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k-inverse-pairs-array | Recursion | Memoization | Bottom Up | Optimal Bottom Up| codestorywithMIK | recursion-memoization-bottom-up-optimal-awwhb | My Youtube Video (Super Detailed) - https://www.youtube.com/watch?v=y9yo1kyW7Bg\n\n/*************************************************************** C++ ******** | mazhar_mik | NORMAL | 2021-06-19T13:21:45.567500+00:00 | 2024-01-27T07:11:04.412912+00:00 | 244 | false | My Youtube Video (Super Detailed) - https://www.youtube.com/watch?v=y9yo1kyW7Bg\n```\n/*************************************************************** C++ ***************************************************************/\n//Approach-1 (Reucr+Memo)\n//T.C : O(n*k*n)\n//S.C : O(n*k) for memo + Recursion call stack\nclas... | 6 | 4 | ['C', 'Java'] | 1 |
k-inverse-pairs-array | Python DP, my progress from recursion to n * k solution | python-dp-my-progress-from-recursion-to-2c012 | The idea is just to add a new number n to an existing array, and the n should be larger than all element of the original array. Therefore we can add up all poss | tonymok97 | NORMAL | 2021-06-19T08:54:15.282188+00:00 | 2021-06-20T08:12:04.668815+00:00 | 326 | false | The idea is just to add a new number n to an existing array, and the n should be larger than all element of the original array. Therefore we can add up all possibilities\n\nBase idea with recursion with memo(TLE)\n```\nclass Solution:\n def kInversePairs(self, n: int, k: int) -> int:\n """"\n given a n... | 6 | 0 | [] | 2 |
k-inverse-pairs-array | Python O(nk) time solution from O(nk) space to O(k) space with explanation. | python-onk-time-solution-from-onk-space-sdfym | Idea:\n\nYou can write down some the cases from 1 to 4 and you will find the pattern.\nLike when i = 4, you insert 4 to all the cases in a proper position when | jedihy | NORMAL | 2017-06-27T01:48:13.834000+00:00 | 2017-06-27T01:48:13.834000+00:00 | 735 | false | #### Idea:\n\nYou can write down some the cases from 1 to 4 and you will find the pattern.\nLike when `i = 4`, you insert 4 to all the cases in a proper position when `i = 3` to get the number of results when `i=4`.\nNote that `i` limits the number of places you can do the insertion. If `i = 5`, there is only `5` plac... | 6 | 0 | [] | 0 |
k-inverse-pairs-array | Recursion, Caching, Tabulation, Space Optimized, Sliding Window+ dp | recursion-caching-tabulation-space-optim-s0iq | One key point in understanding the point\n\nk-i --> means the posibile inversions that number can make with the remaining number\n\n\n\n1. We\'re considering ho | Dixon_N | NORMAL | 2024-08-02T18:30:33.065423+00:00 | 2024-08-02T18:38:07.586857+00:00 | 122 | false | One key point in understanding the point\n\n**k-i --> means the posibile inversions that number can make with the remaining number**\n\n<br/><br/>\n\n1. We\'re considering how many inverse pairs the nth element can create.\n\n2. The nth element can create 0 to n-1 inverse pairs.\ni represents the number of inverse pair... | 5 | 0 | ['Dynamic Programming', 'Java'] | 4 |
k-inverse-pairs-array | C++ Solution | c-solution-by-manoj230322-3b7a | Intuition\n Describe your first thoughts on how to solve this problem. \nThe very first observation here is that if we have an integer x in our array then the m | manoj230322 | NORMAL | 2024-01-27T11:46:53.941878+00:00 | 2024-07-02T07:18:53.774626+00:00 | 277 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe very first observation here is that if we have an integer x in our array then the maximum number of inverse pair [i,j] we can form with integer x at index i is x-1.\nfor eg.\nn=5\n[5, 4, 3, 2, 1]\nhere you can see that with 5 we can f... | 5 | 0 | ['C++'] | 1 |
k-inverse-pairs-array | Proof of recursive relation of O(n*k). Simple DP | proof-of-recursive-relation-of-onk-simpl-h3ai | \n\n# Code\n\nclass Solution {\npublic:\nconst int mod = 1e9+7;\n int kInversePairs(int n, int k) {\n vector<vector<int>> dp(n+1, vector<int>(k+1, 0)) | tryingall | NORMAL | 2024-01-27T01:21:12.507340+00:00 | 2024-01-27T01:21:12.507367+00:00 | 599 | false | \n\n# Code\n```\nclass Solution {\npublic:\nconst int mod = 1e9+7;\n int kInversePairs(int n, int k) {\n vector<vector<int>> dp(n+1, vector<int>(k+1, 0));\n // if k=0 then all possible=1\n for(int i = 1; i < n+1; ++i) {\n dp[i][0] = 1;\n }\n // dp[i][j] = dp[i-1][j]+dp[i... | 5 | 0 | ['C++'] | 0 |
k-inverse-pairs-array | C# Solution - K Inverse Pairs | c-solution-k-inverse-pairs-by-jayflex-jfqs | Intuition\n Describe your first thoughts on how to solve this problem. \nTo solve the problem, use dynamic programming to break it down into smaller subproblems | jayflex | NORMAL | 2024-01-27T00:38:17.388689+00:00 | 2024-01-27T00:38:17.388718+00:00 | 255 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo solve the problem, use dynamic programming to break it down into smaller subproblems. \n\nTo form an array of n numbers with k inverse pairs, add the nth number in n different positions in an array of n-1 numbers. \n# Approach\n<!-- De... | 5 | 0 | ['C#'] | 2 |
