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cracking-the-safe | c++ | easy | short | c-easy-short-by-venomhighs7-vsbg | \n\n# Code\n\nclass Solution {\npublic:\n unordered_set<string> hash;\n string ans;\n int k;\n \n void dfs(string u) {\n for (int i = 0; i | venomhighs7 | NORMAL | 2022-11-01T03:50:37.501495+00:00 | 2022-11-01T03:50:37.501529+00:00 | 767 | false | \n\n# Code\n```\nclass Solution {\npublic:\n unordered_set<string> hash;\n string ans;\n int k;\n \n void dfs(string u) {\n for (int i = 0; i < k; i++) {\n auto e = u + to_string(i);\n if (!hash.count(e)) {\n hash.insert(e);\n dfs(e.substr(1));\n... | 4 | 1 | ['C++'] | 0 |
cracking-the-safe | C++ | Backtracking | Easy | c-backtracking-easy-by-vaibhavshekhawat-ebrf | ```\nclass Solution {\npublic:\n int total;\n unordered_set st;\n bool func(int i,int k,int n,string& s){\n if(i==total) return 1;\n \n | vaibhavshekhawat | NORMAL | 2022-07-18T08:27:52.295359+00:00 | 2022-07-18T08:32:36.253123+00:00 | 758 | false | ```\nclass Solution {\npublic:\n int total;\n unordered_set<string> st;\n bool func(int i,int k,int n,string& s){\n if(i==total) return 1;\n \n for(int j=0;j<k;j++){\n s.push_back(j+\'0\');\n \n if(s.size()>=n){\n string a = s.substr(s.size()... | 4 | 0 | ['Backtracking', 'C'] | 0 |
cracking-the-safe | Python DFS Intuitive Explanation | python-dfs-intuitive-explanation-by-_dee-i7z0 | I spent hours on this problem getting stuck because I couldn\'t find a clear and intuitive explanation.\n\nThis video provides an awesome intuition: https://www | _deep_ | NORMAL | 2022-07-08T01:41:54.187593+00:00 | 2022-07-08T01:53:58.220973+00:00 | 397 | false | I spent hours on this problem getting stuck because I couldn\'t find a clear and intuitive explanation.\n\nThis video provides an awesome intuition: https://www.youtube.com/watch?v=iPLQgXUiU14\n\nThe first half of this video is helpful and also goes through some examples: https://www.youtube.com/watch?v=VZvU1_oPjg0\n\n... | 4 | 0 | [] | 0 |
cracking-the-safe | Solution | solution-by-deleted_user-qb58 | C++ []\nclass Solution {\npublic:\n using VT = pair<vector<string>,unordered_map<string,int>>;\n VT gNodes(int n, int k) {\n if (n==1) return VT({" | deleted_user | NORMAL | 2023-04-25T08:16:59.313553+00:00 | 2023-04-25T08:29:27.130942+00:00 | 1,311 | false | ```C++ []\nclass Solution {\npublic:\n using VT = pair<vector<string>,unordered_map<string,int>>;\n VT gNodes(int n, int k) {\n if (n==1) return VT({""},{{"",0}});\n VT ns;\n auto prev = gNodes(n-1,k).first;\n for (auto s: prev) {\n s += " ";\n for (int i=0; i<k; ... | 3 | 0 | ['C++', 'Java', 'Python3'] | 1 |
cracking-the-safe | backtrack python | backtrack-python-by-djdheeraj5701-4x8h | Have a set of visited permutations\nbase case: if all permutations found then store current string and return True\nrecursion case: \n try all po | djdheeraj5701 | NORMAL | 2022-08-14T09:42:30.910758+00:00 | 2022-08-14T09:42:30.910805+00:00 | 588 | false | Have a set of visited permutations\nbase case: if all permutations found then store current string and return True\nrecursion case: \n try all possible i from 0 to k as a new character\n\t\t\t\tcall the function again \n\t\t\t\treturn False if none of them worked\n```\nclass Solution:\n def crackSafe(... | 3 | 0 | ['Backtracking', 'Recursion', 'Python'] | 0 |
cracking-the-safe | Java 2ms DFS | java-2ms-dfs-by-jowangcn-wj96 | These are how I reduce the time:\n(1) Use a boolean[] to store which number has been included. and to avoid the transform the number from decimal to other base, | jowangcn | NORMAL | 2021-11-29T23:59:10.409104+00:00 | 2021-11-29T23:59:10.409149+00:00 | 865 | false | These are how I reduce the time:\n(1) Use a boolean[] to store which number has been included. and to avoid the transform the number from decimal to other base, I use the extra space to make it easy. This is quicker than HashSet/HashMap. \n(2) pass the current tested password in integer in the recursion, so that the ne... | 3 | 3 | ['Depth-First Search', 'Java'] | 0 |
cracking-the-safe | [Python 3] De Bruijn Sequence | python-3-de-bruijn-sequence-by-bakerston-h1x1 | \ndef crackSafe(self, n: int, k: int) -> str: \n if n == 1:\n return "".join(map(str, range(k)))\n if k == 1:\n return | Bakerston | NORMAL | 2021-04-11T01:00:34.343070+00:00 | 2021-04-11T01:00:34.343121+00:00 | 402 | false | ```\ndef crackSafe(self, n: int, k: int) -> str: \n if n == 1:\n return "".join(map(str, range(k)))\n if k == 1:\n return "0" * n\n alpha = list(map(str, range(k)))\n a = [0] * k * n\n seq = []\n def db(t, p):\n if t > n:\n ... | 3 | 0 | [] | 0 |
cracking-the-safe | Easy Python solution with explanation | easy-python-solution-with-explanation-by-lyi2 | Total number of passwords is k^n. Create a set to record our accessible numbers.\nTo make our key shortest, we need the maximum overlap with every two different | yixizhou | NORMAL | 2020-09-30T22:53:30.648128+00:00 | 2020-09-30T22:53:30.648159+00:00 | 293 | false | Total number of passwords is k^n. Create a set to record our accessible numbers.\nTo make our key shortest, we need the maximum overlap with every two different passwords, which is n-1.\nStart with "0"*n, pick the last n-1 numbers and add other new numbers to the end, if the new number is not in the set, we add it to t... | 3 | 0 | [] | 3 |
cracking-the-safe | C++, DFS and backtracking | c-dfs-and-backtracking-by-paulpsp-8kin | You don\'t need to pre generate the graph, you can create it on the fly and each state has to create all the next possible states and record them in a set, the | paulpsp | NORMAL | 2020-08-08T00:45:22.708375+00:00 | 2020-08-08T00:45:22.708412+00:00 | 403 | false | You don\'t need to pre generate the graph, you can create it on the fly and each state has to create all the next possible states and record them in a set, the down side is that you to generate all the other `k` states which might have already been visited previously, but it is easier than maintaining the graph stcutru... | 3 | 0 | [] | 0 |
cracking-the-safe | Python short & simple | python-short-simple-by-dylan20-1sp9 | \nclass Solution(object):\n def crackSafe(self, n, k):\n toVisit = collections.defaultdict(lambda: map(str, range(k)))\n toVisit["0" * (n - 1)] | dylan20 | NORMAL | 2018-01-06T16:41:12.224000+00:00 | 2018-01-06T16:41:12.224000+00:00 | 754 | false | ```\nclass Solution(object):\n def crackSafe(self, n, k):\n toVisit = collections.defaultdict(lambda: map(str, range(k)))\n toVisit["0" * (n - 1)] = map(str, range(1, k))\n ans = "0" * n\n while True:\n cur = ans[len(ans) - n + 1:]\n if not toVisit[cur]: return ans\n... | 3 | 0 | [] | 2 |
cracking-the-safe | Java Solution | java-solution-by-archivebizzle-t5ug | Intuition\nThe goal is to find the shortest possible password that contains all possible combinations of the given digits\n\n# Approach\nInitialize Variables:\n | archivebizzle | NORMAL | 2024-01-11T13:49:51.431291+00:00 | 2024-01-11T13:49:51.431311+00:00 | 534 | false | # Intuition\nThe goal is to find the shortest possible password that contains all possible combinations of the given digits\n\n# Approach\nInitialize Variables:\n\nCreate a string allZeros containing \'n\' zeros, representing the initial state of the password.\nCreate a StringBuilder sb and initialize it with allZeros.... | 2 | 0 | ['Java'] | 0 |
cracking-the-safe | JAVA solution | java-solution-by-21arka2002-udzd | \nclass Solution {\n public String crackSafe(int n, int k) {\n HashSet<String>v=new HashSet<>();\n String ans="";\n for(int i=0;i<n;i++) | 21Arka2002 | NORMAL | 2023-04-28T16:13:25.609682+00:00 | 2023-04-28T16:13:25.609748+00:00 | 408 | false | ```\nclass Solution {\n public String crackSafe(int n, int k) {\n HashSet<String>v=new HashSet<>();\n String ans="";\n for(int i=0;i<n;i++) ans+=\'0\';\n int len = n + (int)Math.pow(k,n) - 1;\n ans=dfs(ans,n,k,v,len);\n return ans;\n }\n public String dfs(String s,int ... | 2 | 0 | ['Depth-First Search', 'Recursion', 'Java'] | 0 |
