686510+00:00 | 335 | false | # Intuition\nA new bomb explodes if its distance to the current bomb is less than the radius of the current bomb. \nWe can calculate this with a^2 + b^2 = c^2\nWe can use BFS to check which bombs explodes, by adding exploding boms to our queue. \nWe have to check what the maximum number of exploding bombs is for each s... | 4 | 0 | ['JavaScript'] | 3 |
detonate-the-maximum-bombs | Video Explanation - Math to Graph to DFS clear explanation [Java] | video-explanation-math-to-graph-to-dfs-c-9ljo | Approach\nhttps://youtu.be/hTqokf-HCqg\n\n# Similar Problems:\n\n- 463. Island Perimeter\n- 657. Robot Return to Origin\n- 200. Number of Islands\n- Number of I | hridoy100 | NORMAL | 2023-06-02T18:51:32.914083+00:00 | 2023-06-02T19:46:02.602271+00:00 | 978 | false | # Approach\nhttps://youtu.be/hTqokf-HCqg\n\n# Similar Problems:\n\n- [463. Island Perimeter](https://leetcode.com/problems/island-perimeter/)\n- [657. Robot Return to Origin](https://leetcode.com/problems/robot-return-to-origin/)\n- [200. Number of Islands](https://leetcode.com/problems/number-of-islands/)\n- [Number o... | 4 | 0 | ['Depth-First Search', 'Java'] | 1 |
detonate-the-maximum-bombs | BFS SOLUTION | bfs-solution-by-abhai0306-zm3i | \nclass Pair\n{\n int x,y,r;\n Pair(int x,int y,int r)\n {\n this.x = x;\n this.y = y;\n this.r = r;\n }\n}\n\nclass Solution \ | abhai0306 | NORMAL | 2023-06-02T03:06:38.285391+00:00 | 2023-06-02T03:06:38.285436+00:00 | 61 | false | ```\nclass Pair\n{\n int x,y,r;\n Pair(int x,int y,int r)\n {\n this.x = x;\n this.y = y;\n this.r = r;\n }\n}\n\nclass Solution \n{\n int max = 1 ,ans = 1;\n public int maximumDetonation(int[][] bombs) \n {\n \n for(int i=0;i<bombs.length;i++)\n {\n ... | 4 | 0 | ['Breadth-First Search', 'Java'] | 0 |
detonate-the-maximum-bombs | C++ Easy to understand DFS solution | c-easy-to-understand-dfs-solution-by-jay-f0b3 | Intuition\nWe select a bomb Bi to detonate first and then look for other bombs in range (let\'s say it is Bj), it form\'s a graph i.e, there exist a directed ed | jayantsaini0007 | NORMAL | 2023-02-15T14:43:50.472772+00:00 | 2023-02-15T14:43:50.472814+00:00 | 1,552 | false | # Intuition\nWe select a bomb Bi to detonate first and then look for other bombs in range (let\'s say it is Bj), it form\'s a graph i.e, there exist a directed edge between bomb Bi and Bj and so on. Thus, DFS can be applied on each and every bomb in our array and max result can be checked after completion of each DFS.\... | 4 | 0 | ['Graph', 'C++'] | 0 |
detonate-the-maximum-bombs | Python BFS Faster than 96% | python-bfs-faster-than-96-by-welz-vjb7 | Create a dictionary: the key is the index of each bomb, value is the indexes of all the bombs which the key bomb could detonate;\n2. Use BFS to calculate the nu | welz | NORMAL | 2021-12-24T03:57:52.115787+00:00 | 2021-12-24T03:57:52.115814+00:00 | 814 | false | 1. Create a dictionary: the key is the index of each bomb, value is the indexes of all the bombs which the key bomb could detonate;\n2. Use BFS to calculate the number of bombs \n3. **No need to loop for all the keys in the dictionary,** for example:\nif bombs[3] can detonate bombs[4], bombs[5], bombs[6], so there is n... | 4 | 0 | ['Breadth-First Search', 'Python'] | 1 |
detonate-the-maximum-bombs | Simple Java DFS [Faster than 90%] | simple-java-dfs-faster-than-90-by-here-c-qel7 | Here is a brute force solution. TC might be improved. \n\nThis below snippet code improves the runtime to 90%\n\n\nif(k == bombs.length){ //improves runtime.\n | here-comes-the-g | NORMAL | 2021-12-11T16:49:58.928247+00:00 | 2022-01-10T07:45:58.127174+00:00 | 429 | false | Here is a brute force solution. TC might be improved. \n\nThis below snippet code improves the runtime to 90%\n\n```\nif(k == bombs.length){ //improves runtime.\n return k;\n }\n```\n\n```\nclass Solution {\n Map<Integer, List<Integer>> map = new HashMap<>();\n public int maximumDetonation(int[][] bombs) {\n ... | 4 | 0 | [] | 0 |
detonate-the-maximum-bombs | Simple py code explained in detail! || BFS | simple-py-code-explained-in-detail-bfs-b-70cd | Problem Understanding\n\n- This is one of my favourite Math,geometry problem.Let\'s undersatnd the question!\n- They have given a list of list named bombs.It c | arjunprabhakar1910 | NORMAL | 2024-11-05T13:18:42.670605+00:00 | 2024-11-05T13:18:42.670642+00:00 | 335 | false | # Problem Understanding\n\n- This is one of my favourite Math,geometry problem.Let\'s undersatnd the question!\n- They have given a *list of list* named `bombs`.It contains it\'s location and radius as `[x,y,r]`.\n- We can detonate any bomb,and find the `maximum` number of bombs which will be detonated.\n- A `bomb` is... | 3 | 0 | ['Array', 'Math', 'Breadth-First Search', 'Graph', 'Geometry', 'Python3'] | 0 |
detonate-the-maximum-bombs | explaination for why union find not working | explaination-for-why-union-find-not-work-poel | Detonation condition is NOT intersection\n2. 1. BUT intersection should include co-ordinate of other bomb range\n2. UNION FIND WILL GIVE FALSE ANSWER FOR BELOW | youngsam | NORMAL | 2024-03-26T01:48:55.540321+00:00 | 2024-03-26T01:48:55.540367+00:00 | 51 | false | **Detonation condition is NOT intersection\n2. 1. BUT intersection should include co-ordinate of other bomb range\n2. UNION FIND WILL GIVE FALSE ANSWER FOR BELOW CASE**\ngiven A(0,0,5) B (4,0,1) C)(9,0,5)\n\n\... | 3 | 0 | ['Breadth-First Search', 'Graph'] | 0 |
detonate-the-maximum-bombs | Java | BFS | Beats > 99.7% | With Comments | java-bfs-beats-997-with-comments-by-thar-h5x7 | Intuition\n Describe your first thoughts on how to solve this problem. \nWe systematically explore each bomb and its surrounding bombs to determine the maximum | tharunstk2003 | NORMAL | 2023-06-03T03:25:25.967716+00:00 | 2023-06-03T03:25:25.967754+00:00 | 617 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe systematically explore each bomb and its surrounding bombs to determine the maximum number of bombs that can be detonated from each starting point. By performing BFS, we ensure that we consider all reachable bombs and maximize the coun... | 3 | 0 | ['Breadth-First Search', 'Java'] | 2 |
detonate-the-maximum-bombs | An Easy DFS solution approach || C++ | an-easy-dfs-solution-approach-c-by-sazzy-g8zv | Intuition\n Describe your first thoughts on how to solve this problem. \nIf one bomb explodes and the bombs that comes inside its range, it\'s the indication th | sazzysaturn | NORMAL | 2023-06-02T19:25:28.941478+00:00 | 2023-06-02T19:25:28.941515+00:00 | 304 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIf one bomb explodes and the bombs that comes inside its range, it\'s the indication that its a dfs/bfs problem\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nHere, i have discussed the dfs approach, just running ... | 3 | 0 | ['Array', 'Depth-First Search', 'Breadth-First Search', 'C++'] | 0 |
detonate-the-maximum-bombs | Solution in C++ | solution-in-c-by-ashish_madhup-hrsn | 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 | 2023-06-02T15:25:25.508009+00:00 | 2023-06-02T15:25:25.508053+00:00 | 41 | 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 |
detonate-the-maximum-bombs | ✅ 🔥 Python3 || ⚡easy solution | python3-easy-solution-by-maary-ckyy | Code\n\nfrom typing import List\nimport collections\n\nclass Solution:\n def maximumDetonation(self, bombs: List[List[int]]) -> int:\n n2nxt = collect | maary_ | NORMAL | 2023-06-02T14:42:53.671752+00:00 | 2023-06-02T14:42:53.671793+00:00 | 1,237 | false | # Code\n```\nfrom typing import List\nimport collections\n\nclass Solution:\n def maximumDetonation(self, bombs: List[List[int]]) -> int:\n n2nxt = collections.defaultdict(set)\n lb = len(bombs)\n\n for i in range(lb): # i is the source\n xi, yi, ri = bombs[i]\n\n for j in... | 3 | 0 | ['Python3'] | 0 |