k-inverse-pairs-array | PYTHON 95% FASTER EASY BEGINER FRIENDLY Multiple Solutions | python-95-faster-easy-beginer-friendly-m-wskz | DON\'T FORGET TO UPVOTE.\n\n# 1. 95% Faster solution:\n\n\n\t\tclass Solution:\n\t\t\tdef kInversePairs(self, n: int, k: int) -> int:\n\t\t\t\tm = 10 ** 9 + 7\n | anuvabtest | NORMAL | 2022-07-17T05:40:44.966325+00:00 | 2022-07-17T05:47:48.840530+00:00 | 874 | false | # DON\'T FORGET TO UPVOTE.\n\n# 1. 95% Faster solution:\n\n\n\t\tclass Solution:\n\t\t\tdef kInversePairs(self, n: int, k: int) -> int:\n\t\t\t\tm = 10 ** 9 + 7\n\n\t\t\t\tdp0 = [0] * (k + 1)\n\t\t\t\tdp0[0] = 1\n\n\t\t\t\tfor i in range(n):\n\t\t\t\t\tdp1 = []\n\t\t\t\t\ts = 0\n\t\t\t\t\tfor j in range(k + 1):\n\t\t\t... | 5 | 0 | ['Dynamic Programming', 'Python', 'Python3'] | 1 |
k-inverse-pairs-array | C++||Easy & clean code||DP||Tabulation | ceasy-clean-codedptabulation-by-prince_7-rb8s | \nconst int mod = 1e9 + 7, N = 1001;\nclass Solution {\n int dp[N][N] = {};\npublic:\n int kInversePairs(int n, int K) {\n dp[0][0] = 1;\n f | prince_725 | NORMAL | 2022-07-17T04:12:44.520891+00:00 | 2022-07-17T04:12:44.520916+00:00 | 929 | false | ```\nconst int mod = 1e9 + 7, N = 1001;\nclass Solution {\n int dp[N][N] = {};\npublic:\n int kInversePairs(int n, int K) {\n dp[0][0] = 1;\n for (int i = 1; i <= n; ++ i){\n long long x = 0; \n for (int j = 0; j <= K; ++ j){\n x += dp[i - 1][j];\n ... | 5 | 1 | ['Dynamic Programming', 'C'] | 1 |
k-inverse-pairs-array | 🗓️ Daily LeetCoding Challenge July, Day 17 | daily-leetcoding-challenge-july-day-17-b-oad5 | This problem is the Daily LeetCoding Challenge for July, Day 17. Feel free to share anything related to this problem here! You can ask questions, discuss what y | leetcode | OFFICIAL | 2022-07-17T00:00:11.167641+00:00 | 2022-07-17T00:00:11.167687+00:00 | 2,074 | false | This problem is the Daily LeetCoding Challenge for July, Day 17.
Feel free to share anything related to this problem here!
You can ask questions, discuss what you've learned from this problem, or show off how many days of streak you've made!
---
If you'd like to share a detailed solution to the problem, please crea... | 5 | 2 | [] | 17 |
k-inverse-pairs-array | Combinatories solution. | combinatories-solution-by-sorosye-poad | First, we need to introduce new representation of a permution.\nIn Book \u300AThe art of computer programing\u300B vol 4. bring us an inverse num representation | sorosye | NORMAL | 2021-11-17T11:46:28.734294+00:00 | 2022-05-18T11:04:46.279017+00:00 | 315 | false | First, we need to introduce new representation of a permution.\nIn Book \u300AThe art of computer programing\u300B vol 4. bring us an inverse num representation:\nGiven numbers from 1 to n. Sequence m_1,m_2,...,m_n, which m_i mean how many number greater than i and sit on left hand side of i.\neg. 2,0,0 mean 2 number ... | 5 | 0 | ['Combinatorics'] | 0 |
k-inverse-pairs-array | C++ Solution || Using DP | c-solution-using-dp-by-kannu_priya-lzok | \nclass Solution {\npublic:\n int kInversePairs(int n, int k) {\n int dp[n+1][k+1];\n memset(dp, 0, sizeof(dp));\n \n for(int i = | kannu_priya | NORMAL | 2021-06-19T17:40:30.556243+00:00 | 2021-06-19T17:40:30.556286+00:00 | 742 | false | ```\nclass Solution {\npublic:\n int kInversePairs(int n, int k) {\n int dp[n+1][k+1];\n memset(dp, 0, sizeof(dp));\n \n for(int i = 1; i <= n; i++)\n for(int j = 0; j <= k; j++){\n //when j = 0(k = 0) then sorted array is the only solution \n if(j... | 5 | 0 | ['Dynamic Programming', 'C'] | 2 |
k-inverse-pairs-array | Easy to Understand Java Solution, Explanation with Comments with DP | easy-to-understand-java-solution-explana-6fh3 | \nclass Solution {\n \n public int kInversePairs(int n, int k) {\n \n /* Approach. O(n*k)\n \n DP: dp[n][k] -> No. of ar | chiragcodes | NORMAL | 2021-06-19T11:43:55.012534+00:00 | 2021-06-19T11:43:55.012567+00:00 | 487 | false | ```\nclass Solution {\n \n public int kInversePairs(int n, int k) {\n \n /* Approach. O(n*k)\n \n DP: dp[n][k] -> No. of arrays possible with values 1 to n and with k inverse pairs\n \n eg: 1 2 3 4 5\n \n IF we want 0 inverse pair => C... | 5 | 0 | ['Dynamic Programming', 'Java'] | 0 |
k-inverse-pairs-array | JAVA Solution Explained in HINDI | java-solution-explained-in-hindi-by-the_-0f9m | https://youtu.be/Q-Dh8lSxDSU\nFor explanation, please watch the above video and do like, share and subscribe the channel. \u2764\uFE0F Also, please do upvote th | The_elite | NORMAL | 2024-01-27T06:40:08.554329+00:00 | 2024-01-27T06:40:08.554351+00:00 | 393 | false | ERROR: type should be string, got "https://youtu.be/Q-Dh8lSxDSU\\nFor explanation, please watch the above video and do like, share and subscribe the channel. \\u2764\\uFE0F Also, please do upvote the solution if you liked it.\\n\\nSubscribe link:- https://youtube.com/@quantumCode7?si=NzjWYmiCbqfFBUJE\\n\\nSubscribe Goal:- 100\\nCurrent Subscriber:- 79\\n\\n# Code\\n```\\nclass Solution {\\n public int kInversePairs(int n, int k) {\\n \\n int dp[][] = new int[1001][1001];\\n dp[0][0] = 1;\\n\\n for(int i = 1; i <= n; i++) {\\n for(int j = 0; j <= k; j++) {\\n for(int z = 0; z <= Math.min(j, i-1); z++) {\\n if(j - z >= 0) {\\n dp[i][j] = (dp[i][j] + dp[i-1][j-z]) % 1000000007;\\n }\\n }\\n }\\n }\\n return dp[n][k];\\n\\n }\\n}\\n```" | 4 | 0 | ['Java'] | 0 |