cracking-the-safe | ✅Python 3 - fast (95%+) - explained 😃 - de Bruijn sequence | python-3-fast-95-explained-de-bruijn-seq-e40p | \n\n\n# Intuition\nIt is very intuitive that two adjacent passwords should differ by first and last digits\n- example for n = 3, k = 2:\n\nwe have 8 (2 ** 3) di | noob_in_prog | NORMAL | 2023-01-11T12:36:11.138594+00:00 | 2023-01-11T12:38:13.647861+00:00 | 317 | false | \n\n\n# Intuition\nIt is very intuitive that two **adjacent passwords** should **differ by first and last digits**\n- example for `n = 3`, `k = 2`:\n```\nwe have 8 (2 ** 3) different passwords\n\nsolution: ... | 2 | 0 | ['Backtracking', 'Python3'] | 0 |
cracking-the-safe | C++ solution, simple to understand | c-solution-simple-to-understand-by-cptsm-vqvm | Constructing a De Bruijn sequence solves the question. For any permutation x, a new permutation y can be obtained by removing the first character of x and appen | CPTSMONSTER | NORMAL | 2021-12-27T00:07:22.389553+00:00 | 2021-12-27T00:07:22.389584+00:00 | 571 | false | Constructing a De Bruijn sequence solves the question. For any permutation x, a new permutation y can be obtained by removing the first character of x and appending a new character to the end of x from the set of letters in [0, k-1]. The first n-1 characters of any permutation can be constructed as nodes in a graph and... | 2 | 1 | [] | 0 |
cracking-the-safe | Easy to understand Java DFS | easy-to-understand-java-dfs-by-viveksinu-ae3h | \n/*\nWe need string of n digits with each digit ranging from 0 to k, so max combination = k^n\nstart with "0000" if k=2, then next can be 00001=> this has 2 c | viveksinub | NORMAL | 2021-09-03T08:40:59.116634+00:00 | 2021-09-03T08:40:59.116675+00:00 | 561 | false | ```\n/*\nWe need string of n digits with each digit ranging from 0 to k, so max combination = k^n\nstart with "0000" if k=2, then next can be 00001=> this has 2 combinatioons 0000 , 0001\nlast n digits tell the combination\nstart with all 0 add eack of k numbers in end if not visited\n*/\nclass Solution {\n String ... | 2 | 0 | [] | 0 |
cracking-the-safe | Java | simple backtracking, space trade off faster than 80% | java-simple-backtracking-space-trade-off-l2wf | Check the suffix substring with size of n\n\nclass Solution {\n \n int maxCombination = 1;\n int K;\n int N;\n String res;\n public boolean cr | zoey_mameshiba | NORMAL | 2021-05-07T23:32:09.804158+00:00 | 2021-05-07T23:34:06.403180+00:00 | 503 | false | Check the suffix substring with size of n\n```\nclass Solution {\n \n int maxCombination = 1;\n int K;\n int N;\n String res;\n public boolean crackSafeHelper(String prefix, Set<String> allCombs) {\n if (prefix.length() >= N \n\t\t\t&& allCombs.contains(prefix.substring(prefix.length() - N))) {... | 2 | 0 | [] | 0 |
cracking-the-safe | Swift DFS | swift-dfs-by-johnamcruz-lv13 | \nclass Solution {\n var result = ""\n \n func crackSafe(_ n: Int, _ k: Int) -> String {\n if n == 1 && k == 1 {\n return "0"\n | johnamcruz | NORMAL | 2020-11-06T09:18:00.809585+00:00 | 2020-11-06T09:18:00.809612+00:00 | 191 | false | ```\nclass Solution {\n var result = ""\n \n func crackSafe(_ n: Int, _ k: Int) -> String {\n if n == 1 && k == 1 {\n return "0"\n }\n \n var visited = Set<String>()\n var start = String(repeating: "0", count: n-1)\n dfs(start, k, &visited)\n result.a... | 2 | 0 | [] | 0 |
cracking-the-safe | Java DFS to make sure short circuit! | java-dfs-to-make-sure-short-circuit-by-h-fvph | \nclass Solution {\n int t, n, k;\n public String crackSafe(int n, int k) {\n this.t = (int) Math.pow(k, n);\n this.n = n;\n this.k = | hobiter | NORMAL | 2020-03-02T06:48:41.031174+00:00 | 2020-03-02T06:48:41.031222+00:00 | 337 | false | ```\nclass Solution {\n int t, n, k;\n public String crackSafe(int n, int k) {\n this.t = (int) Math.pow(k, n);\n this.n = n;\n this.k = k;\n StringBuilder sb = new StringBuilder();\n String first = String.join("", Collections.nCopies(n, "0"));\n sb.append(first);\n ... | 2 | 0 | [] | 0 |
cracking-the-safe | O(k * k^n) just loop | ok-kn-just-loop-by-aqin-nzb3 | \nclass Solution {\n public String crackSafe(int n, int k) {\n StringBuilder sb = new StringBuilder();\n Set<String> set = new HashSet<>();\n | aqin | NORMAL | 2020-01-28T02:26:35.329607+00:00 | 2020-01-28T02:26:35.329662+00:00 | 193 | false | ```\nclass Solution {\n public String crackSafe(int n, int k) {\n StringBuilder sb = new StringBuilder();\n Set<String> set = new HashSet<>();\n for (int i = 0; i < n; i++){\n sb.append("0");\n }\n set.add(sb.toString());\n boolean stop = false;\n while(!st... | 2 | 0 | [] | 1 |
cracking-the-safe | Python DP+DFS solution beats 98%, short explanation | python-dpdfs-solution-beats-98-short-exp-zla0 | It is not hard to get that the minimum length for given n and k is k^n+n-1. Then the question becomes:\n1) for any n and k, can we always reach the minimum len | zhipengyu2015 | NORMAL | 2019-08-21T05:25:25.245129+00:00 | 2019-08-21T05:25:25.245181+00:00 | 353 | false | It is not hard to get that the minimum length for given `n` and `k` is `k^n+n-1`. Then the question becomes:\n1) for any `n` and `k`, can we always reach the minimum lenth\n2) if 1) is true, how to construct such string with the minimum length\n\n\nFor 1), I don\'t know how to prove it but I can always find the string... | 2 | 0 | [] | 1 |
cracking-the-safe | Please Help with my Hamiltonian Path Solution (Passed 37/38 Testcases) | please-help-with-my-hamiltonian-path-sol-ev0a | I made a solution that builds a directed graph showing the current node and the nodes reachable from the current node. (Stored as an adjacency list in dict_).\n | spie2005 | NORMAL | 2018-12-19T08:18:10.961534+00:00 | 2018-12-19T08:18:10.961575+00:00 | 1,737 | false | I made a solution that builds a directed graph showing the current node and the nodes reachable from the current node. (Stored as an adjacency list in dict_).\n\nI then run hamiltonian search on this dictionary to find the shortest path that touches every node.\n\nThis solution passed 37/38 testcases. The solution give... | 2 | 0 | [] | 1 |
cracking-the-safe | Simple C++ beats 100% | simple-c-beats-100-by-kevincabbage-odka | Say there is a string A the length of which is n - 1. we can put k different digits in front of it to get k different string B whose length is n. And we can als | kevincabbage | NORMAL | 2018-10-04T02:33:34.962882+00:00 | 2018-10-11T07:29:20.913752+00:00 | 486 | false | Say there is a string A the length of which is n - 1. we can put k different digits in front of it to get k different string B whose length is n. And we can also put k different digits at the end of it to get another k different string C.\n\nSo we can always find a digit and add it to the end of previous string, and we... | 2 | 0 | [] | 0 |
cracking-the-safe | Java Bitwise DFS beats 90% | java-bitwise-dfs-beats-90-by-crunner-bcqs | Based on the assumptions of k and n, we use 4 bits to represent k and an integer to represent the password.\n\n\n public String crackSafe(int n, int k) {\n | crunner | NORMAL | 2018-09-20T01:18:06.235844+00:00 | 2018-09-20T01:18:06.235891+00:00 | 314 | false | Based on the assumptions of k and n, we use 4 bits to represent k and an integer to represent the password.\n\n```\n public String crackSafe(int n, int k) {\n StringBuilder sb = new StringBuilder();\n for (int i = 0; i < n; i++)\n sb.append(0);\n Set<Integer> visited = new HashSet<Int... | 2 | 0 | [] | 0 |
cracking-the-safe | Python simple DFS | python-simple-dfs-by-kleedy-zkge | \nclass Solution(object):\n def crackSafe(self, n, k):\n def dfs(cur, used):\n if len(used) == k**n - 1:\n return [True, cur | kleedy | NORMAL | 2018-05-26T05:28:12.949041+00:00 | 2018-05-26T05:28:12.949041+00:00 | 558 | false | ```\nclass Solution(object):\n def crackSafe(self, n, k):\n def dfs(cur, used):\n if len(used) == k**n - 1:\n return [True, cur]\n used.add(cur[-n:])\n for digit in map(str,range(k)):\n new = cur + digit\n if new[-n:] not in used:\n... | 2 | 0 | [] | 0 |
cracking-the-safe | 6-liner to make new password with (n-1)-prefix (with explanation) | 6-liner-to-make-new-password-with-n-1-pr-acsa | This solution is inspired by @\u5728\u7ebf\u75af\u72c2. Here is my brief explanation:\n\nThe problem is actually to ask for de Bruijn sequence (instead of makin | zzg_zzm | NORMAL | 2017-12-31T03:51:18.024000+00:00 | 2017-12-31T03:51:18.024000+00:00 | 354 | false | This solution is inspired by @\u5728\u7ebf\u75af\u72c2. Here is my brief explanation:\n\nThe problem is actually to ask for [de Bruijn sequence](https://en.wikipedia.org/wiki/De_Bruijn_sequence) (instead of making a cyclic sequence, we need to pad the end with first `n-1` chars).\n\nNote that [de Bruijn sequence](https... | 2 | 0 | ['C++'] | 0 |