detonate-the-maximum-bombs | C++ || EASY TO UNDERSTAND || DFS | c-easy-to-understand-dfs-by-chicken_rice-nld2 | 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 | chicken_rice | NORMAL | 2023-06-02T10:52:57.338694+00:00 | 2023-06-02T10:52:57.338736+00:00 | 255 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 0 | ['C++'] | 1 |
detonate-the-maximum-bombs | 🏆 Detailed Explanation 🏆 ( DFS ) with time and space complexity analysis. | detailed-explanation-dfs-with-time-and-s-l8pz | Approach\n Describe your approach to solving the problem. \nBasic idea is to solve this problem is we are considering n bombs from 0 to n-1. We treat this indiv | malav_mevada | NORMAL | 2023-06-02T10:43:37.002326+00:00 | 2023-06-02T10:43:37.002368+00:00 | 619 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nBasic idea is to solve this problem is we are considering n bombs from 0 to n-1. We treat this individuals bomb as node in the graph. If a bomb can reach to the other bombs it means if bomb radius can cover any other bomb then there should be directed... | 3 | 0 | ['Array', 'Math', 'Depth-First Search', 'Graph', 'Java'] | 3 |
detonate-the-maximum-bombs | Simple solution using Java : DFS | simple-solution-using-java-dfs-by-niketh-7yfi | The algorithm is pretty simple \n- First of all you need to create a graph such that it stores the bombs which are located in its range. We have the radius of t | niketh_1234 | NORMAL | 2023-06-02T08:31:11.295761+00:00 | 2023-06-02T08:31:11.295801+00:00 | 295 | false | # The algorithm is pretty simple \n- First of all you need to create a graph such that it stores the bombs which are located in its range. We have the radius of the bomb and we run a for loop iterating over all the bombs calculate the distance between the two bombs and compare with the radius then we will able to decid... | 3 | 0 | ['Depth-First Search', 'Breadth-First Search', 'Graph', 'C++', 'Java'] | 0 |
detonate-the-maximum-bombs | Detonate the Maximum Bombs | Optimized Solution | Daily Leetcode Challenge | detonate-the-maximum-bombs-optimized-sol-z4wf | \nclass Solution {\npublic:\n // Helper function to calculate distance between two points\n int distance(int x, int y, int x1, int y1)\n {\n dou | aditigulati | NORMAL | 2023-06-02T08:29:20.859177+00:00 | 2023-06-02T08:29:20.859218+00:00 | 39 | false | ```\nclass Solution {\npublic:\n // Helper function to calculate distance between two points\n int distance(int x, int y, int x1, int y1)\n {\n double temp = sqrt(pow(x1 - x, 2) + pow(y1 - y, 2)); // Euclidean distance formula\n return ceil(temp); // Round up the distance to the nearest integer... | 3 | 0 | ['Depth-First Search', 'C'] | 0 |
detonate-the-maximum-bombs | C++ DFS/BFS solutions with detonating bomb process beating 90.22% | c-dfsbfs-solutions-with-detonating-bomb-7ksgn | Intuition\n Describe your first thoughts on how to solve this problem. \nThe crucial part is to create the directed graph, i.e. the adjacent list. The bombs are | anwendeng | NORMAL | 2023-06-02T07:05:30.247953+00:00 | 2023-06-04T11:22:00.695507+00:00 | 2,381 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe crucial part is to create the directed graph, i.e. the adjacent list. The bombs are the the vertice. If bomb[j] is within the bomb range of bomb[i], there is a directed edge from bomb[i] to bomb[j]. Then either of DFS or BFS can be ap... | 3 | 0 | ['Depth-First Search', 'Breadth-First Search', 'Graph', 'Geometry', 'C++'] | 1 |
detonate-the-maximum-bombs | C++ solution using DFS | c-solution-using-dfs-by-piyusharmap-vxyv | Complexity\n- Time complexity:\nO(n^3)\n\n- Space complexity:\nO(n^2)to store the graph + O(n)for dfs recursion stack\n\n# Code\n\n class Solution {\npublic:\n | piyusharmap | NORMAL | 2023-06-02T05:21:51.679546+00:00 | 2023-06-02T05:23:58.969375+00:00 | 1,288 | false | # Complexity\n- Time complexity:\nO(n^3)\n\n- Space complexity:\nO(n^2)to store the graph + O(n)for dfs recursion stack\n\n# Code\n```\n class Solution {\npublic:\n //general dfs algorithm to find all the connected bombs (you can use bfs as well)\n void dfs(int node, vector<int> &visited, vector<int> adj[], int &... | 3 | 0 | ['Array', 'Math', 'Depth-First Search', 'Graph', 'C++'] | 1 |
detonate-the-maximum-bombs | [ C++ ] [ DFS ] | c-dfs-by-sosuke23-k6qn | Code\n\nstruct Solution {\n int maximumDetonation(vector<vector<int>>& a) {\n int n = (int) a.size();\n vector<vector<int>> g(n);\n for | Sosuke23 | NORMAL | 2023-06-02T04:44:59.847963+00:00 | 2023-06-02T04:44:59.848020+00:00 | 1,918 | false | # Code\n```\nstruct Solution {\n int maximumDetonation(vector<vector<int>>& a) {\n int n = (int) a.size();\n vector<vector<int>> g(n);\n for (int i = 0; i < n; i++) {\n for (int j = 0; j < n; j++) {\n if (i == j) {\n continue;\n }\n ... | 3 | 0 | ['Depth-First Search', 'C++'] | 0 |
detonate-the-maximum-bombs | Python3 Solution | python3-solution-by-motaharozzaman1996-v8zp | \n\nclass Solution:\n def maximumDetonation(self, bombs: List[List[int]]) -> int:\n graph = [[] for _ in bombs]\n for i, (xi, yi, ri) in enumer | Motaharozzaman1996 | NORMAL | 2023-06-02T02:25:35.205618+00:00 | 2023-06-02T02:25:35.205667+00:00 | 1,377 | false | \n```\nclass Solution:\n def maximumDetonation(self, bombs: List[List[int]]) -> int:\n graph = [[] for _ in bombs]\n for i, (xi, yi, ri) in enumerate(bombs): \n for j, (xj, yj, rj) in enumerate(bombs): \n if i < j: \n dist2 = (xi-xj)**2 + (yi-yj)**2\n ... | 3 | 0 | ['Python', 'Python3'] | 0 |
detonate-the-maximum-bombs | EASY TO UNDERSTAND | C++ | DFS | easy-to-understand-c-dfs-by-romegenix-3nvo | \n\nclass Solution {\npublic:\n int cnt = 0;\n void dfs(vector<int> adj[], vector<int>& v, int i){\n cnt++;\n v[i] = 1;\n for(int x: | romegenix | NORMAL | 2023-06-02T00:38:43.300977+00:00 | 2023-06-02T00:39:37.392516+00:00 | 378 | false | \n```\nclass Solution {\npublic:\n int cnt = 0;\n void dfs(vector<int> adj[], vector<int>& v, int i){\n cnt++;\n v[i] = 1;\n for(int x: adj[i])\n if(!v[x])\n dfs(adj, v, x);\n }\n void makeAdjList(vector<vector<int>>& b, vector<int> adj[]){\n for(int i =... | 3 | 0 | ['Depth-First Search', 'Graph', 'C++'] | 0 |
detonate-the-maximum-bombs | C++,BFS Easy to Understand. | cbfs-easy-to-understand-by-bnb_2001-khov | Approach:-For Every Bomb We will Check how many others Bombs Lies inside it and how many Bombs lies inside the bomb which Lies inside Checking Bomb and this wil | bnb_2001 | NORMAL | 2022-05-04T06:06:56.319634+00:00 | 2022-05-04T06:06:56.319660+00:00 | 167 | false | **Approach**:-For Every Bomb We will Check how many others Bombs Lies inside it and how many Bombs lies inside the bomb which Lies inside Checking Bomb and this will Go on.\n-->This will be done with the help of **BFS.**\n-->We will return the bomb which will cover Maximum number of Bombs..\n```\n#define ll long long i... | 3 | 0 | ['Breadth-First Search'] | 0 |
detonate-the-maximum-bombs | Easy c++ solution || DFS | easy-c-solution-dfs-by-thanoschild-jyfc | \nclass Solution {\npublic:\n void dfs(int i, int &count, vector<bool> &visited, vector<vector<int>> &adj){\n visited[i] = true;\n count++;\n | thanoschild | NORMAL | 2022-02-13T04:37:24.145672+00:00 | 2022-02-13T04:37:24.145714+00:00 | 375 | false | ```\nclass Solution {\npublic:\n void dfs(int i, int &count, vector<bool> &visited, vector<vector<int>> &adj){\n visited[i] = true;\n count++;\n for(auto it : adj[i])\n {\n if(!visited[it])\n dfs(it, count, visited, adj);\n }\n }\n int maximumDetonation(... | 3 | 0 | ['Depth-First Search', 'Graph', 'C'] | 0 |