k-inverse-pairs-array | Easy C++ Solution | Step-wise Approach | Hard problem made easy | DP | easy-c-solution-step-wise-approach-hard-fwfna | Intuition and Approach\n Describe your first thoughts on how to solve this problem. \nThe problem involves counting the number of arrays with exactly k inverse | darkaadityaa | NORMAL | 2024-01-27T04:43:45.697797+00:00 | 2024-01-27T04:43:45.697826+00:00 | 463 | false | # Intuition and Approach\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem involves counting the number of arrays with exactly k inverse pairs. An inverse pair in an array is a pair of indices (i, j) such that i < j and nums[i] > nums[j]. The problem can be solved using dynamic programmi... | 4 | 0 | ['Dynamic Programming', 'C++'] | 0 |
k-inverse-pairs-array | Full explained 🚀 || Easy to Understand ✅|| [Java/C++/Python3/JavaScript] | full-explained-easy-to-understand-javacp-tk4h | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n1. Dynamic Programming Table: The code initializes a 2D array dp of size | Shivansu_7 | NORMAL | 2024-01-27T04:18:41.611539+00:00 | 2024-01-27T17:27:52.153552+00:00 | 1,088 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. **Dynamic Programming Table:** The code initializes a 2D array dp of size (n + 1) x (k + 1) to store intermediate results. dp[i][j] represents the number of permutations of length i with exactly j 2. 2.inverse pairs. \n2.... | 4 | 0 | ['Dynamic Programming', 'C++', 'Java', 'Python3', 'JavaScript'] | 6 |
k-inverse-pairs-array | 1D Go solution | 1d-go-solution-by-evleria-f9em | \nconst Mod = 1_000_000_007\n\nfunc kInversePairs(n int, k int) int {\n\tif k == 0 {\n\t\treturn 1\n\t}\n\n\tdp := make([]int, k+1)\n\ttemp := make([]int, k+1)\ | evleria | NORMAL | 2021-06-19T20:25:42.241177+00:00 | 2021-06-20T18:59:27.429899+00:00 | 143 | false | ```\nconst Mod = 1_000_000_007\n\nfunc kInversePairs(n int, k int) int {\n\tif k == 0 {\n\t\treturn 1\n\t}\n\n\tdp := make([]int, k+1)\n\ttemp := make([]int, k+1)\n\tfor i := 1; i <= n; i++ {\n\t\ttemp[0] = 1\n\t\tfor j := 1; j <= k; j++ {\n\t\t\tx := 0\n\t\t\tif j >= i {\n\t\t\t\tx = dp[j-i]\n\t\t\t}\n\t\t\ttemp[j] = ... | 4 | 0 | ['Dynamic Programming', 'Go'] | 2 |
k-inverse-pairs-array | K Inverse Pairs Array Python, DP+window slice Solution, with explanation | k-inverse-pairs-array-python-dpwindow-sl-c88v | python\nclass Solution:\n\n def kInversePairs(self, n: int, k: int) -> int:\n # max pairs is m = (n-1)+...+1\n m = n*(n-1)//2\n \n | haotianzhu | NORMAL | 2021-06-19T10:32:27.061887+00:00 | 2021-06-19T10:55:33.349772+00:00 | 316 | false | ```python\nclass Solution:\n\n def kInversePairs(self, n: int, k: int) -> int:\n # max pairs is m = (n-1)+...+1\n m = n*(n-1)//2\n \n # {n, k} = {n, m-k} ex: {n=4, k=0} <=> {n=4, k=6} <=> 1; since m=3+2+1=6\n\t\t# {n, k} = sum of [ {n-1, k-n}..... {n-1, k} ]\n\t\t# consider we want n=6 ... | 4 | 1 | [] | 0 |
k-inverse-pairs-array | 5 Best Solutions from Beginner to Advanced 🚀 || Top down || Bottom up || LC Daily || JAN 27 | 5-best-solutions-from-beginner-to-advanc-hwro | Code -> Time : O(nk) -> Space : O(k)\n\n// MOST OPTIMIZED\n// SLIGHTLY OPTIMIZED + SPACE OPTIMIZATIONS\n\nclass Solution {\npublic:\n int mod=1e9+7;\n int | kushagra_2k24 | NORMAL | 2024-01-27T16:38:30.694441+00:00 | 2024-01-27T16:38:30.694470+00:00 | 238 | false | # Code -> Time : O(n*k) -> Space : O(k)\n```\n// MOST OPTIMIZED\n// SLIGHTLY OPTIMIZED + SPACE OPTIMIZATIONS\n\nclass Solution {\npublic:\n int mod=1e9+7;\n int kInversePairs(int n, int k) {\n vector<long long int>prev(k+1,0),curr(k+1,0);\n prev[0]=1;\n for(int i=1;i<=n;i++)\n {\n ... | 3 | 0 | ['Array', 'Math', 'Dynamic Programming', 'Recursion', 'Memoization', 'Sliding Window', 'Matrix', 'C++'] | 0 |
k-inverse-pairs-array | Tried to explained in the most simple words.... | tried-to-explained-in-the-most-simple-wo-8n6s | Approach\n- Here, dp[i][j] number of arrays of length i with j inversions\n\n- let us try to find dp[i][j] constructively,\n\n- now, we must appreciate the fact | __ARYAN1__ | NORMAL | 2024-01-27T13:29:27.125858+00:00 | 2024-01-27T13:29:27.125881+00:00 | 38 | false | # Approach\n- Here, dp[i][j] number of arrays of length i with j inversions\n\n- let us try to find dp[i][j] constructively,\n\n- now, we must appreciate the fact dp[i-1][x] is basically number of arrays of length i-1 with x inversions\n- let the count of such arrays be n\n- now, one of these n arrays would look like:\... | 3 | 0 | ['C++'] | 1 |
k-inverse-pairs-array | Recursive solution with explanation Python 3 | recursive-solution-with-explanation-pyth-kvvw | Explanation of the problem\nImagine you have some numbers from 1 to n, and you want to arrange them in different ways. You also want to count how many times a b | AlexanderBogdan | NORMAL | 2024-01-27T13:18:48.267743+00:00 | 2024-01-27T13:20:07.596093+00:00 | 130 | false | # Explanation of the problem\nImagine you have some numbers from 1 to n, and you want to arrange them in different ways. You also want to count how many times a bigger number comes before a smaller number in each arrangement. For example, if n is 3, you have these possible arrangements:\n\n1, 2, 3 -> no bigger number c... | 3 | 0 | ['Math', 'Dynamic Programming', 'Recursion', 'Memoization', 'Python3'] | 0 |