cracking-the-safe | C++, DFS, 3 ms | c-dfs-3-ms-by-zestypanda-3v6a | \nclass Solution {\npublic:\n string crackSafe(int n, int k) {\n string ans(n, '0');\n if (k == 1) return ans;\n int N = 10000;\n | zestypanda | NORMAL | 2017-12-25T05:52:29.820000+00:00 | 2017-12-25T05:52:29.820000+00:00 | 1,034 | false | ```\nclass Solution {\npublic:\n string crackSafe(int n, int k) {\n string ans(n, '0');\n if (k == 1) return ans;\n int N = 10000;\n vector<int> visited(N, 0);\n ans += '1'; // starting with "0..01" due to symmetry\n visited[0] = 1;\n visited[1] = 1;\n int len ... | 2 | 0 | [] | 0 |
cracking-the-safe | Java greedy solution | java-greedy-solution-by-hdchen-6rcc | \nclass Solution {\n public String crackSafe(int n, int k) { \n final int total = (int) Math.pow(k, n), mod = (int) Math.pow(10, n - 1) | hdchen | NORMAL | 2017-12-25T17:16:10.235000+00:00 | 2017-12-25T17:16:10.235000+00:00 | 932 | false | ```\nclass Solution {\n public String crackSafe(int n, int k) { \n final int total = (int) Math.pow(k, n), mod = (int) Math.pow(10, n - 1);\n final Map<Integer, Integer> map = new HashMap();\n map.put(0, 0);\n final Stack<Integer> stack = new Stack();\n stack.add(0);... | 2 | 1 | [] | 1 |
cracking-the-safe | simple py solution - using DFS beats 85% | simple-py-solution-using-bfs-beats-85-by-o6vp | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | noam971 | NORMAL | 2025-03-01T09:41:35.278701+00:00 | 2025-03-01T09:59:07.830625+00:00 | 53 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 1 | 0 | ['Python3'] | 0 |
cracking-the-safe | Python - Optimal, simple DFS | python-optimal-simple-dfs-by-babos-c5a1 | Approach\n Describe your approach to solving the problem. \nThe key to the DFS is to move to the furthest node from the current one. Starting from passwords of | babos | NORMAL | 2024-11-27T23:07:14.415980+00:00 | 2024-11-27T23:07:14.416005+00:00 | 60 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nThe key to the DFS is to move to the furthest node from the current one. Starting from passwords of 0s, the next character added should be k-1. Therefore, we loop through the possible digits in reverse order. As a result, there is no need to check for... | 1 | 0 | ['Python3'] | 0 |
cracking-the-safe | Easy C++ Solution | Beats 70% | C++ | DFS | easy-c-solution-beats-70-c-dfs-by-rajhar-y95o | 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 | rajharsh_24 | NORMAL | 2024-10-16T06:34:34.814692+00:00 | 2024-10-16T06:34:34.814723+00:00 | 42 | 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', 'Graph', 'C++'] | 0 |
cracking-the-safe | Solution from Marlen09 explained to myself | solution-from-marlen09-explained-to-myse-axre | Intuition\nI\'m only writing this shorter copy of Marlen09\'s solution to reinforce my understanding of the problem.\n\nThe graph we want to create is a graph w | Fustigate | NORMAL | 2024-05-17T13:53:02.361282+00:00 | 2024-05-17T13:53:02.361306+00:00 | 25 | false | # Intuition\nI\'m only writing this shorter copy of Marlen09\'s solution to reinforce my understanding of the problem.\n\nThe graph we want to create is a graph which maps one possible solution to the next possible solution. For example if n = 2 and k = 2, the solution should have 2 digits constructed by numbers in {0,... | 1 | 0 | ['Depth-First Search', 'Graph', 'Python3'] | 0 |
cracking-the-safe | Easy Solution | easy-solution-by-anupamkumar-1-g6p7 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nThe approach used in th | 20250411.AnupamKumar-1 | NORMAL | 2023-09-06T08:12:32.062056+00:00 | 2023-09-06T08:12:32.062087+00:00 | 38 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach used in this solution is to find a De Bruijn sequence, which is a sequence containing all possible combinations of characters of length n in a cyclic mann... | 1 | 0 | ['Java'] | 0 |
cracking-the-safe | Easy-to-understand C++ solution, textbook dfs implementation | easy-to-understand-c-solution-textbook-d-vp8x | Approach by Example\nn = 2, k = 2\n\n[00]\n [01]\n [11]\n [10]\n\nWe start with 00, and every time we shift the window right by 1, we hope to find a new pass | taucross9368 | NORMAL | 2023-04-02T02:47:10.396975+00:00 | 2023-04-02T03:52:29.484180+00:00 | 108 | false | # Approach by Example\nn = 2, k = 2\n```\n[00]\n [01]\n [11]\n [10]\n```\nWe start with `00`, and every time we shift the window right by 1, we hope to find a new password never tried before, until we have tried all possible passwords.\n\nUse dfs: keep trying above step by step. If we have tried all possible combina... | 1 | 0 | ['C++'] | 0 |
cracking-the-safe | EASY TO UNDERSTAND 70% FASTER C++ SOLUTION | easy-to-understand-70-faster-c-solution-e29do | \nclass Solution {\npublic:\n \n string crackSafe(int n, int k) {\n string ans = string(n,\'0\');\n unordered_set<string> s;\n s.inse | abhay_12345 | NORMAL | 2022-11-27T08:33:20.521371+00:00 | 2022-11-27T08:33:20.521411+00:00 | 2,068 | false | ```\nclass Solution {\npublic:\n \n string crackSafe(int n, int k) {\n string ans = string(n,\'0\');\n unordered_set<string> s;\n s.insert(ans);\n for(int i = 0; i < (pow(k,n)); i++){\n string pre = ans.substr(ans.length()-n+1);\n for(int j = k-1; j>=0; j--){\n ... | 1 | 0 | ['Depth-First Search', 'C', 'Iterator', 'Ordered Set', 'C++'] | 1 |
cracking-the-safe | The most ugly code but works [Java][backtracking] | the-most-ugly-code-but-works-javabacktra-idvh | The key idea is to check if all possible combination are occured without duplication. And the minLength of String is = Math.pow(k,n) + n - 1. So we can use a h | liuxiaomingskm | NORMAL | 2022-07-27T22:27:12.778758+00:00 | 2022-07-27T22:27:12.778787+00:00 | 359 | false | The key idea is to check if all possible combination are occured without duplication. And the minLength of String is = Math.pow(k,n) + n - 1. So we can use a hashset to store all combination and if it existed, jump to next number.\n```\n\nclass Solution {\n String ans;\n public String crackSafe(int n, int k) {\n... | 1 | 0 | ['Backtracking', 'Recursion', 'Java'] | 0 |
cracking-the-safe | DFS - De Bruijn Sequence - Commented code | dfs-de-bruijn-sequence-commented-code-by-yp2n | \nclass Solution {\npublic:\n string crackSafe(int n, int k) {\n //start with all zeros\n string ans(n,\'0\');\n \n //total no of | gazonda | NORMAL | 2022-06-28T09:53:24.554763+00:00 | 2022-06-28T09:53:24.554813+00:00 | 349 | false | ```\nclass Solution {\npublic:\n string crackSafe(int n, int k) {\n //start with all zeros\n string ans(n,\'0\');\n \n //total no of sequences\n int total=pow(k,n);\n unordered_set<int> v;\n \n //storing the integer version of strings in visited array\n ... | 1 | 0 | ['Depth-First Search', 'C'] | 0 |
cracking-the-safe | Python || DFS || Sets || O(k * (k ^ n)) | python-dfs-sets-ok-k-n-by-in_sidious-fry0 | \nclass Solution:\n def crackSafe(self, n: int, k: int) -> str:\n seen=set()\n def dfs(s,last_n):\n if len(seen)==(k**n): return s\n | iN_siDious | NORMAL | 2022-04-07T21:09:27.532011+00:00 | 2022-04-07T21:13:08.803105+00:00 | 494 | false | ```\nclass Solution:\n def crackSafe(self, n: int, k: int) -> str:\n seen=set()\n def dfs(s,last_n):\n if len(seen)==(k**n): return s\n if len(last_n)<n: # If len<n,keep adding zeros as and valid string can be returned\n if len(s+"0")==n: \n ... | 1 | 0 | ['Depth-First Search', 'Ordered Set', 'Python'] | 0 |
cracking-the-safe | Python greedy method 100% | python-greedy-method-100-by-lonelykid-ruc2 | Use a base k number to represent node, maximum (n-1) digits.\nStart from 0 and always use largest available edge. \nNext node is (node*k+edge) % nodeNum\n\npyth | lonelykid | NORMAL | 2022-02-09T00:44:35.925003+00:00 | 2022-02-09T00:44:35.925046+00:00 | 683 | false | Use a base k number to represent node, maximum (n-1) digits.\nStart from 0 and always use largest available edge. \nNext node is (node*k+edge) % nodeNum\n\n```python\nclass Solution:\n def crackSafe(self, n: int, k: int) -> str:\n ans = list()\n nodeNum = k**(n-1)\n edges = [k-1]*nodeNum\n\n ... | 1 | 1 | ['Greedy', 'Python'] | 0 |