detonate-the-maximum-bombs | Union Find 146 / 160 test cases passed. Why? | union-find-146-160-test-cases-passed-why-683a | Any buddy give me some advise? Why Union Find does\'t work!\n\n\nclass Union:\n def __init__ (self,n):\n self.root = [i for i in range(n)]\n se | AndrewHou | NORMAL | 2021-12-12T22:56:11.506873+00:00 | 2021-12-12T22:56:11.506903+00:00 | 947 | false | Any buddy give me some advise? Why Union Find does\'t work!\n```\n\nclass Union:\n def __init__ (self,n):\n self.root = [i for i in range(n)]\n self.rank = [1 for i in range(n)]\n self.count = n\n \n def union(self,x,y):\n rootX = self.find(x)\n rootY = self.find(y)\n... | 3 | 0 | ['Union Find'] | 2 |
detonate-the-maximum-bombs | c++(28ms 100%) Hamilton path with BFS | c28ms-100-hamilton-path-with-bfs-by-zx00-7rbr | Runtime: 28 ms, faster than 100.00% of C++ online submissions for Detonate the Maximum Bombs.\nMemory Usage: 20.4 MB, less than 53.85% of C++ online submissions | zx007pi | NORMAL | 2021-12-12T07:21:40.505602+00:00 | 2021-12-12T07:28:53.961417+00:00 | 145 | false | Runtime: 28 ms, faster than 100.00% of C++ online submissions for Detonate the Maximum Bombs.\nMemory Usage: 20.4 MB, less than 53.85% of C++ online submissions for Detonate the Maximum Bombs.\n```\nclass Solution {\npublic:\n int maximumDetonation(vector<vector<int>>& b) {\n int n = b.size(), answer = 1;\n vect... | 3 | 0 | ['C', 'C++'] | 0 |
detonate-the-maximum-bombs | JavaScript simple BFS explained | javascript-simple-bfs-explained-by-svolk-1d85 | For every bomb we build a list of bombs that will detonate if current bomb will detonante.\nStart dfs for every bomb and count how many bombs will finally deton | svolkovichs | NORMAL | 2021-12-11T18:29:20.456453+00:00 | 2021-12-12T10:46:44.066698+00:00 | 382 | false | For every bomb we build a list of bombs that will detonate if current bomb will detonante.\nStart dfs for every bomb and count how many bombs will finally detonate if current bomb will detonate first.\nStore the max achived number of bombs.\n```\nvar maximumDetonation = function(bombs) {\n const n = bombs.length\n ... | 3 | 0 | ['Breadth-First Search', 'JavaScript'] | 0 |
detonate-the-maximum-bombs | [C++] Graph | Thought Process | c-graph-thought-process-by-user1908v-ti9c | Thought Process in Brief: \n\nI thought about the relation between any two bombs. It appeared every (bomb1,bom2) pair can be represented with true/false. i.e do | user1908v | NORMAL | 2021-12-11T16:53:57.891341+00:00 | 2021-12-11T17:27:50.468471+00:00 | 355 | false | Thought Process in Brief: \n\nI thought about the relation between any two bombs. It appeared every (bomb1,bom2) pair can be represented with true/false. i.e does triggering bomb2 triggers bomb1? So I created a matrix with **(i,j)==true**, if: **j**th bomb triggers **i**th bomb\n\nI thought about triggering every one o... | 3 | 0 | ['Depth-First Search', 'Graph'] | 1 |
detonate-the-maximum-bombs | BFS, C++ | bfs-c-by-1mknown-hqos | First make a graph in such a way that there will be a directed edge from u to v if and only if the distance between u and v is less than or equal to the radius | 1mknown | NORMAL | 2021-12-11T16:14:38.437935+00:00 | 2021-12-11T16:23:05.381945+00:00 | 325 | false | First make a graph in such a way that there will be a directed edge from u to v if and only if the distance between u and v is less than or equal to the radius of u.\n\nThen do Bfs one by one from each bomb and find the maximum ans.\n\ncode:\n```\nint maximumDetonation(vector<vector<int>>& nums) {\n vector<vect... | 3 | 0 | ['Breadth-First Search', 'C'] | 0 |
detonate-the-maximum-bombs | Optimal DFS (Detailed Explanation) | 27ms | optimal-dfs-detailed-explanation-27ms-by-5n0z | Explanation\nThis problem can be approached from the perspective of the bombs being associated with each other in a directed manner (i.e. for a pair of bombs i | ttaylor27 | NORMAL | 2023-11-29T21:21:51.332070+00:00 | 2023-12-19T20:36:49.350428+00:00 | 353 | false | # Explanation\nThis problem can be approached from the perspective of the bombs being associated with each other in a directed manner (i.e. for a pair of bombs `i` and `j`, there can be a case where bomb `i` can blow up bomb `j`, but bomb `j` cannot blow up bomb `i`, and vice versa). Because of this, the bombs have a 1... | 2 | 0 | ['Depth-First Search', 'C++'] | 0 |
detonate-the-maximum-bombs | [python] Union find *can* work but not in the way your thinking | python-union-find-can-work-but-not-in-th-mjqf | Intuition\nIn order to understand the following I assume you have a strong grasp of union find and the default solution to this problem. \n\nWhy doesnt vanilla | ada8 | NORMAL | 2023-10-25T00:19:22.739821+00:00 | 2023-10-29T00:20:33.128578+00:00 | 214 | false | # Intuition\nIn order to understand the following I assume you have a strong grasp of union find and the default solution to this problem. \n\nWhy doesnt vanilla union find work here? Because edges can be directional or bidirectional in this problem. \ni.e If the graph had all bi-directional edges union find would work... | 2 | 0 | ['Union Find', 'Graph', 'Biconnected Component', 'Strongly Connected Component', 'Python3'] | 1 |
detonate-the-maximum-bombs | C++ Simple DFS Solution | c-simple-dfs-solution-by-h_wan8h-e1ab | 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\n\n# Code\n\nclass S | h_WaN8H_ | NORMAL | 2023-09-06T04:49:54.939690+00:00 | 2023-09-06T04:49:54.939709+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\n\n# Code\n```\nclass Solution {\npublic:\n void dfs (unordered_map<int,vector<int>>&adj, int u, vector<bool>&visited, int &ans){\n\n visited[u]=true;\n ... | 2 | 0 | ['Math', 'Depth-First Search', 'Graph', 'Geometry', 'C++'] | 0 |
detonate-the-maximum-bombs | VERY Intuitive SOLUTION!! AMAZING DSA CONCEPTS USED!!! MUST SEE!!! | very-intuitive-solution-amazing-dsa-conc-4lx8 | This question deserves more attention as it is one of the most beautiful problems I\'ve seen! Absolute JOY !!\n# Intuition\n Describe your first thoughts on how | iamsuteerth | NORMAL | 2023-06-13T20:38:56.519307+00:00 | 2023-06-13T20:38:56.519323+00:00 | 55 | false | ## This question deserves more attention as it is one of the most beautiful problems I\'ve seen! Absolute JOY !!\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nNow, the biggest question in anyone\'s mind when they first see this question is "How is this a graph\'s question"? And that... | 2 | 0 | ['Array', 'Math', 'Depth-First Search', 'Graph', 'C++'] | 0 |
detonate-the-maximum-bombs | Easy to understand, Straight forward C++ solution || DFS ✈️✈️✈️✈️✈️✈️✈️ | easy-to-understand-straight-forward-c-so-xk2r | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ajay_1134 | NORMAL | 2023-06-07T08:26:02.436663+00:00 | 2023-06-07T08:26:02.436715+00:00 | 19 | 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 | ['Depth-First Search', 'Graph', 'C++'] | 0 |
detonate-the-maximum-bombs | ✅ || C++ || Beginner Friendly || Easy to Understand || O(N^3) Time Complexity || | c-beginner-friendly-easy-to-understand-o-nmlk | Complexity\n- ## Time complexity:\n1. The outer loop iterates N times, where N is the number of bombs (for(long long i=0; i<bombs.size(); i++)).\n\n Time com | ananttater | NORMAL | 2023-06-03T07:21:20.787925+00:00 | 2023-06-03T07:21:20.787969+00:00 | 23 | false | # Complexity\n- ## Time complexity:\n1. The outer loop iterates N times, where N is the number of bombs (for(long long i=0; i<bombs.size(); i++)).\n\n Time complexity: O(N)\n2. Inside the outer loop, there is a while loop that iterates until the stack is empty.\n\n3. The while loop can potentially iterate N times... | 2 | 0 | ['C++'] | 0 |