k-inverse-pairs-array | DP | Python | Easy | dp-python-easy-by-khosiyat-3jvc | see the Successfully Accepted Submission\n\npython\nclass Solution(object):\n def kInversePairs(self, n, k):\n MOD = 10**9 + 7\n dp = [[0] * (k | Khosiyat | NORMAL | 2024-01-27T08:48:08.398899+00:00 | 2024-01-27T08:48:08.398919+00:00 | 191 | false | [see the Successfully Accepted Submission](https://leetcode.com/submissions/detail/1158100692/)\n\n```python\nclass Solution(object):\n def kInversePairs(self, n, k):\n MOD = 10**9 + 7\n dp = [[0] * (k + 1) for _ in range(n + 1)]\n dp[0][0] = 1\n\n for i in range(1, n + 1):\n p... | 3 | 0 | ['Dynamic Programming', 'Python'] | 0 |
k-inverse-pairs-array | Kotlin | Rust | kotlin-rust-by-samoylenkodmitry-dixz | \nhttps://youtu.be/M1umaleU75w\n#### Join me on Telegram\n\nhttps://t.me/leetcode_daily_unstoppable/485\n\n#### Problem TLDR\n\nNumber of arrays of 1..n with k | SamoylenkoDmitry | NORMAL | 2024-01-27T08:41:53.177302+00:00 | 2024-01-27T08:41:53.177323+00:00 | 61 | false | \nhttps://youtu.be/M1umaleU75w\n#### Join me on Telegram\n\nhttps://t.me/leetcode_daily_unstoppable/485\n\n#### Problem TLDR\n\nNumber of arrays of 1..n with k reversed order pairs.\n\n#### Intuition\n\nFir... | 3 | 0 | ['Dynamic Programming', 'Rust', 'Kotlin'] | 0 |
k-inverse-pairs-array | Explaning different approaches O(nk^2) and O(nk) both space O(k) | explaning-different-approaches-onk2-and-ukh7c | Question\nGiven n and k we need to find the number of arrays [1, n] such that there are exactly k inverse pairs.\n\n# Intuition 1\n\nfor k = 0; the answer is al | abhaybhat784 | NORMAL | 2024-01-27T08:12:14.339858+00:00 | 2024-01-27T08:12:14.339889+00:00 | 88 | false | # Question\nGiven n and k we need to find the number of arrays [1, n] such that there are exactly k inverse pairs.\n\n# Intuition 1\n\nfor `k = 0`; the answer is always `1` since no inverse pairs essentially means the array is sorted and there can be only one sorted order for any array `[1, n]`.\n\nalso since there can... | 3 | 0 | ['Java'] | 1 |
k-inverse-pairs-array | Java - Recursion + Memoization | java-recursion-memoization-by-iamvineett-ztlr | \nclass Solution {\n private final int MOD = (int) (1e9 + 7);\n private Integer dp[][];\n\n public int kInversePairs(int n, int k) {\n dp = new | iamvineettiwari | NORMAL | 2024-01-27T08:07:02.581280+00:00 | 2024-01-27T08:07:02.581319+00:00 | 473 | false | ```\nclass Solution {\n private final int MOD = (int) (1e9 + 7);\n private Integer dp[][];\n\n public int kInversePairs(int n, int k) {\n dp = new Integer[n + 1][k + 1];\n\n return getInversions(n, k);\n }\n\n private int getInversions(int n, int k) {\n if (n == 0) {\n ret... | 3 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'Java'] | 0 |
k-inverse-pairs-array | DP and prefix sum Solution | dp-and-prefix-sum-solution-by-techtinker-3ihd | Intuition\nThe intuition is to find a recorrence relation\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nAfter dry running a few t | TechTinkerer | NORMAL | 2024-01-27T07:01:54.177570+00:00 | 2024-01-27T07:01:54.177596+00:00 | 19 | false | # Intuition\nThe intuition is to find a recorrence relation\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nAfter dry running a few test cases we can find out that \nF(n,k)=F(n-1,k)+F(n-1,k-1)+.............+F(n-1,k-n+1)\nAt each recursive step, we cannot find out the function for\nal... | 3 | 0 | ['Dynamic Programming', 'Prefix Sum', 'C++'] | 0 |
k-inverse-pairs-array | Simple solutionUsing DP(Memoization) | simple-solutionusing-dpmemoization-by-ze-680s | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n1. Initialization: The | Zeeshan_2512 | NORMAL | 2024-01-27T03:54:29.672730+00:00 | 2024-01-27T03:54:29.672766+00:00 | 848 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. ```Initialization```: The dp vector is initialized with -1, indicating that no subproblem has been solved yet. The size of dp is 1001 x 1001, which is the maximum c... | 3 | 0 | ['C++'] | 1 |
k-inverse-pairs-array | Easy to Understand Java Code || Beats 100% | easy-to-understand-java-code-beats-100-b-ezw5 | Complexity\n- Time complexity:\nO(n)\n- Space complexity:\nO(n)\n# Code\n\nclass Solution {\n public int kInversePairs(int n, int k) {\n int MOD = 100 | Saurabh_Mishra06 | NORMAL | 2024-01-27T03:22:08.526796+00:00 | 2024-01-27T03:28:46.630689+00:00 | 263 | false | # Complexity\n- Time complexity:\nO(n)\n- Space complexity:\nO(n)\n# Code\n```\nclass Solution {\n public int kInversePairs(int n, int k) {\n int MOD = 1000000007;\n int[][] dp = new int[k + 1][n];\n for (int i = 0; i <= k; i++) {\n for (int j = 0; j < n; j++) {\n if (i... | 3 | 0 | ['Dynamic Programming', 'Java'] | 0 |
k-inverse-pairs-array | Typescript solution that beats 100% | typescript-solution-that-beats-100-by-da-kgi2 | Intuition\nThe problem is asking for the number of different arrays of length n that have exactly k inverse pairs. We can solve this problem by using dynamic pr | daniilboyarinkov | NORMAL | 2024-01-27T01:53:24.583795+00:00 | 2024-01-27T01:53:24.583817+00:00 | 125 | false | # Intuition\nThe problem is asking for the number of different arrays of length `n` that have exactly `k` inverse pairs. We can solve this problem by using dynamic programming, where we systematically consider all possible ways to form arrays with a given number of inverse pairs.\n\n# Approach\nWe initialize a two-dime... | 3 | 0 | ['TypeScript'] | 1 |