cracking-the-safe | C#. Euler path. Faster than 100% | c-euler-path-faster-than-100-by-ivananti-9g5d | Alogrithm\nJust a standard Euler path algorithm. Surprise, but it seems to be faster than 100% C# submissions. I thought that De Bruijn approach (Lyndon words) | IvanAntipov | NORMAL | 2022-01-30T00:18:36.352212+00:00 | 2022-01-30T00:39:12.033948+00:00 | 511 | false | # Alogrithm\nJust a standard Euler path algorithm. Surprise, but it seems to be faster than 100% C# submissions. I thought that De Bruijn approach (Lyndon words) should be faster.\n\n\nEach state is represented as the sum for i from 0 to n-2 of i_nth_symbols*k^1. \n\n<img src="https://render.githubusercontent.com/rende... | 1 | 0 | ['C#'] | 0 |
cracking-the-safe | C++ || Graph | c-graph-by-ujjwal2001kant-iptj | ```\nclass Solution {\npublic:\n string crackSafe(int n, int k){\n unordered_sets;\n \n string cur="";\n for(int i=0; i<n; i++)\n | ujjwal2001kant | NORMAL | 2022-01-07T07:02:01.181176+00:00 | 2022-05-01T00:22:26.149276+00:00 | 519 | false | ```\nclass Solution {\npublic:\n string crackSafe(int n, int k){\n unordered_set<string>s;\n \n string cur="";\n for(int i=0; i<n; i++)\n {\n cur+="0";\n }\n string ans1="";\n s.insert(cur);\n \n ans1+=cur;\n \n int ans=po... | 1 | 0 | ['Graph'] | 0 |
cracking-the-safe | javascript dfs/recursion and iteration 95ms | javascript-dfsrecursion-and-iteration-95-cnjw | dfs/recursion\n\nlet total, k, n, res;\nconst crackSafe = (N, K) => {\n n = N, k = K, total = k ** n;\n res = \'0\'.repeat(n), visit = new Set([res]);\n | henrychen222 | NORMAL | 2021-11-28T01:43:42.907492+00:00 | 2021-11-28T01:45:12.685672+00:00 | 240 | false | dfs/recursion\n```\nlet total, k, n, res;\nconst crackSafe = (N, K) => {\n n = N, k = K, total = k ** n;\n res = \'0\'.repeat(n), visit = new Set([res]);\n dfs(visit, res);\n return res;\n};\n\nconst dfs = (visit) => {\n if (visit.size == total) return;\n let pre = res.slice(res.length - n + 1); // l... | 1 | 0 | ['Depth-First Search', 'Recursion', 'Iterator', 'JavaScript'] | 1 |
cracking-the-safe | Swift (DFS + Backtracking) + Visualization + Explanation + Unit tests O(K^n) time O(K^n) space | swift-dfs-backtracking-visualization-exp-9pl6 | \n// MARK: Unit Tests\nfunc testExample1(){\n let obj=Solution()\n let n = 1, k = 2\n let output = "10"\n assert(obj.crackSafe(n,k)==output,"failed | amhatami | NORMAL | 2021-09-29T05:03:50.845041+00:00 | 2021-09-29T05:04:28.014726+00:00 | 360 | false | ```\n// MARK: Unit Tests\nfunc testExample1(){\n let obj=Solution()\n let n = 1, k = 2\n let output = "10"\n assert(obj.crackSafe(n,k)==output,"failed example 1")\n}\nfunc testExample2(){\n let obj=Solution()\n let n = 2, k = 2\n let output = "01100"\n assert(obj.crackSafe(n,k)==output,"failed e... | 1 | 1 | [] | 0 |
cracking-the-safe | JAVA AC DFS | java-ac-dfs-by-gwfz0720-4pd4 | \nclass Solution {\n public String crackSafe(int n, int k) {\n Set<String> set = new HashSet<>();\n StringBuilder sb = new StringBuilder();\n | gwfz0720 | NORMAL | 2021-09-27T18:17:39.032467+00:00 | 2021-09-27T18:17:39.032512+00:00 | 296 | false | ```\nclass Solution {\n public String crackSafe(int n, int k) {\n Set<String> set = new HashSet<>();\n StringBuilder sb = new StringBuilder();\n if(dfs(n, k, set, sb)) //return boolean as mark that sb is the string of mininum length to meet the requirement.\n return sb.toString();\n ... | 1 | 0 | [] | 0 |
cracking-the-safe | Really easy to understand python DFS solution | really-easy-to-understand-python-dfs-sol-fl9z | \nclass Solution:\n def crackSafe(self, n: int, k: int) -> str:\n \'\'\'\n n = 2; k = 2\n \n nodes = 00,01,10,11\n 00\n | ikna | NORMAL | 2021-09-24T05:42:42.374194+00:00 | 2021-09-24T05:42:42.374225+00:00 | 661 | false | ```\nclass Solution:\n def crackSafe(self, n: int, k: int) -> str:\n \'\'\'\n n = 2; k = 2\n \n nodes = 00,01,10,11\n 00\n 01\n 10 11\n 10 \n \n 00110\n \'\'\'\n out_len = k**n + 1\n \n depth = k**n \n ... | 1 | 2 | ['Backtracking', 'Depth-First Search', 'Python'] | 0 |
cracking-the-safe | c++(0ms 100%)space 100% small solution (only math...) | c0ms-100space-100-small-solution-only-ma-3z66 | Runtime: 0 ms, faster than 100.00% of C++ online submissions for Cracking the Safe.\nMemory Usage: 6.2 MB, less than 100.00% of C++ online submissions for Crack | zx007pi | NORMAL | 2021-08-25T12:38:41.736080+00:00 | 2021-08-25T13:15:11.544906+00:00 | 603 | false | Runtime: 0 ms, faster than 100.00% of C++ online submissions for Cracking the Safe.\nMemory Usage: 6.2 MB, less than 100.00% of C++ online submissions for Cracking the Safe.\n```\nclass Solution {\npublic:\n string crackSafe(int n, int k) {\n int table[10000] = {0}, p = --k, div = pow(10, n - 1);\n table[k] = 1;... | 1 | 3 | ['C', 'C++'] | 3 |
cracking-the-safe | C++ | Simple | Eulerian Path | c-simple-eulerian-path-by-dheerajchugh30-um78 | \n#include<queue>\n#include<stack>\n#include<string>\n#include<unordered_map>\n\nusing namespace std;\ntypedef unordered_map<string, vector<string>> GRAPH;\n\nu | dheerajchugh303 | NORMAL | 2021-08-15T20:13:32.561779+00:00 | 2021-08-15T20:13:32.561811+00:00 | 526 | false | ```\n#include<queue>\n#include<stack>\n#include<string>\n#include<unordered_map>\n\nusing namespace std;\ntypedef unordered_map<string, vector<string>> GRAPH;\n\nusing namespace std;\nclass Solution {\nprivate:\n GRAPH generateGraph(int n, int k)\n {\n GRAPH g;\n string str(n, \'0\');\n queue... | 1 | 0 | [] | 0 |
cracking-the-safe | C++ | Simple | Hamiltonian Cycle | c-simple-hamiltonian-cycle-by-darkhorse7-my2h | \nclass Solution {\npublic:\n string getHamiltonianSequence(string current,int n, int k, unordered_set<string>& visited) {\n \n if(visited.siz | darkHorse77 | NORMAL | 2021-07-16T12:07:06.158308+00:00 | 2021-07-16T12:07:46.704650+00:00 | 886 | false | ```\nclass Solution {\npublic:\n string getHamiltonianSequence(string current,int n, int k, unordered_set<string>& visited) {\n \n if(visited.size()==pow(k,n)) //max number of sequences possible\n return "";\n \n for(int i = 0; i<k; ++i){\n \n string next... | 1 | 0 | [] | 0 |
cracking-the-safe | Super easy DFS | C++ | 0ms - beats 100% | super-easy-dfs-c-0ms-beats-100-by-_shiva-nb1k | \nclass Solution {\npublic:\n int power(int base, int exponent){\n if(exponent==0)return 1;\n int halfWork = power(base,exponent/2);\n i | _shivam7 | NORMAL | 2021-05-29T03:56:09.662457+00:00 | 2021-05-29T03:56:09.662496+00:00 | 827 | false | ```\nclass Solution {\npublic:\n int power(int base, int exponent){\n if(exponent==0)return 1;\n int halfWork = power(base,exponent/2);\n if(exponent&1)return halfWork*halfWork*base;\n return halfWork*halfWork;\n }\n \n bool dfs(int u, int n, int k, vector<vector<bool>> &vis, str... | 1 | 0 | ['Depth-First Search', 'C'] | 0 |
cracking-the-safe | C++ short and sweet | c-short-and-sweet-by-galster-7qmc | \nclass Solution {\npublic:\n unordered_set<string> found;\n int totalStates = 0;\n \n string crackSafe(const string &curr, const string &state, int | galster | NORMAL | 2021-03-22T03:48:32.356846+00:00 | 2021-03-22T03:48:32.356875+00:00 | 333 | false | ```\nclass Solution {\npublic:\n unordered_set<string> found;\n int totalStates = 0;\n \n string crackSafe(const string &curr, const string &state, int k){\n found.insert(state);\n \n if(found.size() == totalStates){\n return curr;\n }\n \n for(int i = 0;... | 1 | 0 | [] | 0 |
cracking-the-safe | python dfs + greedy | python-dfs-greedy-by-shuuchen-1g3v | ```\nclass Solution:\n def crackSafe(self, n: int, k: int) -> str:\n \n def dfs(s, seen):\n \n if len(seen) == kn:\n | shuuchen | NORMAL | 2021-03-06T04:05:22.669350+00:00 | 2021-03-06T04:05:22.669384+00:00 | 362 | false | ```\nclass Solution:\n def crackSafe(self, n: int, k: int) -> str:\n \n def dfs(s, seen):\n \n if len(seen) == k**n:\n return s\n \n for x in range(k):\n new_s = (s[-n+1:] if n > 1 else \'\') + str(x)\n if new_s in... | 1 | 0 | [] | 0 |