detonate-the-maximum-bombs | Kosaraju's Algorithm + Topological Sort | kosarajus-algorithm-topological-sort-by-g1ms2 | Intuition\n Describe your first thoughts on how to solve this problem. \nThis is an attempt to arrive at an O(n^2) solution by using the concept of Strongly Con | sinclaire | NORMAL | 2023-06-03T05:03:41.920738+00:00 | 2023-06-03T05:03:41.920779+00:00 | 430 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis is an attempt to arrive at an $$O(n^2)$$ solution by using the concept of Strongly Connected Components. This approach still runs at $$O(n^3)$$ time.\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1) Creati... | 2 | 0 | ['Breadth-First Search', 'Topological Sort', 'Strongly Connected Component', 'Python3'] | 2 |
detonate-the-maximum-bombs | Easy to Understand with video solution | easy-to-understand-with-video-solution-b-zikr | Intuition\n Describe your first thoughts on how to solve this problem. \nWe have a radius/range for each bomb and every bomb in that radius will explode that wi | mrgokuji | NORMAL | 2023-06-02T17:42:46.880274+00:00 | 2023-06-02T18:55:54.912751+00:00 | 136 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe have a radius/range for each bomb and every bomb in that radius will explode that will start a chain reaction. This implies that we need to constuct a graph where adjecent nodes will be the bomb index in the range.\n\n# Approach\n<!-- ... | 2 | 0 | ['Math', 'Depth-First Search', 'Graph', 'Geometry', 'C++'] | 0 |
detonate-the-maximum-bombs | Maybe fastest Rust solution (100% beat by runtime and memory) | maybe-fastest-rust-solution-100-beat-by-nwaap | Intuition\nFirst thought was to sort by influence to neighbor bombs, then link top influencers. But I\'ve quickly infered it\'s a wrong way.\n\nSecond thought w | zlumyo | NORMAL | 2023-06-02T16:43:24.700486+00:00 | 2023-06-02T16:47:23.464388+00:00 | 64 | false | # Intuition\nFirst thought was to sort by influence to neighbor bombs, then link top influencers. But I\'ve quickly infered it\'s a wrong way.\n\nSecond thought was describe bombs and ther "victims" as graph. Then there is need to extract separated clusters of connected nodes. Cluster with biggest number of nodes is th... | 2 | 0 | ['Graph', 'Geometry', 'Rust'] | 0 |
detonate-the-maximum-bombs | 🔥🔥Easy To Understand Solution- C++ With Comments🔥🔥 | easy-to-understand-solution-c-with-comme-c5de | Intuition\nIn the given 2D vector bombs they are giving a co-ordinate with radius.\nSo intuition is to build an adjacency list in which we can see how \nmany mo | yuvrajkarna27 | NORMAL | 2023-06-02T15:33:14.009708+00:00 | 2023-06-02T15:46:53.158407+00:00 | 531 | false | # Intuition\nIn the given 2D vector bombs they are giving a co-ordinate with radius.\nSo intuition is to build an adjacency list in which we can see how \nmany more co-ordinates we can access at a particular index.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nSo the Approach is to... | 2 | 0 | ['Depth-First Search', 'Graph', 'Recursion', 'C++'] | 2 |
detonate-the-maximum-bombs | Java BFS Solution | java-bfs-solution-by-tbekpro-5cvp | Code\n\nclass Solution {\n public int maximumDetonation(int[][] bombs) {\n Arrays.sort(bombs, (o1, o2) -> o2[2] - o1[2]);\n int res = 0;\n | tbekpro | NORMAL | 2023-06-02T15:31:21.203843+00:00 | 2023-06-02T15:31:21.203888+00:00 | 382 | false | # Code\n```\nclass Solution {\n public int maximumDetonation(int[][] bombs) {\n Arrays.sort(bombs, (o1, o2) -> o2[2] - o1[2]);\n int res = 0;\n for (int i = 0; i < bombs.length; i++) {\n int[][] copy = Arrays.copyOf(bombs, bombs.length);\n int count = 1;\n Queue<... | 2 | 0 | ['Java'] | 1 |
detonate-the-maximum-bombs | C++ || DFS || Intuitive || Clean Code | c-dfs-intuitive-clean-code-by-adi1707-834v | Intuition\nThe bombs will detonate all of the bombs which are in its range. So form groups which connects all bombs which are linked to each other.\nCount the n | adi1707 | NORMAL | 2023-06-02T15:01:04.881847+00:00 | 2023-06-02T15:01:04.881886+00:00 | 48 | false | # Intuition\nThe bombs will detonate all of the bombs which are in its range. So form groups which connects all bombs which are linked to each other.\nCount the number of bombs connected together in a group. The group with maximum number of bombs will be our answer.\n\n# Approach\nForm the a directed graph which connec... | 2 | 0 | ['Depth-First Search', 'Graph', 'Geometry', 'C++'] | 0 |
detonate-the-maximum-bombs | ✨🔥 Python: DFS Solution 🔥✨ | python-dfs-solution-by-patilsantosh-yusp | Approach\n Describe your approach to solving the problem. \nDFS Solution\n\n# Complexity\n- Time complexity: O(N^2)\n Add your time complexity here, e.g. O(n) \ | patilsantosh | NORMAL | 2023-06-02T13:51:36.252139+00:00 | 2023-06-02T13:51:36.252178+00:00 | 103 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nDFS Solution\n\n# Complexity\n- Time complexity: O(N^2)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N + E + V)\n where\n - N : Length of array\n - E : Number of Edges\n - V : Number of Nodes\n<!-- Add yo... | 2 | 0 | ['Python3'] | 0 |
detonate-the-maximum-bombs | DFS || Cinch Solution | dfs-cinch-solution-by-bhumit_joshi-ol42 | Intuition \n- Determine if a bomb is placed within the destroyable range of other bombs \n- Recall equation of a circle- \n X^2 + Y^2 = R^2\n- Equation for | Bhumit_Joshi | NORMAL | 2023-06-02T12:56:40.845594+00:00 | 2023-06-02T12:56:40.845635+00:00 | 147 | false | # Intuition \n- Determine if a bomb is placed within the destroyable range of other bombs \n- **Recall equation of a circle- \n X^2 + Y^2 = R^2**\n- Equation for bomb Within the Range\n**(X-x)^2 + (Y-y)^2 <= R^2** \n- If there is a bomb within range, establish a connection.\n- Detonate each bomb and identify the on... | 2 | 0 | ['Depth-First Search', 'C++'] | 1 |
detonate-the-maximum-bombs | ✅ [Solution][Swift] DFS | solutionswift-dfs-by-adanilyak-0fwx | TC: O(n * n)\nSC: O(n * n)\n\nclass Solution {\n func maximumDetonation(_ bombs: [[Int]]) -> Int {\n func check(\n point: (x: Int, y: Int), | adanilyak | NORMAL | 2023-06-02T12:16:44.287850+00:00 | 2023-06-02T12:16:44.287893+00:00 | 94 | false | **TC:** O(n * n)\n**SC:** O(n * n)\n```\nclass Solution {\n func maximumDetonation(_ bombs: [[Int]]) -> Int {\n func check(\n point: (x: Int, y: Int),\n isInside circle: (x: Int, y: Int, r: Int)\n ) -> Bool {\n let distance = sqrt(\n pow(Double(point.x - ... | 2 | 0 | ['Depth-First Search', 'Swift'] | 0 |
detonate-the-maximum-bombs | Java || 3 ms 100% || Build graph, then DFS from each bomb | java-3-ms-100-build-graph-then-dfs-from-nmg1k | This code converts the bomb location and radius to a graph where each bomb is a node in the graph, and the directional edges of the graph are the other bombs th | dudeandcat | NORMAL | 2023-06-02T09:56:26.261787+00:00 | 2023-06-02T21:53:20.756174+00:00 | 446 | false | This code converts the bomb location and radius to a graph where each bomb is a node in the graph, and the directional edges of the graph are the other bombs that the current bomb can reach with its blast radius. After the graph is build into the variable `links[][]`, a depth-first-search (DFS) is performed starting f... | 2 | 0 | ['Java'] | 0 |
detonate-the-maximum-bombs | C++ || DFS || EASY TO UNDERSTAND | c-dfs-easy-to-understand-by-coder_shaile-6tz0 | 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 | coder_shailesh411 | NORMAL | 2023-06-02T09:34:34.016290+00:00 | 2023-06-02T09:34:34.016319+00:00 | 1,390 | 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 | ['Depth-First Search', 'Graph', 'C++'] | 1 |