k-inverse-pairs-array | 🔥 Clean Javascript Solution - DP - 100% 100% | clean-javascript-solution-dp-100-100-by-sfndu | \n\n\nfunction kInversePairs(n, k, MOD = 10 ** 9 + 7) {\n const [dp, pr] = [new Array(k + 2).fill(0), new Array(k + 2).fill(0)]\n \n for (let i = 1; i | joenix | NORMAL | 2022-07-17T20:32:45.159147+00:00 | 2022-07-17T20:45:16.280659+00:00 | 145 | false | <img width="640" src="https://assets.leetcode.com/users/images/bc94ac7c-ac1c-4f97-8c4b-a0bf817ba99f_1658090051.072736.png" />\n\n```\nfunction kInversePairs(n, k, MOD = 10 ** 9 + 7) {\n const [dp, pr] = [new Array(k + 2).fill(0), new Array(k + 2).fill(0)]\n \n for (let i = 1; i <= n; i++) {\n dp[1] = 1\... | 3 | 0 | ['JavaScript'] | 0 |
k-inverse-pairs-array | [Python] O(N^2 * K) DP to O(N * K); how I simplified my DP approach | python-on2-k-dp-to-on-k-how-i-simplified-1ata | There are some DP-based problems on this site, like this one, that are difficult because their "straightforward" DP approach is too slow to pass. In this post, | grawlixes | NORMAL | 2022-07-17T16:46:18.782879+00:00 | 2022-07-17T16:53:13.828737+00:00 | 217 | false | There are some DP-based problems on this site, like this one, that are difficult because their "straightforward" DP approach is too slow to pass. In this post, I\'ll show you how to simplify it to be fast enough to pass, and I\'ll give you some helpful tips to make it easier for you to do so for other problems as well.... | 3 | 0 | ['Python'] | 0 |
k-inverse-pairs-array | Python || Four Methods || Memoization & Dynamic Programming | python-four-methods-memoization-dynamic-rdhp7 | \nMOD = 10 ** 9 + 7\n\n\nclass Solution:\n\n def kInversePairs(self, N: int, K: int) -> int:\n """\n @see https://leetcode.com/problems/k-inver | shivkj | NORMAL | 2022-07-17T13:02:51.683178+00:00 | 2022-07-17T18:24:20.866469+00:00 | 283 | false | ```\nMOD = 10 ** 9 + 7\n\n\nclass Solution:\n\n def kInversePairs(self, N: int, K: int) -> int:\n """\n @see https://leetcode.com/problems/k-inverse-pairs-array/discuss/104815/Java-DP-O(nk)-solution\n\n let dp[n, k] = number of different arrays containing numbers 1, 2, ..., n with exactly k inve... | 3 | 0 | ['Dynamic Programming', 'Memoization', 'Sliding Window', 'Python'] | 0 |
k-inverse-pairs-array | Easy Understanding C++ Solution using DP | easy-understanding-c-solution-using-dp-b-ak0l | Thinking:\nLet\'s Construct the array having numbers 1 to n from start:\n For \'1\' we have one choice to place it and the array is [1], k=0.\n For \'2\' we hav | anonymous_st | NORMAL | 2022-07-17T06:45:36.972969+00:00 | 2022-07-17T06:45:36.973009+00:00 | 296 | false | **Thinking:**\nLet\'s Construct the array having numbers 1 to n from start:\n* For \'1\' we have one choice to place it and the array is [1], k=0.\n* For \'2\' we have 2 choices (prev array [1])\n\t* Either place it at the end and k would be same as previous, i.e. [1,2], k=0.\n\t* Or place it at the starting and k woul... | 3 | 1 | ['Dynamic Programming', 'C', 'C++'] | 1 |
k-inverse-pairs-array | Java easy solution | java-easy-solution-by-parthhsoni7-j1r1 | \nclass Solution {\n public int kInversePairs(int n, int k) {\n long mod = 1000000007L;\n long dp[] = new long[k + 1];\n dp[0] = 1;\n | parthhsoni7 | NORMAL | 2022-07-17T05:17:15.899975+00:00 | 2022-07-17T05:17:15.900016+00:00 | 201 | false | ```\nclass Solution {\n public int kInversePairs(int n, int k) {\n long mod = 1000000007L;\n long dp[] = new long[k + 1];\n dp[0] = 1;\n for (int i = 2; i <= n; i++) {\n long curr[] = new long[k + 1];\n for (int j = 0; j < curr.length; j++) {\n long currVa... | 3 | 0 | ['Java'] | 0 |
k-inverse-pairs-array | C++ O(nk) solution dp with explanation | c-onk-solution-dp-with-explanation-by-ut-ihxr | \t lets say we have 4, 2 (n = 4 and k = 2)\n\t\t1 2 3 4 (4 is the new comer)\n\n now carefully observe, that if 4 comes somewhere in the middle, it create | uttu_dce | NORMAL | 2021-11-06T19:07:11.646875+00:00 | 2021-11-06T19:46:40.126192+00:00 | 225 | false | \t lets say we have 4, 2 (n = 4 and k = 2)\n\t\t1 2 3 4 (4 is the new comer)\n\n now carefully observe, that if 4 comes somewhere in the middle, it creates number of inversions = number of elements coming after it (because all the elements other than 4 are smaller, and if smaller number comes after a bigger it wi... | 3 | 0 | [] | 1 |
k-inverse-pairs-array | Python VERY simple + short solution with explanation | python-very-simple-short-solution-with-e-xsz0 | Probably the problem I am most proud for having solvd so far. The soution has 2 steps.\n\n1. Notice that f(n,k) = f(n-1,k)+f(n-1,k-1)...f(n-1,k-n+1).\ne.g. for | ko082528 | NORMAL | 2021-09-24T21:38:06.895845+00:00 | 2021-09-24T21:47:22.225198+00:00 | 208 | false | Probably the problem I am most proud for having solvd so far. The soution has 2 steps.\n\n1. Notice that `f(n,k) = f(n-1,k)+f(n-1,k-1)...f(n-1,k-n+1)`.\ne.g. for `n=3, k=1`, the solutions are `[1,3,2]` and `[2,1,3]`. These solutions can be calculated using values of `f` for any value of `k` `f(n=2,k=...)`. That can b... | 3 | 0 | [] | 0 |