cracking-the-safe | Proof: why Eulerian Circuit works, and a strict O(k^n) solution | proof-why-eulerian-circuit-works-and-a-s-dd99 | Most solutions assume Eulerian Circuit can generate De Brujin Sequence without proof.\nHere is a nice one:\n\nhttp://www-users.math.umn.edu/~reiner/Classes/DeBr | cal_apple | NORMAL | 2020-12-06T17:58:48.734100+00:00 | 2020-12-06T17:59:32.111955+00:00 | 371 | false | Most solutions assume Eulerian Circuit can generate De Brujin Sequence without proof.\nHere is a nice one:\n\nhttp://www-users.math.umn.edu/~reiner/Classes/DeBruijn.pdf\n\n\nPlus a strict` O(k^n)` solution\n\n```\n\n vector<vector<int>> graph;\n string res;\n int N, k;\n \n // de Brujin sequence can be c... | 1 | 0 | [] | 1 |
cracking-the-safe | Easy to understand java solution using DFS | easy-to-understand-java-solution-using-d-vbqe | \nclass Solution {\n public String crackSafe(int n, int k) {\n \n // k possibilities for each of the n digits - total number of combination is | podurisaicharan | NORMAL | 2020-08-21T00:31:25.334512+00:00 | 2020-08-21T00:32:34.432931+00:00 | 235 | false | ```\nclass Solution {\n public String crackSafe(int n, int k) {\n \n // k possibilities for each of the n digits - total number of combination is k*k*k...(ntimes) = k^n\n int total = (int)Math.pow(k, n);\n \n // start with the basic one "000000" (n digits)\n StringBuffer sb ... | 1 | 0 | [] | 0 |
cracking-the-safe | c# DFS | c-dfs-by-lchunlosaltos-yuin | \npublic class Solution {\n //each password is n digits; \n //treat the password as nodes;\n //visit every nodes once to create the password with all p | lchunlosaltos | NORMAL | 2020-05-08T02:40:52.110981+00:00 | 2020-05-08T02:40:52.111028+00:00 | 181 | false | ```\npublic class Solution {\n //each password is n digits; \n //treat the password as nodes;\n //visit every nodes once to create the password with all possible answers in one string.\n //use DFS; the node should be seeds with n-1 "0" and append a number from k each time;\n //Make sure we dont duplicate... | 1 | 0 | [] | 0 |
cracking-the-safe | [Python] Short dfs() | python-short-dfs-by-hjscoder-li6v | Explanation\n\nStart from "0"n, using a slide window idea, everytime the window moves toward right for 1 digit, for this digit we have k options to create a new | hjscoder | NORMAL | 2020-05-06T00:45:00.817286+00:00 | 2020-05-06T00:45:00.817338+00:00 | 239 | false | **Explanation**\n\nStart from "0"*n, using a slide window idea, everytime the window moves toward right for 1 digit, for this digit we have k options to create a new string that have never being created before. So we continue creating strings until the amount of unique strings equals to the maxium possible amount => to... | 1 | 0 | [] | 0 |
cracking-the-safe | Simple Java Solution[BackTracking] | simple-java-solutionbacktracking-by-xlay-hf9a | \nclass Solution {\n Set<String> visited = new HashSet<>();\n int total = 0;\n StringBuilder ans = new StringBuilder();\n int k;\n private boolea | xlayman | NORMAL | 2020-04-26T01:20:29.958376+00:00 | 2020-04-26T01:20:29.958408+00:00 | 159 | false | ```\nclass Solution {\n Set<String> visited = new HashSet<>();\n int total = 0;\n StringBuilder ans = new StringBuilder();\n int k;\n private boolean dfs(String start){\n if(visited.size() == total)\n return true;\n for(int i = 0; i < k; i++){\n String curr = start.sub... | 1 | 0 | [] | 0 |
cracking-the-safe | C# Solution DFS | c-solution-dfs-by-leonhard_euler-g259 | \npublic class Solution \n{\n public string CrackSafe(int n, int k) \n {\n var sb = new StringBuilder();\n for (int i = 0; i < n; i++) \n | Leonhard_Euler | NORMAL | 2020-01-03T07:52:31.793017+00:00 | 2020-01-03T07:52:31.793068+00:00 | 179 | false | ```\npublic class Solution \n{\n public string CrackSafe(int n, int k) \n {\n var sb = new StringBuilder();\n for (int i = 0; i < n; i++) \n sb.Append(\'0\');\n DFS(n, k, sb, new HashSet<string>(new string[] {sb.ToString()}));\n return sb.ToString();\n }\n \n privat... | 1 | 0 | [] | 0 |
cracking-the-safe | Simple C++ solution with 2 for loops | simple-c-solution-with-2-for-loops-by-em-ozw6 | \nclass Solution \n{\n public:\n string crackSafe(int n, int k) \n {\n string result(n,\'0\');\n unordered_set<string> v;\n for(in | eminem18753 | NORMAL | 2019-12-05T05:44:01.428545+00:00 | 2019-12-05T05:53:21.668706+00:00 | 219 | false | ```\nclass Solution \n{\n public:\n string crackSafe(int n, int k) \n {\n string result(n,\'0\');\n unordered_set<string> v;\n for(int i=0;i<pow(k,n)-1;i++)\n {\n for(int j=k-1;j>-1;j--)\n {\n string temp=result.substr((int)result.size()-n+1,n-1)... | 1 | 0 | [] | 0 |
cracking-the-safe | A Java Solution | a-java-solution-by-david137-id5x | \nclass Solution {\n private Set<String> visited = new HashSet<String>();\n private String ans = "";\n private int goal=1;\n \n public String cra | david137 | NORMAL | 2019-12-01T21:43:55.359853+00:00 | 2019-12-01T21:43:55.359892+00:00 | 256 | false | ```\nclass Solution {\n private Set<String> visited = new HashSet<String>();\n private String ans = "";\n private int goal=1;\n \n public String crackSafe(int n, int k) {\n visited.clear();\n goal = 1;\n ans="";\n for(int i=0; i < n; i++) {\n goal = goal*k;\n }\n\t... | 1 | 0 | [] | 0 |
cracking-the-safe | DFS Hamilton Cycle with explanation - JavaScript | dfs-hamilton-cycle-with-explanation-java-jy0q | \n 1. Get all perms of 0 ... k - 1 of length n (these are nodes of the graph)\n 2. Make a graph where each connection represents an overlap (Hamilton Cycle). A | auwdish | NORMAL | 2019-11-02T14:23:24.135234+00:00 | 2019-11-02T14:28:13.668430+00:00 | 1,622 | false | \n 1. Get all perms of 0 ... k - 1 of length n (these are nodes of the graph)\n 2. Make a graph where each connection represents an overlap (Hamilton Cycle). An overlap means that if we can add one number to the end of a permutation and another permutation is made in the process, then these two permutations overlap. F... | 1 | 0 | [] | 0 |
cracking-the-safe | Python, easy to understand DFS | python-easy-to-understand-dfs-by-roozbeh-lvdf | My approach was calculate all the possible sequence for K and N, and then try to find the shortest one. Each node in the DFS is one sequence, and the search is | roozbeh3 | NORMAL | 2019-10-08T01:48:49.704144+00:00 | 2019-10-08T01:48:49.704189+00:00 | 197 | false | My approach was calculate all the possible sequence for K and N, and then try to find the shortest one. Each node in the DFS is one sequence, and the search is trying to find the next sequence that has N-1 comming characters with it.\n\nThe runtime was slow, but the memory usage was better than 100% of the submissions.... | 1 | 0 | [] | 1 |
cracking-the-safe | No Code, Proof of Correctness for Greedy Dequeue Approach | no-code-proof-of-correctness-for-greedy-uudbs | This is basically a re-explanation and expansion of the very excellent proof here.\n\nFirst we define a directed graph structure where the nodes are every possi | supermatthew | NORMAL | 2019-08-27T15:37:47.001280+00:00 | 2019-11-20T16:58:14.458499+00:00 | 214 | false | This is basically a re-explanation and expansion of [the very excellent proof here](https://leetcode.com/problems/cracking-the-safe/discuss/110261/Deque-solution-with-proof).\n\nFirst we define a directed graph structure where the nodes are every possible length-n sequence made from the alphabet {0,1,...,k-1}. Two nod... | 1 | 0 | [] | 0 |
cracking-the-safe | N-ary solution [C++] | n-ary-solution-c-by-timd000-kctg | Consider traversing a n-ary tree. There are k^n times of unique traverses starting from the root. The tree has n level. And since each parent node has exact k c | timd000 | NORMAL | 2019-07-16T05:23:30.976283+00:00 | 2019-07-16T05:23:30.976328+00:00 | 186 | false | Consider traversing a n-ary tree. There are k^n times of unique traverses starting from the root. The tree has n level. And since each parent node has exact k childs, including the root node, each parent node of a leaf node will be visited k times from other n-1 level paths. That means all k^n paths could be fulfilled ... | 1 | 0 | [] | 1 |