detonate-the-maximum-bombs | Python Approach by using Euclidean distance and DFS | python-approach-by-using-euclidean-dista-mntf | \n# Code\n\nclass Solution:\n # DFS ...!\n def dfs(self, i):\n self.visit[i] = 1\n \n for node in self.d[i]:\n if self.vis | Bala_1543 | NORMAL | 2023-06-02T09:14:12.342159+00:00 | 2023-06-02T09:14:12.342203+00:00 | 152 | false | \n# Code\n```\nclass Solution:\n # DFS ...!\n def dfs(self, i):\n self.visit[i] = 1\n \n for node in self.d[i]:\n if self.visit[node] == 1:\n continue\n self.dfs(node)\n\n def maximumDetonation(self, bombs: List[List[int]]) -> int:\n n = len(bomb... | 2 | 0 | ['Python3'] | 0 |
detonate-the-maximum-bombs | Python short and clean. DFS. Functional programming. | python-short-and-clean-dfs-functional-pr-zeez | Approach\nTL;DR, Similar to Editorial Solution but shorter and cleaner.\n\n# Complexity\n- Time complexity: O(n ^ 3)\n\n- Space complexity: O(n ^ 2)\n\n# Code\n | darshan-as | NORMAL | 2023-06-02T04:35:12.450237+00:00 | 2023-06-02T04:35:12.450283+00:00 | 605 | false | # Approach\nTL;DR, Similar to [Editorial Solution](https://leetcode.com/problems/detonate-the-maximum-bombs/editorial/) but shorter and cleaner.\n\n# Complexity\n- Time complexity: $$O(n ^ 3)$$\n\n- Space complexity: $$O(n ^ 2)$$\n\n# Code\n```python\nclass Solution:\n def maximumDetonation(self, bombs: list[list[in... | 2 | 0 | ['Depth-First Search', 'Graph', 'Geometry', 'Python', 'Python3'] | 1 |
detonate-the-maximum-bombs | JAVA || DFS || C++ | java-dfs-c-by-deepakpatel4115-t220 | 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 | deepakpatel4115 | NORMAL | 2023-06-02T03:33:36.959567+00:00 | 2023-06-02T03:33:36.959613+00:00 | 252 | 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 | ['Depth-First Search', 'C++', 'Java'] | 0 |
detonate-the-maximum-bombs | Swift | Minimal BFS | swift-minimal-bfs-by-upvotethispls-567u | BFS (accepted answer)\n\nclass Solution {\n func maximumDetonation(_ bombs: [[Int]]) -> Int {\n\t\n func contains(_ a:[Int], _ b:[Int]) -> Bool {\n | UpvoteThisPls | NORMAL | 2023-06-02T02:43:35.659159+00:00 | 2023-06-02T05:29:24.637862+00:00 | 728 | false | **BFS (accepted answer)**\n```\nclass Solution {\n func maximumDetonation(_ bombs: [[Int]]) -> Int {\n\t\n func contains(_ a:[Int], _ b:[Int]) -> Bool {\n let (deltaX, deltaY) = (a[0]-b[0], a[1]-b[1])\n return deltaX * deltaX + deltaY * deltaY <= a[2] * a[2]\n }\n \n ... | 2 | 0 | ['Swift'] | 0 |
detonate-the-maximum-bombs | C++ || using graph | c-using-graph-by-_biranjay_kumar-xm05 | 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 | _Biranjay_kumar_ | NORMAL | 2023-06-02T01:41:00.680701+00:00 | 2023-06-02T01:41:00.680750+00:00 | 2,207 | 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 |
detonate-the-maximum-bombs | Ruby solution with BFS (100%/100%) | ruby-solution-with-bfs-100100-by-dtkalla-knw8 | Intuition\nFor each bomb, find the bombs it immediately causes to explode, then BFS to find how many it explodes in total.\n\n# Approach\n1. Create a hash of bo | dtkalla | NORMAL | 2023-06-02T00:14:30.835239+00:00 | 2023-06-02T00:19:13.511789+00:00 | 77 | false | # Intuition\nFor each bomb, find the bombs it immediately causes to explode, then BFS to find how many it explodes in total.\n\n# Approach\n1. Create a hash of bombs that each bomb will directly explode (each key is an index, each value is an array of indices of other bombs).\n2. Iterate through every pair of bombs. F... | 2 | 0 | ['Ruby'] | 1 |
detonate-the-maximum-bombs | 🗓️ Daily LeetCoding Challenge June, Day 2 | daily-leetcoding-challenge-june-day-2-by-e1d0 | This problem is the Daily LeetCoding Challenge for June, Day 2. Feel free to share anything related to this problem here! You can ask questions, discuss what yo | leetcode | OFFICIAL | 2023-06-02T00:00:18.667680+00:00 | 2023-06-02T00:00:18.667731+00:00 | 4,455 | false | This problem is the Daily LeetCoding Challenge for June, Day 2.
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If you'd like to share a detailed solution to the problem, please creat... | 2 | 0 | [] | 16 |
detonate-the-maximum-bombs | [C#] BFS Graph Traversal | c-bfs-graph-traversal-by-sh0wmet3hc0de-dz5l | Intuition\nPorted from Java from: https://leetcode.com/problems/detonate-the-maximum-bombs/solutions/1623563/simple-java-bfs/?page=3\n# Approach\nDue to its rad | sh0wMet3hC0de | NORMAL | 2022-12-28T07:53:02.381653+00:00 | 2022-12-28T07:53:02.381700+00:00 | 554 | false | # Intuition\nPorted from Java from: https://leetcode.com/problems/detonate-the-maximum-bombs/solutions/1623563/simple-java-bfs/?page=3\n# Approach\nDue to its radial exploration nature, BFS almost litteraly fits this problem. I confess I did try first Union-Find, but could only get 125 of LeetCode\'s test cases to pass... | 2 | 0 | ['Breadth-First Search', 'C#'] | 0 |
detonate-the-maximum-bombs | C | c-by-tinachien-zy17 | \nint maximumDetonation(int** bombs, int bombsSize, int* bombsColSize){\n bool* check = calloc(bombsSize, sizeof(bool));\n bool** attach = malloc(bombsSiz | TinaChien | NORMAL | 2022-11-05T08:01:23.569812+00:00 | 2022-11-05T08:01:23.569844+00:00 | 71 | false | ```\nint maximumDetonation(int** bombs, int bombsSize, int* bombsColSize){\n bool* check = calloc(bombsSize, sizeof(bool));\n bool** attach = malloc(bombsSize * sizeof(bool*));\n for(int i = 0; i < bombsSize; i++){\n attach[i] = calloc(bombsSize , sizeof(bool));\n }\n\n for(int i = 0; i < bombsSiz... | 2 | 0 | ['Breadth-First Search'] | 0 |
detonate-the-maximum-bombs | C++ || Using DFS || Neat code with comments | c-using-dfs-neat-code-with-comments-by-a-bc94 | \nclass Solution {\npublic:\n void dfs(int node, vector<bool>&visited, vector<int>adj[], int &currDiffused){\n \n currDiffused += 1;// increasi | AvnUtk_26 | NORMAL | 2022-10-11T21:32:46.799614+00:00 | 2022-10-11T21:32:46.799638+00:00 | 656 | false | ```\nclass Solution {\npublic:\n void dfs(int node, vector<bool>&visited, vector<int>adj[], int &currDiffused){\n \n currDiffused += 1;// increasing the number of bombs getting detonated as we do dfs traversal\n \n for(auto child : adj[node]){\n if(!visited[child]){\n ... | 2 | 0 | ['Depth-First Search', 'Graph', 'Geometry', 'C++'] | 0 |
detonate-the-maximum-bombs | Easy to understand Java solution using BFS | easy-to-understand-java-solution-using-b-3f4s | See the solution here:\nhttps://github.com/Freeze777/SDE-Interviewer-Notes/blob/main/LeetCodeJava/src/main/java/leetcode/medium/graph/DetonateMaxBombs.java | freeze_francis | NORMAL | 2022-10-08T17:30:39.116167+00:00 | 2022-10-08T17:30:39.116211+00:00 | 663 | false | See the solution here:\nhttps://github.com/Freeze777/SDE-Interviewer-Notes/blob/main/LeetCodeJava/src/main/java/leetcode/medium/graph/DetonateMaxBombs.java | 2 | 0 | ['Breadth-First Search'] | 0 |
detonate-the-maximum-bombs | Easy-understanding || Python || Faster than 85% | easy-understanding-python-faster-than-85-sb7f | Don\'t be afraid about the size of my code, I just break this down in classes and methods! What I did was converte the bombs in a graph, when the ditance betwee | duduita | NORMAL | 2022-09-22T22:51:12.822282+00:00 | 2022-09-22T22:51:49.321540+00:00 | 1,022 | false | Don\'t be afraid about the size of my code, I just break this down in classes and methods! What I did was converte the bombs in a graph, when the ditance between one bomb and another is smaller than the radius of the source bomb, we connect those two bombs on the graph. \n\nAfter this, we can do a DFS approach and use ... | 2 | 0 | ['Depth-First Search', 'Graph', 'Python', 'Python3'] | 1 |