k-inverse-pairs-array | [C++] DP | c-dp-by-codedayday-3erw | Approach 1: DP [1]\nIteration equation:\ndp[n][k] = dp[n - 1][k] + dp[n - 1][k-1] + ... + dp[n - 1][k - n + 1]\n\nclass Solution {\npublic:\n int kInversePai | codedayday | NORMAL | 2021-06-19T18:56:29.196636+00:00 | 2021-06-20T01:17:09.636921+00:00 | 468 | false | Approach 1: DP [1]\nIteration equation:\ndp[n][k] = dp[n - 1][k] + dp[n - 1][k-1] + ... + dp[n - 1][k - n + 1]\n```\nclass Solution {\npublic:\n int kInversePairs(int n, int k) {\n int M = 1000000007;\n // dp[i][m]:the number of different arrays consist of numbers from 1 to i such that there are exactl... | 3 | 1 | ['C'] | 0 |
k-inverse-pairs-array | DP and recursive solution C++ | dp-and-recursive-solution-c-by-peeyushbh-hhl0 | \nclass Solution {\npublic:\n /*long long fun(int n,int k,vector<vector<int> > &dp,long long m){\n if(k==0) return 1;\n if(n==0 || k<0) return | peeyushbh1998 | NORMAL | 2020-04-22T08:19:18.009435+00:00 | 2020-04-22T08:19:18.009489+00:00 | 466 | false | ```\nclass Solution {\npublic:\n /*long long fun(int n,int k,vector<vector<int> > &dp,long long m){\n if(k==0) return 1;\n if(n==0 || k<0) return 0;\n if((n*(n-1))/2==k) return 1;\n if((n*(n-1))/2<k) return 0;\n if(dp[n][k]!=-1) return dp[n][k]%m;\n long long sum=fun(n-1,k,d... | 3 | 3 | ['Dynamic Programming', 'C', 'C++'] | 0 |
k-inverse-pairs-array | C# Solution | c-solution-by-leonhard_euler-c3ti | \npublic class Solution \n{\n public int KInversePairs(int n, int k) \n {\n var dp = new int[n + 1, k + 1];\n for(int i = 0; i <= n; i++)\n | Leonhard_Euler | NORMAL | 2019-12-13T01:45:00.701865+00:00 | 2019-12-13T01:55:06.566438+00:00 | 171 | false | ```\npublic class Solution \n{\n public int KInversePairs(int n, int k) \n {\n var dp = new int[n + 1, k + 1];\n for(int i = 0; i <= n; i++)\n for(int j = 0; j <= k; j++)\n dp[i,j] = -1;\n return KInversePairs(n, k, dp);\n }\n \n public int KInversePairs(int... | 3 | 0 | [] | 1 |
k-inverse-pairs-array | intuition plus dp | intuition-plus-dp-by-je390-a1q9 | my intuition was I am using the last element of [1,2,3,4,5,6,..., n] \nI want exactly k inverse pairs\n\n2 options:\n1. either I leave it where it is its f(n-1, | je390 | NORMAL | 2019-02-23T18:52:38.163113+00:00 | 2019-02-23T18:52:38.163213+00:00 | 304 | false | my intuition was I am using the last element of [1,2,3,4,5,6,..., **n**] \nI want exactly k inverse pairs\n\n2 options:\n1. either I leave it where it is its ```f(n-1,k) ``` permutations that have exactly k inverse paris\n1. either I choose to permute it with somebody, I have (n-1) elements to permute it with, \n\nexa... | 3 | 0 | [] | 0 |
k-inverse-pairs-array | Optimization to O(klogk) time complexity for all cases where k < n+3 or k > n*(n-1)/2 - n - 3. | optimization-to-oklogk-time-complexity-f-72a4 | Preface (skip if you prefer to end of preface if you prefer to get right into math/algo approach and proof)The standard dp approach for this problem has Time co | NickUlman | NORMAL | 2025-02-28T04:32:37.811179+00:00 | 2025-02-28T06:50:48.996298+00:00 | 11 | false | Preface (skip if you prefer to end of preface if you prefer to get right into math/algo approach and proof)
The standard dp approach for this problem has Time complexity $$O(nk)$$. I felt there must be some more efficient approach to reduce this, yet couldn't find any better solutions on leetcode. So I worked on the f... | 2 | 0 | ['Math', 'Binary Search', 'Counting', 'Number Theory', 'Java'] | 0 |
k-inverse-pairs-array | Recursive + Memo -> BottomUp -> Optimized BottomUp (Detailed Explanation) | recursive-memo-bottomup-optimized-bottom-fpup | Recursive + Memoization:\n\n# Intuition\n\n- The problem involves finding the number of arrays of size n that satisfy a condition related to the number of inver | harshilsharma2020 | NORMAL | 2024-01-28T16:13:42.720527+00:00 | 2024-01-28T16:13:42.720558+00:00 | 93 | false | # Recursive + Memoization:\n\n# Intuition\n\n- The problem involves finding the number of arrays of size **`n`** that satisfy a condition related to the number of inverse pairs.\n- Utilizes a recursive approach with memoization to avoid redundant calculations and optimize the solution.\n\n# Approach\n\n- **Recursive Fu... | 2 | 0 | ['Dynamic Programming', 'Java'] | 0 |
k-inverse-pairs-array | 🔥 BEATS 100% | ✅ [ JAVA] | beats-100-java-by-enoch1974-2xof | \n# Code\n\npublic class Solution {\n public int kInversePairs(int n, int k) {\n int MOD= 1000000007;\n int[] dp = new int[k+1];\n dp[0] | enoch1974 | NORMAL | 2024-01-27T16:28:06.120434+00:00 | 2024-01-27T16:28:06.120463+00:00 | 277 | false | \n# Code\n```\npublic class Solution {\n public int kInversePairs(int n, int k) {\n int MOD= 1000000007;\n int[] dp = new int[k+1];\n dp[0] = 1;\n for(int i = 2;i <= n;i++){\n for(int j = 1;j <= k;j++){\n dp[j] = (dp[j] + dp[j-1]) % MOD;\n }\n ... | 2 | 0 | ['Java'] | 10 |
k-inverse-pairs-array | C++ || Dp Tabulation solution✅ | c-dp-tabulation-solution-by-051_pooja_ku-mye0 | \n\n# Code\n\n/*\nProblem : Given n, find the count of the arrays(Permutation) of arr[1,2....n] such that there are exactly k inverse pair in the array.\n\n-> A | 051_Pooja_Kumari | NORMAL | 2024-01-27T13:16:47.052182+00:00 | 2024-01-27T13:16:47.052204+00:00 | 49 | false | \n\n# Code\n```\n/*\nProblem : Given n, find the count of the arrays(Permutation) of arr[1,2....n] such that there are exactly k inverse pair in the array.\n\n-> As the answer can be huge return the ans with modulo(10^9+7).\n\ndp[i][j]=> Maximum number of valid array possible till index i having j number of pairs.\n\n*... | 2 | 0 | ['Dynamic Programming', 'C++'] | 0 |