find-indices-with-index-and-value-difference-ii | [Java/C++/Python] One Pass, O(1) Space | javacpython-one-pass-o1-space-by-lee215-bjzp | Intuition\nThe brute force solution is two for loop to check all pairs.\n\nIf index difference d = 1,\nit will be more straight forward,\nwe can iterate the arr | lee215 | NORMAL | 2023-10-15T04:02:34.999875+00:00 | 2023-10-15T04:02:34.999904+00:00 | 4,767 | false | # **Intuition**\nThe brute force solution is two `for` loop to check all pairs.\n\nIf index difference `d = 1`,\nit will be more straight forward,\nwe can iterate the array `A`,\nand keep minimum and maximum in the prefix of `A`.\n<br>\n\n# **Explanation**\nFor `d > 1`, we can do the same thing.\n`mini` is the index of... | 104 | 3 | ['C', 'Python', 'Java'] | 18 |
find-indices-with-index-and-value-difference-ii | Map vs. Min/Max | map-vs-minmax-by-votrubac-fut1 | Min/Max\nI initially came up with the map solution, only then to realize that we can just store min and max value.\nC++\ncpp\nvector<int> findIndices(vector<int | votrubac | NORMAL | 2023-10-15T04:00:41.794081+00:00 | 2023-10-15T04:12:09.214759+00:00 | 1,739 | false | ## Min/Max\nI initially came up with the map solution, only then to realize that we can just store min and max value.\n**C++**\n```cpp\nvector<int> findIndices(vector<int>& nums, int indexDifference, int valueDifference) {\n int min_j = 0, max_j = 0;\n for (int i = indexDifference, j = 0; i < nums.size(); ++i, ++... | 22 | 1 | ['C'] | 5 |
find-indices-with-index-and-value-difference-ii | Using priority queue || Very simple and easy to understand solution | using-priority-queue-very-simple-and-eas-5bg2 | \n# Approach \n- Find the largest and smallest number which is away from i by indDiff distance. Then check if the smallest and ith or larget and ith having a di | kreakEmp | NORMAL | 2023-10-15T04:01:35.004781+00:00 | 2023-10-15T04:23:30.254747+00:00 | 2,864 | false | \n# Approach \n- Find the largest and smallest number which is away from i by indDiff distance. Then check if the smallest and ith or larget and ith having a diff greater than valueDiff\n- To do so we use min heap and max heap to find max and min after (i+ indDiff)th element\n\n# Code\n```\nclass Solution {\npublic:\n ... | 18 | 2 | ['C++'] | 3 |
find-indices-with-index-and-value-difference-ii | Easy Video Solution (Brute->Optimal) 🔥 || Suffix Max-Min 🔥 || C++,JAVA,PYTHON | easy-video-solution-brute-optimal-suffix-kyfv | Intuition\n Describe your first thoughts on how to solve this problem. \nTry to think about prefix and suffix\n\n# Detailed and easy Video Solution\n\nhttps://y | ayushnemmaniwar12 | NORMAL | 2023-10-15T05:12:02.280461+00:00 | 2023-10-19T14:55:47.942978+00:00 | 1,687 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTry to think about prefix and suffix\n\n# ***Detailed and easy Video Solution***\n\nhttps://youtu.be/neaCFDgBBn8\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe code finds indices in an array where the absolute... | 17 | 1 | ['Suffix Array', 'C++', 'Java', 'Python3'] | 4 |
find-indices-with-index-and-value-difference-ii | c++||Using Set || Very Easy | cusing-set-very-easy-by-baibhavkr143-79lf | 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 | baibhavkr143 | NORMAL | 2023-10-15T04:02:41.840096+00:00 | 2023-10-15T04:02:41.840118+00:00 | 1,559 | 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. $$O(n)$$ --... | 13 | 1 | ['C++'] | 3 |
find-indices-with-index-and-value-difference-ii | [Python3] greedy | python3-greedy-by-ye15-ro0f | \n\nclass Solution:\n def findIndices(self, nums: List[int], indexDifference: int, valueDifference: int) -> List[int]:\n mn = mx = -1 \n for i, | ye15 | NORMAL | 2023-10-15T04:01:13.256444+00:00 | 2023-10-15T04:01:13.256465+00:00 | 556 | false | \n```\nclass Solution:\n def findIndices(self, nums: List[int], indexDifference: int, valueDifference: int) -> List[int]:\n mn = mx = -1 \n for i, x in enumerate(nums): \n if i >= indexDifference: \n if mn == -1 or nums[i-indexDifference] < nums[mn]: mn = i-indexDifference\n ... | 10 | 0 | ['Python3'] | 0 |
find-indices-with-index-and-value-difference-ii | Best solution.✅NO Priority queue,NO sliding window. Just plane logic✅✅ | best-solutionno-priority-queueno-sliding-c7on | Intuition\nIf we carefully see the problem, it may look like we need to store a lot many things if we are at index i, then we need to know every damn thing abou | adiityyya | NORMAL | 2023-10-16T20:37:21.232906+00:00 | 2023-10-16T20:38:22.676159+00:00 | 298 | false | # Intuition\nIf we carefully see the problem, it may look like we need to store a lot many things if we are at index i, then we need to know every damn thing about the values at indexes i - id, but that\'s not the case.\n\n# Approach\nWe will just keep a record of maximum, and minimum element in the range [0,i-id] if w... | 8 | 1 | ['Array', 'C++'] | 1 |
find-indices-with-index-and-value-difference-ii | ✅☑[C++/C/Java/Python/JavaScript] || EXPLAINED🔥 | ccjavapythonjavascript-explained-by-mark-z385 | PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n(Also explained in the code)\n1. The function findIndices takes three parameters: nums (a vector of integer | MarkSPhilip31 | NORMAL | 2023-10-15T07:25:32.310951+00:00 | 2023-10-15T07:25:32.310974+00:00 | 465 | false | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n1. The function `findIndices` takes three parameters: `nums` (a vector of integers), `indexDifference` (an integer), and `valueDifference` (an integer). It aims to find a pair of indices in the vector nums such that the absolute di... | 6 | 0 | ['C', 'Ordered Set', 'C++', 'Java', 'Python3', 'JavaScript'] | 1 |
find-indices-with-index-and-value-difference-ii | ✅✔️Easy to understand || clean code || C++ Solution ✈️✈️✈️✈️✈️ | easy-to-understand-clean-code-c-solution-ly3w | 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 | 2023-10-15T04:06:03.131650+00:00 | 2023-10-15T04:06:03.131668+00:00 | 763 | 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. $$O(n)$$ --... | 6 | 0 | ['Array', 'Sliding Window', 'Suffix Array', 'C++'] | 0 |
find-indices-with-index-and-value-difference-ii | Simple prefix sum | simple-prefix-sum-by-joe54-s1tx | 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 | joe54 | NORMAL | 2023-10-15T04:01:01.460758+00:00 | 2023-10-15T04:01:01.460779+00:00 | 571 | 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 $$O(n)$$\n\n- Space complexity:\n $$O(n)$$\n\n# Code\n```\nclass Solution {\npublic:\n vector<int> findIndices(vector<int>& num... | 5 | 0 | ['C++'] | 2 |
find-indices-with-index-and-value-difference-ii | Easy 4 liner explained O(n) time O(1) space, beats 100%, 100% | easy-4-liner-explained-on-time-o1-space-175df | Intuition\n Describe your first thoughts on how to solve this problem. \nSince we have to find an value which is greater than a given valueDifference which are | flame_blitz | NORMAL | 2023-10-15T06:20:05.196305+00:00 | 2023-10-15T06:21:57.142017+00:00 | 155 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSince we have to find an value which is greater than a given valueDifference which are atleast indexDifference apart, we can just keep track of max and min values as we come across them and then just evaulate them with the current elemnt ... | 4 | 0 | ['Array', 'Two Pointers', 'JavaScript'] | 3 |
find-indices-with-index-and-value-difference-ii | Java Easy solution - O(N) | java-easy-solution-on-by-abhishek1208-cxyd | \nclass Solution {\n public int[] findIndices(int[] nums, int indexDifference, int valueDifference) {\n int[] ans = new int[]{-1,-1};\n \n | lcabhishek | NORMAL | 2023-10-15T04:01:58.953451+00:00 | 2023-10-15T04:54:57.091641+00:00 | 454 | false | ```\nclass Solution {\n public int[] findIndices(int[] nums, int indexDifference, int valueDifference) {\n int[] ans = new int[]{-1,-1};\n \n List<int[]> list = new ArrayList<>();\n int min = 0;\n int max = 0;\n \n \n //store the max and min till the index i\n ... | 4 | 1 | ['Java'] | 2 |
find-indices-with-index-and-value-difference-ii | C++ very easy and optimised solution | c-very-easy-and-optimised-solution-by-aa-pdge | Intuition\n Describe your first thoughts on how to solve this problem. \n\njust use 2 stack and use logic\n\n# Approach\n Describe your approach to solving the | AAA_code | NORMAL | 2023-10-17T16:43:06.209498+00:00 | 2023-10-17T16:43:06.209522+00:00 | 206 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\njust use 2 stack and use logic\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\nmaking a two- way sliding window type that is sorted\n\n# Complexity\n- Time complexity:\n- o(n);\n<!-- Add your time complexity he... | 3 | 1 | ['Stack', 'C++'] | 0 |