detonate-the-maximum-bombs | Java DFS | java-dfs-by-java_programmer_ketan-werr | \nclass Solution {\n public int maximumDetonation(int[][] bombs) {\n int n = bombs.length;\n Circle[] circles = new Circle[n];\n for(int | Java_Programmer_Ketan | NORMAL | 2022-09-21T15:13:53.963310+00:00 | 2022-09-21T15:13:53.963347+00:00 | 844 | false | ```\nclass Solution {\n public int maximumDetonation(int[][] bombs) {\n int n = bombs.length;\n Circle[] circles = new Circle[n];\n for(int i=0;i<n;i++) circles[i] = new Circle(bombs[i][0],bombs[i][1],bombs[i][2]);\n List<List<Integer>> graph = new ArrayList<>();\n for(int i=0;i<n;... | 2 | 0 | ['Depth-First Search', 'Java'] | 1 |
detonate-the-maximum-bombs | [Java] Easy and intuitive with explaination || Graph || DFS || Geometry | java-easy-and-intuitive-with-explainatio-rj7x | \nclass Solution {\n /*\n Make directed graph\n u -> v means, v is in the range of u\n check from which node maximum nodes can be reached and return | khushalabrol | NORMAL | 2022-07-21T10:16:07.929147+00:00 | 2022-07-21T10:16:07.929199+00:00 | 730 | false | ```\nclass Solution {\n /*\n Make directed graph\n u -> v means, v is in the range of u\n check from which node maximum nodes can be reached and return the number of nodes reached\n */\n public int maximumDetonation(int[][] bombs) {\n Map<Integer, List<Integer>> graph = new HashMap<>();\n \n... | 2 | 0 | ['Depth-First Search', 'Graph', 'Java'] | 0 |
detonate-the-maximum-bombs | Java with comments 11ms beats 100% DFS | java-with-comments-11ms-beats-100-dfs-by-y9bn | graph is a adjacency list.\n\nclass Solution {\n public int maximumDetonation(int[][] bombs) {\n int n = bombs.length;\n List<Integer>[] graph | ar9 | NORMAL | 2022-04-16T17:59:23.766573+00:00 | 2022-04-28T12:44:45.984575+00:00 | 420 | false | graph is a adjacency list.\n```\nclass Solution {\n public int maximumDetonation(int[][] bombs) {\n int n = bombs.length;\n List<Integer>[] graph = new List[n];\n for (int i = 0; i < n; i++) graph[i] = new ArrayList<Integer>();\n\t\t// O(n^2 /2) by using j = i+1\n\t\t for (int i = 0; i < n; i+... | 2 | 0 | ['Depth-First Search', 'Java'] | 0 |
detonate-the-maximum-bombs | Python Solution that you want : | python-solution-that-you-want-by-goxy_co-jvj1 | \nclass Solution:\n def maximumDetonation(self, bombs: List[List[int]]) -> int:\n if len(bombs)==1:\n return 1\n \n adlist={i | goxy_coder | NORMAL | 2022-03-21T13:20:28.075911+00:00 | 2022-03-21T13:20:28.075956+00:00 | 846 | false | ```\nclass Solution:\n def maximumDetonation(self, bombs: List[List[int]]) -> int:\n if len(bombs)==1:\n return 1\n \n adlist={i:[] for i in range(len(bombs))}\n \n for i in range(len(bombs)):\n x1,y1,r1=bombs[i]\n for j in range(i+1,len(bombs)):\n ... | 2 | 0 | ['Depth-First Search', 'Python', 'Python3'] | 0 |
detonate-the-maximum-bombs | c# : Easy Solution | c-easy-solution-by-rahul89798-1vj2 | \tpublic class Solution\n\t{\n\t\tpublic int MaximumDetonation(int[][] bombs)\n\t\t{\n\t\t\tint max = 0;\n\t\t\tDictionary> graph = new Dictionary>();\n\n\t\t\t | rahul89798 | NORMAL | 2022-02-26T13:54:55.500765+00:00 | 2022-02-26T13:55:11.166363+00:00 | 102 | false | \tpublic class Solution\n\t{\n\t\tpublic int MaximumDetonation(int[][] bombs)\n\t\t{\n\t\t\tint max = 0;\n\t\t\tDictionary<int, List<int>> graph = new Dictionary<int, List<int>>();\n\n\t\t\tfor (int i = 0; i < bombs.GetLength(0); i++)\n\t\t\t{\n\t\t\t\tif (!graph.ContainsKey(i))\n\t\t\t\t\tgraph[i] = new List<int>();\n... | 2 | 0 | ['Breadth-First Search', 'Graph'] | 0 |
detonate-the-maximum-bombs | Java DFS with memorization, faster than 97% | java-dfs-with-memorization-faster-than-9-bkil | Basic idea is simmilar to highest voted posts, just optimized a little with caching.\nHere I used a Set for marking visited for simplicity and it can also be re | shibainulol | NORMAL | 2022-02-24T00:21:36.004792+00:00 | 2022-02-24T00:21:36.004823+00:00 | 153 | false | Basic idea is simmilar to highest voted posts, just optimized a little with caching.\nHere I used a Set for marking visited for simplicity and it can also be reused later, since it contains info about all bombs that will explode if we start at bomb[i].\n\n````\nclass Solution {\n public int maximumDetonation(int[][]... | 2 | 0 | [] | 2 |
detonate-the-maximum-bombs | Simple C++ code | Both BFS and DFS Approaches | simple-c-code-both-bfs-and-dfs-approache-2k3y | 1. DFS Approach : \n\n\n// DFS Approach : ---------->\nclass Solution {\npublic:\n \n // find if point p2(x2,y2) will be inside the circle of point p1(x1, | HustlerNitin | NORMAL | 2022-02-11T11:54:06.668559+00:00 | 2022-02-11T11:54:06.668592+00:00 | 155 | false | **1. DFS Approach :** \n```\n\n// DFS Approach : ---------->\nclass Solution {\npublic:\n \n // find if point p2(x2,y2) will be inside the circle of point p1(x1,y1) \n bool insideCircle(int x1, int y1, int r, int x2, int y2){\n\t // euclidean distance\n int dist = ceil(sqrt( pow(abs(x2-x1),2) + pow(a... | 2 | 1 | ['C', 'C++'] | 0 |
lexicographically-smallest-palindrome | [Java/C++/Python] Two Pointers | javacpython-two-pointers-by-lee215-oay0 | Explanation\nCompare each s[i] with its symmetrical postion in palindrome,\nwhich is s[n - 1 - i].\n\nTo make the lexicographically smallest palindrome,\nwe mak | lee215 | NORMAL | 2023-05-21T04:01:48.905500+00:00 | 2023-05-21T04:01:48.905548+00:00 | 3,936 | false | # **Explanation**\nCompare each `s[i]` with its symmetrical postion in palindrome,\nwhich is `s[n - 1 - i]`.\n\nTo make the lexicographically smallest palindrome,\nwe make `s[i] = s[n - 1 - i] = min(s[i], s[n - i - 1])`\n<br>\n\n# **Complexity**\nTime `O(n)`\nSpace `O(n)`\n<br>\n\n**Java**\n```java\n public String m... | 50 | 0 | ['C', 'Python', 'Java'] | 4 |
lexicographically-smallest-palindrome | Python Elegant & Short | O(n) | 4 Lines | python-elegant-short-on-4-lines-by-kyryl-vf5h | Complexity\n- Time complexity: O(n)\n- Space complexity: O(n)\n\n# Code\n\nclass Solution:\n def makeSmallestPalindrome(self, s: str) -> str:\n letter | Kyrylo-Ktl | NORMAL | 2023-05-21T16:26:32.453582+00:00 | 2023-05-21T16:26:32.453618+00:00 | 1,490 | false | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n```\nclass Solution:\n def makeSmallestPalindrome(self, s: str) -> str:\n letters = list(s)\n\n for i in range(len(s) // 2):\n letters[i] = letters[~i] = min(letters[i], letters[~i])\n\n return \'\'.joi... | 17 | 0 | ['Python', 'Python3'] | 0 |
lexicographically-smallest-palindrome | Java | Easy solution | 8 lines | java-easy-solution-8-lines-by-judgementd-ita9 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | judgementdey | NORMAL | 2023-05-21T04:01:47.176242+00:00 | 2023-05-21T04:04:34.868753+00:00 | 2,061 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here... | 16 | 1 | ['Java'] | 2 |
lexicographically-smallest-palindrome | Take min || Very simple and easy to understand solution | take-min-very-simple-and-easy-to-underst-rv5e | Up vote if this solution helps\n# Approach\nTake two iterator one from front and one from back.\nThen take the min of s(front) & s(back) and set both s(front) | kreakEmp | NORMAL | 2023-05-21T04:04:17.946790+00:00 | 2023-05-21T04:56:10.212035+00:00 | 2,560 | false | <b> Up vote if this solution helps</b>\n# Approach\nTake two iterator one from front and one from back.\nThen take the min of s(front) & s(back) and set both s(front) and s(back) to the min value.\n \n# Code\n```\n string makeSmallestPalindrome(string s) {\n int front = 0, back = s.size()-1;\n while(fr... | 14 | 1 | ['C++'] | 4 |