k-inverse-pairs-array | ✅C++ || Explained || Recursion -> Memoisation || w/ Intuition and Comments | c-explained-recursion-memoisation-w-intu-fp47 | you basically have to observe one thing here:\n\nif you move any greater number to i index behind you will generate i new pairs:\nlet\'s take an example: 1,2,3, | yashsachan | NORMAL | 2024-01-27T10:28:54.616147+00:00 | 2024-01-27T10:28:54.616174+00:00 | 27 | false | you basically have to observe one thing here:\n\nif you move any greater number to i index behind you will generate i new pairs:\nlet\'s take an example: `1,2,3,4` if you move 4, 2 index behind\n`1,4,2,3` you now have 2 more inverse pairs i.e. `4>3 and 4>2`\nthis condition holds true even for your next move let\'s move... | 2 | 0 | ['Recursion', 'Memoization'] | 0 |
k-inverse-pairs-array | C++ easy to understand dp memoization code | c-easy-to-understand-dp-memoization-code-or95 | 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 | gauravbisht126 | NORMAL | 2024-01-27T10:00:37.581405+00:00 | 2024-01-27T10:00:37.581450+00:00 | 436 | 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 | ['Dynamic Programming', 'Memoization', 'C++'] | 0 |
k-inverse-pairs-array | C# Solution for K Inverse Pairs Array Problem | c-solution-for-k-inverse-pairs-array-pro-j6q2 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe dynamic programming approach aims to find the number of arrays with a specific coun | Aman_Raj_Sinha | NORMAL | 2024-01-27T06:38:59.969916+00:00 | 2024-01-27T06:38:59.969951+00:00 | 72 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe dynamic programming approach aims to find the number of arrays with a specific count of inverse pairs. It breaks down the problem into subproblems by using a 2D array, where dp[i][j] represents the number of arrays with j inverse pair... | 2 | 0 | ['C#'] | 0 |
k-inverse-pairs-array | Simple || DP || Easy to Understand | simple-dp-easy-to-understand-by-allenpau-j950 | 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 | AllenPaul | NORMAL | 2024-01-27T06:06:25.224730+00:00 | 2024-01-27T06:06:25.224771+00:00 | 30 | 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 | ['Python3'] | 0 |
k-inverse-pairs-array | [C++]✅3 Easy Solutions | Recursion | Memoization | Tabulation | Formula | c3-easy-solutions-recursion-memoization-ri1u4 | Approach : Recursion\n Describe your approach to solving the problem. \n#### Using the Simple formula : \n- k < 0, there is no way, return 0\n- k == 0, there | balajirai | NORMAL | 2024-01-27T05:48:00.631016+00:00 | 2024-02-05T16:38:28.392356+00:00 | 502 | false | # Approach : Recursion\n<!-- Describe your approach to solving the problem. -->\n#### Using the Simple formula : \n- `k < 0`, there is no way, `return 0`\n- `k == 0`, there is only one way possible, `return 1`\n- $$f(n,k)=f(n\u22121,k)+f(n\u22121,k\u22121)+f(n\u22121,k\u22122)+\u22EF+f(n\u22121,k\u2212n+1)$$\n\n# Com... | 2 | 0 | ['Math', 'Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 0 |
k-inverse-pairs-array | Kotlin fast 🚀 & clean ✨ iterative solution 1D DP O(k) memory | kotlin-fast-clean-iterative-solution-1d-8r655 | Intuition\n Describe your first thoughts on how to solve this problem. \nAn array of size n is a distinct positive integer sequence 1..n. So the question asks f | justodiaz | NORMAL | 2024-01-27T05:08:29.960898+00:00 | 2024-01-27T05:08:29.960923+00:00 | 66 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAn array of size `n` is a distinct positive integer sequence `1..n`. So the question asks for the number of permutations that have `k` inverse pairs.\nWe should start by finding a pattern when permuting the numbers `1..n`\nWhen `n=2` all ... | 2 | 0 | ['Kotlin'] | 0 |
k-inverse-pairs-array | Efficient JS Solution - Precomputed DP (Beat 100% both time and memory) | efficient-js-solution-precomputed-dp-bea-07s6 | \n\nPro tip 1: Love lil Xi\xE8 \uD83D\uDC9C \u7231\u5C0F\u8C22 \uD83D\uDC9C\nPro tip 2: Think of dp[n][k] as the number of sequence of distinct numbers of lengt | CuteTN | NORMAL | 2024-01-12T09:56:35.126668+00:00 | 2024-01-27T11:19:35.384461+00:00 | 70 | false | \n\nPro tip 1: Love lil Xi\xE8 \uD83D\uDC9C \u7231\u5C0F\u8C22 \uD83D\uDC9C\nPro tip 2: Think of `dp[n][k]` as the number of sequence of **distinct numbers** of length `n` whose number of inverse is `k`.\nP... | 2 | 0 | ['Dynamic Programming', 'JavaScript'] | 1 |
k-inverse-pairs-array | Easy Solution || Beats 99% || Python | easy-solution-beats-99-python-by-twist_t-s187 | 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 | twist_turn | NORMAL | 2023-11-28T08:42:40.299231+00:00 | 2023-11-28T08:42:40.299252+00:00 | 345 | 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 | ['Python3'] | 0 |