find-indices-with-index-and-value-difference-ii | CPP || Two pointer || 100 % || TC:O(n) SC:O(1) | cpp-two-pointer-100-tcon-sco1-by-kgaurav-ey3h | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem is about finding two indices in an array such that the absolute difference | kgaurav8026 | NORMAL | 2023-10-15T19:31:11.434436+00:00 | 2023-10-15T19:31:11.434454+00:00 | 220 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem is about finding two indices in an array such that the absolute difference between their values is greater than or equal to a given value and the absolute difference between the indices is greater than or equal to another give... | 3 | 0 | ['Two Pointers', 'C++'] | 0 |
find-indices-with-index-and-value-difference-ii | ✅ ✅ Indices with index and value difference | Beginner friendly | Min array & Max array ✅ ✅ | indices-with-index-and-value-difference-j8gci | Hi,\n\nApproach: I utilized two arrays to store the minimum and maximum values, then used them to efficiently find indices with the desired value difference.\n\ | Surendaar | NORMAL | 2023-10-15T17:45:26.909420+00:00 | 2023-10-15T17:47:39.014407+00:00 | 106 | false | Hi,\n\n**Approach:** I utilized two arrays to store the minimum and maximum values, then used them to efficiently find indices with the desired value difference.\n\n**Intuition:** Instead of a direct brute force approach which would result in TLE, I optimized by iterating through the loop fewer than n^2 times.\n\n**Ste... | 3 | 0 | ['Array', 'Java'] | 0 |
find-indices-with-index-and-value-difference-ii | [C++] Insert the candidates {value, index} into an ordered set | c-insert-the-candidates-value-index-into-fq3n | Intuition\n Describe your first thoughts on how to solve this problem. \nFor each index i, try the minimum and maximum values in the indices j = [i + indexDIffe | pepe-the-frog | NORMAL | 2023-10-15T07:22:01.786216+00:00 | 2023-10-15T07:22:01.786242+00:00 | 180 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFor each index `i`, try the minimum and maximum values in the indices `j = [i + indexDIfference, n)`\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- for each index `i`, only indices in `[i + indexDifference, n)` ... | 3 | 1 | ['Ordered Set', 'C++'] | 0 |
find-indices-with-index-and-value-difference-ii | Python3 | Binary Search | python3-binary-search-by-tkr_6-xmte | \n\n# Code\n\nclass Solution:\n def findIndices(self, nums: List[int], ind: int, val: int) -> List[int]:\n \n \n \n n=len(nums)\n | TKR_6 | NORMAL | 2023-10-15T05:44:48.546467+00:00 | 2023-10-15T05:44:48.546500+00:00 | 160 | false | \n\n# Code\n```\nclass Solution:\n def findIndices(self, nums: List[int], ind: int, val: int) -> List[int]:\n \n \n \n n=len(nums)\n sl=[] #SortedList\n \n for i in range(ind,n):\n insort(sl,[nums[i-ind],i-ind])\n \n a,i1=sl[0]\n ... | 3 | 0 | ['Binary Search', 'Python', 'Python3'] | 0 |
find-indices-with-index-and-value-difference-ii | C++ - Binary Search Solution | c-binary-search-solution-by-omkamble0800-k7jp | Complexity\n- Time complexity: O(nlogn)\n\n- Space complexity: O(n)\n\n# Code\n\nclass Solution {\npublic:\n vector<int> findIndices(vector<int>& nums, int i | omkamble0800 | NORMAL | 2023-10-15T04:19:17.271597+00:00 | 2023-10-15T04:19:17.271616+00:00 | 382 | false | # Complexity\n- Time complexity: $$O(nlogn)$$\n\n- Space complexity: $$O(n)$$\n\n# Code\n```\nclass Solution {\npublic:\n vector<int> findIndices(vector<int>& nums, int id, int vd) {\n int n = nums.size();\n set<pair<int, int>> st;\n for(int i = id; i < n; i++){\n st.insert({nums[i - ... | 3 | 1 | ['Array', 'Binary Search', 'Ordered Set', 'C++'] | 1 |
find-indices-with-index-and-value-difference-ii | Simple solution with comments which is easy to understand | simple-solution-with-comments-which-is-e-hj4x | 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 | Code_Of_Duty | NORMAL | 2023-10-15T04:06:05.842691+00:00 | 2023-10-15T04:06:05.842709+00:00 | 270 | 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 | ['Array', 'Prefix Sum', 'C++'] | 0 |
find-indices-with-index-and-value-difference-ii | 🔥GREEDY : O(n) TC || ✅⚡Clean & Best Code | greedy-on-tc-clean-best-code-by-adish_21-19sk | Connect with me on LinkedIN : https://www.linkedin.com/in/aditya-jhunjhunwala-51b586195/\n\n# Complexity\n\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n | aDish_21 | NORMAL | 2023-10-15T04:03:20.886742+00:00 | 2023-10-15T04:05:39.405052+00:00 | 307 | false | ## Connect with me on LinkedIN : https://www.linkedin.com/in/aditya-jhunjhunwala-51b586195/\n\n# Complexity\n```\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n```\n\n# Code\n# Please Upvote if it helps\uD83E\uDD17\n```\nclass Solution {\npublic:\n vector<int> findIndices(vector<int>& nums, int indexDiffere... | 3 | 0 | ['Greedy', 'C++'] | 0 |
find-indices-with-index-and-value-difference-ii | ❇ find-indices-with-index-and-value-difference-ii Images👌 🏆O(N)❤️ Javascript🎯 Memory👀95%🕕 ++Exp | find-indices-with-index-and-value-differ-9zsb | Time Complexity: O(N)\uD83D\uDD55\nSpace Complexity: O(N)\n\n\nvar findIndices = function (nums, indexDifference, valueDifference) {\n if (indexDifference == | anurag-sindhu | NORMAL | 2024-09-26T18:28:08.827261+00:00 | 2024-09-26T18:30:39.017381+00:00 | 37 | false | **Time Complexity: O(N)\uD83D\uDD55\nSpace Complexity: O(N)**\n\n```\nvar findIndices = function (nums, indexDifference, valueDifference) {\n if (indexDifference == 0) {\n if (valueDifference === 0) {\n return [0, 0];\n }\n }\n\n const bigNumArr = [];\n const bigNumArrIndexMapping =... | 2 | 0 | ['JavaScript'] | 0 |
find-indices-with-index-and-value-difference-ii | Java Simple Two Pointer Solution || With Detail Explanation | java-simple-two-pointer-solution-with-de-kgug | Intuition\n- We take 2 variables, min and max, which stores the indices of min and max element respectively.\n- We start a loop with 2 pointers, one at 0 and th | Shree_Govind_Jee | NORMAL | 2024-07-03T15:41:38.580740+00:00 | 2024-07-03T15:41:38.580773+00:00 | 73 | false | # Intuition\n- We take 2 variables, min and max, which stores the indices of **$$min$$** and **$$max$$** element respectively.\n- We start a loop with **$$2 pointers$$**, one at $$0$$ and the other indexDifference apart from it so we dont have to worry about the condition `Math.abs(i - j) >= indexDifference` as it will... | 2 | 0 | ['Array', 'Two Pointers', 'Greedy', 'Java'] | 0 |
find-indices-with-index-and-value-difference-ii | C++ || Segment Tree || Unique Solution 🔥✅💯 | c-segment-tree-unique-solution-by-_kvsch-33mz | \n# Code\n\nclass Solution {\npublic:\n // SEGMENT TREES :- vector<int> seg(1000), vector<int>arr(1000), i = 0, l = 0, h = n-1, x = idx, v = val, (x,y) = (l, | _kvschandra_2234 | NORMAL | 2024-04-30T14:57:19.610680+00:00 | 2024-04-30T14:57:38.803954+00:00 | 20 | false | \n# Code\n```\nclass Solution {\npublic:\n // SEGMENT TREES :- vector<int> seg(1000), vector<int>arr(1000), i = 0, l = 0, h = n-1, x = idx, v = val, (x,y) = (l,r).\n void buildSeg1(vector<int>& seg, const vector<int>& arr, int i, int l, int h) {\n if (l == h) {\n seg[i] = arr[l];\n re... | 2 | 0 | ['Segment Tree', 'C++'] | 0 |
find-indices-with-index-and-value-difference-ii | [Java] 2ms, 88%, Sliding Window + CLEAN CODE | java-2ms-88-sliding-window-clean-code-by-f0hf | Approach\n1. Fail fast: if indexDiff >= nums.length, then return {-1,-1} as you cannot have such distant indices\n2. Looking at the indexDiff, you realise that | StefanelStan | NORMAL | 2023-11-30T21:33:02.370370+00:00 | 2023-11-30T21:33:02.370391+00:00 | 50 | false | # Approach\n1. Fail fast: if indexDiff >= nums.length, then return {-1,-1} as you cannot have such distant indices\n2. Looking at the indexDiff, you realise that if you pick i [0..n-indexDiff], then the next available number you can chose if after i + indexDiff.\n - Thus, we observe a gap/window: the other index mus... | 2 | 0 | ['Two Pointers', 'Java'] | 0 |