lexicographically-smallest-palindrome | Python 3 || 4 lines, w/ explanation an example || T/S: 58% / 72% | python-3-4-lines-w-explanation-an-exampl-847r | \nclass Solution:\n def makeSmallestPalindrome(self, s: str) -> str: # Example: s = \'sdnvnfe\'\n\n ans, n = list(s), len(s) | Spaulding_ | NORMAL | 2023-05-21T17:25:08.397632+00:00 | 2024-06-20T19:18:41.140726+00:00 | 1,089 | false | ```\nclass Solution:\n def makeSmallestPalindrome(self, s: str) -> str: # Example: s = \'sdnvnfe\'\n\n ans, n = list(s), len(s) # n = 7 , n//2 = 3\n # ans = [s, d, n, v, n, f, e]\n for i in range(n//2): \n ... | 13 | 0 | ['Python3'] | 1 |
lexicographically-smallest-palindrome | ✔💯 DAY 416 | TWO pointers | 0ms 100% | | [PYTHON/JAVA/C++] | EXPLAINED 🆙🆙🆙 | day-416-two-pointers-0ms-100-pythonjavac-ez94 | Intuition & Approach\n Describe your approach to solving the problem. \n##### \u2022\tThe string is converted to a char array so that the characters can be acce | ManojKumarPatnaik | NORMAL | 2023-05-21T04:01:59.985541+00:00 | 2023-05-21T04:14:22.978269+00:00 | 1,522 | false | # Intuition & Approach\n<!-- Describe your approach to solving the problem. -->\n##### \u2022\tThe string is converted to a char array so that the characters can be accessed more easily.\n##### \u2022\tTwo pointers are initialized to the beginning and end of the array.\n##### \u2022\tA variable is initialized to keep t... | 13 | 0 | ['Two Pointers', 'Python', 'C++', 'Java', 'Python3'] | 2 |
lexicographically-smallest-palindrome | C++ || TWO POINTER || EASY TO UNDERSTAND | c-two-pointer-easy-to-understand-by-gane-bion | \n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n\n// <!-- UPVOTE IF THIS CODE IS HELP FULL FOR YOU\n// IF ANY SUGGETION YOU CAN | ganeshkumawat8740 | NORMAL | 2023-05-21T04:28:02.001411+00:00 | 2023-05-21T05:11:50.421420+00:00 | 1,395 | false | \n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n```\n// <!-- UPVOTE IF THIS CODE IS HELP FULL FOR YOU\n// IF ANY SUGGETION YOU CAN COMMENT HERE. -->\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int i = 0, j = s.length()-1;//make pointer index\n ... | 10 | 0 | ['Two Pointers', 'C++'] | 1 |
lexicographically-smallest-palindrome | || in java | in-java-by-2manas1-mq9l | Intuition\n Describe your first thoughts on how to solve this problem. \n\nRemoved the use of StringBuilder s1, as it was not necessary for the task. Instead, I | 2manas1 | NORMAL | 2023-08-11T14:33:08.309541+00:00 | 2023-08-11T14:33:08.309575+00:00 | 145 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nRemoved the use of StringBuilder s1, as it was not necessary for the task. Instead, I directly manipulated the char array ch.\nRemoved the unnecessary sorting of s1, as you only need to replace characters with the smaller of the two.\nU... | 5 | 0 | ['Java'] | 0 |
lexicographically-smallest-palindrome | ✅[Python] Simple and Clean, beats 88%✅ | python-simple-and-clean-beats-88-by-_tan-erp5 | Please upvote if you find this helpful. \u270C\n\n\nThis is an NFT\n\n# Intuition\nThe problem asks us to modify a given string s by performing operations on it | _Tanmay | NORMAL | 2023-05-29T07:56:04.687662+00:00 | 2023-05-29T07:56:04.687703+00:00 | 467 | false | ### Please upvote if you find this helpful. \u270C\n<img src="https://assets.leetcode.com/users/images/b8e25620-d320-420a-ae09-94c7453bd033_1678818986.7001078.jpeg" alt="Cute Robot - Stable diffusion" width="200"/>\n\n*This is an NFT*\n\n# Intuition\nThe problem asks us to modify a given string `s` by performing operat... | 4 | 0 | ['Two Pointers', 'String', 'Python', 'Python3'] | 0 |
lexicographically-smallest-palindrome | Python3, Two Lines, Use Smaller characters | python3-two-lines-use-smaller-characters-jr0p | Intuition\nWe are checking pairs of characters: s[i] and s[-i-1] and replace them with the lexicograficaly smaller one of them.\n\n# Complexity\n- Time complexi | silvia42 | NORMAL | 2023-05-21T04:39:19.970508+00:00 | 2023-05-21T04:41:00.764962+00:00 | 986 | false | # Intuition\nWe are checking pairs of characters: `s[i]` and `s[-i-1]` and replace them with the lexicograficaly smaller one of them.\n\n# Complexity\n- Time complexity:\n`O(N)`\n\n- Space complexity:\n`O(N)`\n\n# Code\n```\nclass Solution:\n def makeSmallestPalindrome(self, s: str) -> str:\n pal=\'\'.join([m... | 4 | 0 | ['Python3'] | 0 |
lexicographically-smallest-palindrome | c++ solution || easy to understand | c-solution-easy-to-understand-by-harshil-i5f5 | Code\n\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int left = 0;\n int right = s.length() - 1;\n\n while (l | harshil_sutariya | NORMAL | 2023-05-21T04:03:45.328823+00:00 | 2023-05-21T04:03:45.328852+00:00 | 952 | false | # Code\n```\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int left = 0;\n int right = s.length() - 1;\n\n while (left < right) {\n if (s[left] != s[right]) {\n string modified1 = s;\n modified1[right] = s[left];\n\n ... | 4 | 0 | ['C++'] | 1 |
lexicographically-smallest-palindrome | Easy cpp solution | easy-cpp-solution-by-inderjeet09-htvs | \n\n# Code\n\n#include <string>\n#include <algorithm>\n\nclass Solution {\npublic:\n string makeSmallestPalindrome(string str) {\n int start = 0;\n | inderjeet09 | NORMAL | 2023-05-21T04:02:39.429652+00:00 | 2023-05-21T04:02:39.429686+00:00 | 50 | false | \n\n# Code\n```\n#include <string>\n#include <algorithm>\n\nclass Solution {\npublic:\n string makeSmallestPalindrome(string str) {\n int start = 0;\n int end = str.length() - 1;\n char ch[str.length()];\n copy(str.begin(), str.end(), ch);\n \n while (start <= end) {\n ... | 4 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | SIMPLE TWO-POINTER C++ SOLUTION | simple-two-pointer-c-solution-by-jeffrin-p1iz | 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 | Jeffrin2005 | NORMAL | 2024-07-19T12:46:45.976205+00:00 | 2024-08-12T17:10:32.239754+00:00 | 175 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:o(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:o(1)\n<!-- Add your space complexity here, e.g. $$O... | 3 | 0 | ['C++'] | 0 |
lexicographically-smallest-palindrome | Lexicographically Smallest Palindrome (JavaScript | Two Pointers) | lexicographically-smallest-palindrome-ja-9w4p | Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\n\n/**\n * @param {string} s\n * @return {string}\n */\nvar makeSmallestPalindrome = | TheCodeLord | NORMAL | 2024-04-10T01:04:07.594046+00:00 | 2024-04-10T01:04:07.594074+00:00 | 61 | false | # Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(n)$$\n\n# Code\n```\n/**\n * @param {string} s\n * @return {string}\n */\nvar makeSmallestPalindrome = function(s) {\n let left = 0;\n let right = s.length - 1;\n let str = s.split(\'\');\n\n while(left < right) {\n if(str[left] !... | 3 | 0 | ['Two Pointers', 'JavaScript'] | 0 |
lexicographically-smallest-palindrome | Easy to Understand Java Code || Beats 100% | easy-to-understand-java-code-beats-100-b-4y0d | Complexity\n- Time complexity:\nO(n)\n- Space complexity:\nO(1)\n# Code\n\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n char[ | Saurabh_Mishra06 | NORMAL | 2024-04-08T03:26:04.987994+00:00 | 2024-04-08T03:26:04.988039+00:00 | 373 | false | # Complexity\n- Time complexity:\nO(n)\n- Space complexity:\nO(1)\n# Code\n```\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n char[] c = s.toCharArray();\n int i = 0, j = s.length()-1;\n\n while(i < j){\n if(c[i] <c[j]){\n c[j--] = c[i++];\n ... | 3 | 0 | ['Java'] | 1 |
lexicographically-smallest-palindrome | Simple - Easy to Understand Solution || Beats - 100% | simple-easy-to-understand-solution-beats-73t9 | \n\n# Code\n\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n char str[] = s.toCharArray();\n int i=0, j=s.length()-1;\n | tauqueeralam42 | NORMAL | 2023-07-17T09:16:49.586100+00:00 | 2023-07-17T09:16:49.586125+00:00 | 314 | false | \n\n# Code\n```\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n char str[] = s.toCharArray();\n int i=0, j=s.length()-1;\n while(i<j){\n str[i] = (char)Math.min(str[i],str[j]);\n str[j]= str[i];\n i++;\n j--;\n }\n ... | 3 | 0 | ['Two Pointers', 'String', 'C++', 'Java'] | 1 |