find-indices-with-index-and-value-difference-ii | Best Java Solution || Beats 100% | best-java-solution-beats-100-by-ravikuma-bghn | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ravikumar50 | NORMAL | 2023-10-15T07:34:32.044780+00:00 | 2023-10-15T07:34:32.044799+00:00 | 158 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['Java'] | 0 |
find-indices-with-index-and-value-difference-ii | 🔥🔥🔥 Python Simple Solution 🔥🔥🔥 | python-simple-solution-by-hululu0405-hk8i | Brute force\n# Intuition\n Describe your first thoughts on how to solve this problem. \nCheck all combination in array.\n# Approach\n Describe your approach to | hululu0405 | NORMAL | 2023-10-15T06:31:21.656206+00:00 | 2023-10-16T02:11:56.108761+00:00 | 167 | false | # Brute force\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nCheck all combination in array.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nGo through the array.\n# Complexity\n- Time complexity: O(n^2)\n- Space complexity: O(1)\n# Code\n```\nclass Solution:\n ... | 2 | 0 | ['Python3'] | 1 |
find-indices-with-index-and-value-difference-ii | One pass by record left_min index and left_max index | one-pass-by-record-left_min-index-and-le-a6gn | 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 | pohanchi | NORMAL | 2023-10-15T04:07:46.007385+00:00 | 2023-10-15T04:07:46.007403+00:00 | 212 | 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. $... | 2 | 1 | ['Python3'] | 0 |
find-indices-with-index-and-value-difference-ii | Python/Rust/Go, O(n) time, O(1) space with explanation | pythonrustgo-on-time-o1-space-with-expla-d6ym | Intuition\n\nLets look at some array $v_0, v_1, v_2, ... v_n$ and take a look at some position $v_i$. How can we tell that index $i$ will be a part of the resul | salvadordali | NORMAL | 2023-10-15T04:07:37.006743+00:00 | 2023-10-16T23:07:24.634075+00:00 | 168 | false | # Intuition\n\nLets look at some array $v_0, v_1, v_2, ... v_n$ and take a look at some position $v_i$. How can we tell that index $i$ will be a part of the result? For this to happen, there should be some position `j <= i - diff_i` for which `|nums[j] - nums[i]| >= diff_v`.\n\nTo not check for every index, we can just... | 2 | 0 | ['Go', 'Python3', 'Rust'] | 0 |
find-indices-with-index-and-value-difference-ii | [c++] one pass - O(N) time O(1) space | c-one-pass-on-time-o1-space-by-akshay4gu-tfxx | Intuition\n Describe your first thoughts on how to solve this problem. \nyou just need the min or max value till i - indexDifference\n\n# Approach\n Describe yo | Akshay4Gupta | NORMAL | 2023-10-15T04:06:08.014607+00:00 | 2023-10-15T04:06:34.723566+00:00 | 149 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nyou just need the min or max value till `i - indexDifference`\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- for each i\n - keep finding the min and max for `i - indexDifference`\n - and check if abs diffe... | 2 | 0 | ['C++'] | 0 |
find-indices-with-index-and-value-difference-ii | Beats 100% | Python 3 | beats-100-python-3-by-alpha2404-7o81 | Please UpvoteCode | Alpha2404 | NORMAL | 2025-02-15T08:29:58.939559+00:00 | 2025-02-15T08:29:58.939559+00:00 | 35 | false | # Please Upvote
# Code
```python3 []
__import__("atexit").register(lambda: open("display_runtime.txt", "w").write("0"))
class Solution:
def findIndices(self, nums: List[int], indexDifference: int, valueDifference: int) -> List[int]:
mn = mx = -1
for i, x in enumerate(nums):
if i>= index... | 1 | 0 | ['Python3'] | 0 |
find-indices-with-index-and-value-difference-ii | python solution | python-solution-by-ayoublaar-phyi | 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 | AyoubLaar | NORMAL | 2024-09-07T18:30:50.517488+00:00 | 2024-09-07T18:30:50.517523+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. $$O(n)$$ -->\nO(n^2)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n... | 1 | 0 | ['Python3'] | 0 |
find-indices-with-index-and-value-difference-ii | Detailed Explanation | detailed-explanation-by-rengubareva-rbkw | Intuition\n1. For i-th element check elements with indexes j: abs(j-i) >= indexDifference\n2. If there is elements which is: abs(nums[i] - nums[j]) >= valueDiff | rengubareva | NORMAL | 2024-06-25T13:09:29.602797+00:00 | 2024-06-25T13:09:29.602821+00:00 | 134 | false | # Intuition\n1. For i-th element check elements with indexes j: ```abs(j-i) >= indexDifference```\n2. If there is elements which is: ```abs(nums[i] - nums[j]) >= valueDifference``` ```return {i, j}```, otherwise continue to check each element till the end of the array.\n3. if there is no such elements, ```return {-1, -... | 1 | 0 | ['Sliding Window', 'Java'] | 1 |
find-indices-with-index-and-value-difference-ii | Python | O(nlogn) | Priority Queue | python-onlogn-priority-queue-by-mandeepg-nz0i | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | mandeepgoyal | NORMAL | 2023-10-24T16:03:16.447839+00:00 | 2023-10-24T16:03:16.447866+00:00 | 22 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(NlogN)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n\n- Space complexity: O(N)\n<!-- Add your space complexity here, ... | 1 | 0 | ['Heap (Priority Queue)', 'Python3'] | 0 |
find-indices-with-index-and-value-difference-ii | Find Indices With Index and Value Difference II - Easy Java Solution | find-indices-with-index-and-value-differ-4pva | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | whopiyushanand | NORMAL | 2023-10-18T07:59:01.775956+00:00 | 2023-10-18T07:59:01.775978+00:00 | 54 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $... | 1 | 0 | ['Array', 'Java'] | 0 |
find-indices-with-index-and-value-difference-ii | Same as problem 2903 | java O(n) | Prefix sum | same-as-problem-2903-java-on-prefix-sum-e6x0s | Intuition\nSee\uFF1A https://leetcode.com/problems/find-indices-with-index-and-value-difference-i/solutions/4170252/java-o-n-prefix-sum-100/\n Describe your fir | ly2015cntj | NORMAL | 2023-10-17T03:07:57.302927+00:00 | 2023-10-17T03:08:47.943094+00:00 | 21 | false | # Intuition\nSee\uFF1A https://leetcode.com/problems/find-indices-with-index-and-value-difference-i/solutions/4170252/java-o-n-prefix-sum-100/\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... | 1 | 0 | ['Java'] | 0 |
find-indices-with-index-and-value-difference-ii | Simple and easy || min-max || O(n) | simple-and-easy-min-max-on-by-spats7-lbkz | ```\nclass Solution {\n public int[] findIndices(int[] nums, int indexDifference, int valueDifference) {\n int n = nums.length;\n int[][] minMa | spats7 | NORMAL | 2023-10-16T21:26:09.227671+00:00 | 2023-10-16T21:26:09.227695+00:00 | 299 | false | ```\nclass Solution {\n public int[] findIndices(int[] nums, int indexDifference, int valueDifference) {\n int n = nums.length;\n int[][] minMax = new int[n][4];\n \n minMax[n-1][0] = nums[n-1];\n minMax[n-1][1] = n-1;\n \n minMax[n-1][2] = nums[n-1];\n minMax[... | 1 | 0 | ['Array', 'Java'] | 0 |
find-indices-with-index-and-value-difference-ii | C#: O(n) | c-on-by-igor0-5kmy | Intuition\nGo from the end to the begin of the array nums. Keep the min and max.\n\n# Approach\nGo from the end to the indexDifference position of the array num | igor0 | NORMAL | 2023-10-16T16:16:40.169338+00:00 | 2023-10-16T16:16:40.169365+00:00 | 19 | false | # Intuition\nGo from the end to the begin of the array `nums`. Keep the min and max.\n\n# Approach\nGo from the end to the `indexDifference` position of the array `nums`\n```\nfor (int i = nums.Length - 1; i >= indexDifference; i--)\n{\n```\nBy each iteration change the min and max:\n```\n min = Math.Min(min, nums[i... | 1 | 0 | ['Prefix Sum', 'C#'] | 0 |
find-indices-with-index-and-value-difference-ii | C++ || Two Pointer || without set or heap | c-two-pointer-without-set-or-heap-by-moh-8j1m | Intuition\nMaintaining minimum and maximum element in range [0 i-indexDifference]\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n# Code | mohaldarprakash | NORMAL | 2023-10-15T10:13:57.643193+00:00 | 2023-10-15T10:13:57.643225+00:00 | 277 | false | # Intuition\nMaintaining minimum and maximum element in range [0 i-indexDifference]\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n# Code\n```\nfunc abs(n int) int {\n if n < 0 {\n return -n\n }\n return n\n}\n\nfunc findIndices(nums []int, indexDifference int, valueDifference i... | 1 | 0 | ['Two Pointers', 'Go'] | 0 |
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