lexicographically-smallest-palindrome | Transform | transform-by-votrubac-22sd | C++\ncpp\nstring makeSmallestPalindrome(string s) {\n transform(begin(s), end(s), rbegin(s), begin(s), [](char a, char b){\n return min(a, b);\n }) | votrubac | NORMAL | 2023-05-24T20:20:35.528265+00:00 | 2023-05-24T20:20:35.528300+00:00 | 166 | false | **C++**\n```cpp\nstring makeSmallestPalindrome(string s) {\n transform(begin(s), end(s), rbegin(s), begin(s), [](char a, char b){\n return min(a, b);\n });\n return s;\n}\n``` | 3 | 0 | ['C'] | 1 |
lexicographically-smallest-palindrome | Python3 Solution | python3-solution-by-motaharozzaman1996-m431 | \n\nclass Solution:\n def makeSmallestPalindrome(self, s: str) -> str:\n return \'\'.join(map(min,zip(s,s[::-1])))\n | Motaharozzaman1996 | NORMAL | 2023-05-22T01:30:15.063773+00:00 | 2023-05-22T01:30:15.063811+00:00 | 292 | false | \n```\nclass Solution:\n def makeSmallestPalindrome(self, s: str) -> str:\n return \'\'.join(map(min,zip(s,s[::-1])))\n``` | 3 | 0 | ['Python', 'Python3'] | 0 |
lexicographically-smallest-palindrome | Diagram & Image Explaination🥇 C++ Full Optimized🔥2 PTR | Well Explained | diagram-image-explaination-c-full-optimi-an8m | Diagram\n Describe your first thoughts on how to solve this problem. \n\n\n\n# Approach\n Describe your approach to solving the problem. \nint i=0,j=s.length()- | 7mm | NORMAL | 2023-05-21T06:38:00.973777+00:00 | 2023-05-21T06:38:00.973819+00:00 | 322 | false | # Diagram\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nint i=0,j=s.length()-1;: Initialize ... | 3 | 0 | ['String', 'String Matching', 'C++'] | 1 |
lexicographically-smallest-palindrome | Two Pointers | C++ | two-pointers-c-by-tusharbhart-cehe | \nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int l = 0, r = s.size() - 1;\n while(l < r) {\n if(s[l] != | TusharBhart | NORMAL | 2023-05-21T04:04:53.727734+00:00 | 2023-05-21T04:04:53.727779+00:00 | 590 | false | ```\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int l = 0, r = s.size() - 1;\n while(l < r) {\n if(s[l] != s[r]) {\n char c = min(s[l], s[r]);\n s[l] = s[r] = c;\n }\n l++, r--;\n }\n return s;\n ... | 3 | 0 | ['Two Pointers', 'C++'] | 0 |
lexicographically-smallest-palindrome | Easy Java Solution | easy-java-solution-by-codehunter01-gzty | Approach\n Describe your approach to solving the problem. \nJust check forward and Backword elements are equal or Not. If Not then change bigger one to smaller | codeHunter01 | NORMAL | 2023-05-21T04:04:17.328788+00:00 | 2023-05-22T12:45:15.824215+00:00 | 816 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nJust check forward and Backword elements are equal or Not. If Not then change bigger one to smaller one.\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 s... | 3 | 0 | ['String', 'Java'] | 0 |
lexicographically-smallest-palindrome | Short || Clean || Simple || Java Solution | short-clean-simple-java-solution-by-hima-g747 | \njava []\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n for(int i=0; i<s.length()/2; i++){\n char c = (char)Math.m | HimanshuBhoir | NORMAL | 2023-05-21T04:02:04.975301+00:00 | 2023-05-21T04:29:56.643585+00:00 | 2,505 | false | \n```java []\nclass Solution {\n public String makeSmallestPalindrome(String s) {\n for(int i=0; i<s.length()/2; i++){\n char c = (char)Math.min((int)s.charAt(i),(int)s.charAt(s.length()-1-i));\n s = s.substring(0,i) + c + s.substring(i+1,s.length()-i-1) + c + s.substring(s.length()-i);\... | 3 | 0 | ['Java'] | 5 |
lexicographically-smallest-palindrome | C++ ✅ || EASY ✅ || 3 Lines | c-easy-3-lines-by-dheeraj3220-lqd4 | \n<<<<UpVote\n\n\n\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int i=0,j=s.size()-1;\n while(i<j){\n | Dheeraj3220 | NORMAL | 2023-05-21T04:01:00.834506+00:00 | 2023-05-21T04:08:36.282301+00:00 | 388 | false | \n**<<<<UpVote**\n\n\n```\nclass Solution {\npublic:\n string makeSmallestPalindrome(string s) {\n int i=0,j=s.size()-1;\n while(i<j){\n if(s[i]<s[j]) s[j--]=s[i++];\n else s[i++]=s[j--];\n }\n return s;\n }\n};\n``` | 3 | 0 | ['Two Pointers', 'C'] | 1 |
lexicographically-smallest-palindrome | ⬆️🆙✅ Easy to understand & Best solution || 💯 Beats 100% of users with Java || O(n)💥👏🔥 | up-easy-to-understand-best-solution-beat-8o1m | Please upvote if my solution and efforts helped you.\n***\n\n\n## Approach - 2 Pointers Method\n1 pointer (i) from starting of the string\nanother pointer (j) f | SumitMittal | NORMAL | 2024-06-09T16:35:43.095669+00:00 | 2024-06-09T16:35:43.095692+00:00 | 58 | false | # Please upvote if my solution and efforts helped you.\n***\n\n\n## Approach - 2 Pointers Method\n1 pointer (i) from starting of the string\nanother pointer (j) from the last of the string\nPut minimum elem... | 2 | 0 | ['Array', 'Two Pointers', 'String', 'Java'] | 0 |
lexicographically-smallest-palindrome | Easy JavaScript solution | easy-javascript-solution-by-navyatjacob-8cyy | 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 | navyatjacob | NORMAL | 2024-02-05T12:32:51.961152+00:00 | 2024-02-05T12:32:51.961170+00:00 | 65 | 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)$$ --... | 2 | 0 | ['JavaScript'] | 0 |
lexicographically-smallest-palindrome | 97 ms | 97-ms-by-satvik_yewale-1qes | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Satvik_Yewale | NORMAL | 2023-12-18T19:11:55.179331+00:00 | 2023-12-18T19:11:55.179360+00:00 | 19 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['Python'] | 0 |
lexicographically-smallest-palindrome | Beginner-friendly || Simple solution with Two Pointer in Python3 / TypeScript | beginner-friendly-simple-solution-with-t-juh0 | Intuition\nLet\'s briefly explain what the problem is:\n- there\'s a string s with lowercase English letters\n- our goal is to modify s to be the lexicographica | subscriber6436 | NORMAL | 2023-10-23T17:56:28.807623+00:00 | 2024-01-11T04:42:35.429332+00:00 | 211 | false | # Intuition\nLet\'s briefly explain what the problem is:\n- there\'s a string `s` with lowercase English letters\n- our goal is to modify `s` to be **the lexicographically smallest palindrome**\n\n**A palindrome** is a string, when an original word is equal to its **reversed version**.\nThe simplest way to build this p... | 2 | 0 | ['Two Pointers', 'String', 'TypeScript', 'Python3'] | 1 |
lexicographically-smallest-palindrome | [JAVA] easy solution 95% faster | java-easy-solution-95-faster-by-jugantar-9x8b | \n\n# Approach\n Describe your approach to solving the problem. \nJust check whether elements are equal in the first half and second half of the string. If not | Jugantar2020 | NORMAL | 2023-07-19T17:02:42.866416+00:00 | 2023-07-19T17:02:42.866436+00:00 | 158 | false | \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nJust check whether elements are equal in the first half and second half of the string. If not then change lexicographically bigger characters one to smaller ones.\n\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $... | 2 | 0 | ['Two Pointers', 'String', 'Java'] | 0 |
lexicographically-smallest-palindrome | Simple JAVA Solution for beginners. 9ms. Beats 94.80%. | simple-java-solution-for-beginners-9ms-b-3x77 | 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 | sohaebAhmed | NORMAL | 2023-05-27T09:53:01.394076+00:00 | 2023-05-27T09:53:01.394106+00:00 | 867 | 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 | ['Two Pointers', 'String', 'Java'] | 0 |
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