k-inverse-pairs-array | 629: Solution with step by step explanation | 629-solution-with-step-by-step-explanati-74yr | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n\nWe can use dynamic programming to solve this problem.\n1. Define mod as | Marlen09 | NORMAL | 2023-03-18T06:08:40.624692+00:00 | 2023-03-18T06:08:40.624716+00:00 | 71 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n\nWe can use dynamic programming to solve this problem.\n1. Define mod as 10^9 + 7 to avoid overflow.\n2. Create a 2D list \'dp\' with dimensions (n+1) x (k+1) and initialize all elements to 0.\n3. Set dp[0][0] to 1 as there... | 2 | 0 | ['Dynamic Programming', 'Python', 'Python3'] | 0 |
k-inverse-pairs-array | C++ || Recursion+ Memoization+ Optimized || O(NK) | c-recursion-memoization-optimized-onk-by-vpir | Time Complexity: O(N*K)\n Intution: use Brute-force to print sample dp array\n then find a very Complex pattern to optimize DP\'s for loop\n \n \n\n\n\nint M=1e | naveengarg1136 | NORMAL | 2022-07-17T21:51:04.691473+00:00 | 2022-07-17T21:51:04.691501+00:00 | 161 | false | **Time Complexity: O(N*K)**\n Intution: use Brute-force to print sample dp array\n then find a very Complex pattern to optimize DP\'s for loop\n \n \n\n\n```\nint M=1e9+7;\n int dp[1005][1005];\n \n in... | 2 | 0 | ['Memoization'] | 0 |
k-inverse-pairs-array | 629. K Inverse Pairs Array [ python3] 91 % faster | 629-k-inverse-pairs-array-python3-91-fas-cz5v | \nclass Solution:\n def kInversePairs(self, n: int, k: int) -> int:\n if k == 0 : return 1\n count = [1] + [0 for x in range(k)]\n for j | deannos | NORMAL | 2022-07-17T19:36:16.427563+00:00 | 2022-07-17T19:36:16.427598+00:00 | 75 | false | ```\nclass Solution:\n def kInversePairs(self, n: int, k: int) -> int:\n if k == 0 : return 1\n count = [1] + [0 for x in range(k)]\n for j in range(1, n):\n new_count = [1] + [0 for x in range(k)]\n tmp = 1\n for i in range(1, k+1):\n tmp += count... | 2 | 0 | ['Dynamic Programming'] | 2 |
k-inverse-pairs-array | Python | Top Down DP | Clean Code | O(nk) | python-top-down-dp-clean-code-onk-by-ary-xkif | dfs(i, j) the number of permutations of 1...i with j inversions\nprefix(i,j): dfs(i,j)+dfs(i,j-1)+dfs(i,j-2)+...+dfs(i,max(j-i+1,0))\n```\nfrom functools import | aryonbe | NORMAL | 2022-07-17T13:05:03.978592+00:00 | 2022-07-17T13:05:03.978632+00:00 | 161 | false | dfs(i, j) the number of permutations of 1...i with j inversions\nprefix(i,j): dfs(i,j)+dfs(i,j-1)+dfs(i,j-2)+...+dfs(i,max(j-i+1,0))\n```\nfrom functools import cache\nclass Solution:\n def kInversePairs(self, n: int, k: int) -> int:\n M = 10**9 + 7\n @cache\n def prefix(i, j):\n if j... | 2 | 0 | ['Dynamic Programming', 'Python'] | 0 |
k-inverse-pairs-array | JAVA || DP | java-dp-by-aditibhagat23-o39k | The naive approach(TLE)\nclass Solution {\n public int kInversePairs(int n, int k) {\n if(k > n * (n - 1) / 2) // n numbers can generate at most n * ( | aditibhagat23 | NORMAL | 2022-07-17T12:41:51.684413+00:00 | 2022-07-17T12:41:51.684456+00:00 | 174 | false | The naive approach(TLE)\n```class Solution {\n public int kInversePairs(int n, int k) {\n if(k > n * (n - 1) / 2) // n numbers can generate at most n * (n - 1) / 2 inverse pairs\n return 0;\n \n if(k == n * (n - 1) / 2 || k == 0)\n return 1;\n int mod = 1000000007;\n... | 2 | 0 | ['Dynamic Programming', 'Java'] | 0 |
k-inverse-pairs-array | C++ thought/optimization process from TLE to T/M better than 95% | c-thoughtoptimization-process-from-tle-t-zm74 | When I tried to solve the problem, I just listed all permutations of the array from 1 to n and tried to observe if there\'re some interesting patterns to exploi | sc420 | NORMAL | 2022-07-17T11:36:03.632006+00:00 | 2022-07-17T12:40:16.857353+00:00 | 148 | false | When I tried to solve the problem, I just listed all permutations of the array from 1 to n and tried to observe if there\'re some interesting patterns to exploit.\n\nFor example, if we focus on the leftmost number in the 6 permutations of `n=3`, we can see when we change the leftmost number from `1->2`, it will contrib... | 2 | 0 | ['C', 'C++'] | 1 |
k-inverse-pairs-array | C++ DP | Faster Than 99.36% | c-dp-faster-than-9936-by-visualtaggy-0wu8 | \n\n\nclass Solution {\npublic:\n\tint kInversePairs(int n, int k) {\n\tint dp[2][1001] = { 1 };\n\tfor (int N = 1; N <= n; ++N)\n\t\tfor (int K = 0; K <= k; ++ | visualtaggy | NORMAL | 2022-07-17T09:54:13.232962+00:00 | 2022-07-17T09:54:58.272800+00:00 | 241 | false | \n\n```\nclass Solution {\npublic:\n\tint kInversePairs(int n, int k) {\n\tint dp[2][1001] = { 1 };\n\tfor (int N = 1; N <= n; ++N)\n\t\tfor (int K = 0; K <= k; ++K) {\n\t\t\tdp[N % 2][K] = (dp[(N - 1) % 2][K] ... | 2 | 0 | ['C'] | 0 |
k-inverse-pairs-array | Intuition || Visualisation of time optimisation || C++ | intuition-visualisation-of-time-optimisa-uhyu | \n\n```\n\nint dp[1001][1001] = { 1 };\nclass Solution {\npublic:\n int kInversePairs(int n, int k) {\n if (dp[n][k])\n return dp[n][k];\n | Get_the_Guns | NORMAL | 2022-07-17T09:18:46.396797+00:00 | 2022-07-17T09:18:46.396834+00:00 | 158 | false | \n\n```\n\nint dp[1001][1001] = { 1 };\nclass Solution {\npublic:\n int kInversePairs(int n, int k) {\n if (dp[n][k])\n return dp[n][k];\n for (int N = 1; N <= n; ++N)\n f... | 2 | 1 | ['C++'] | 1 |
k-inverse-pairs-array | ✅ C++ || Simple Code || 95% better solution | c-simple-code-95-better-solution-by-psrl-qp5n | ```\n\nclass Solution {\npublic:\n int kInversePairs(int n, int k) {\n int m=1000000007;\n vector> dp(n+1,vector(k+1,0));\n for(int i=1; | psrlucifer | NORMAL | 2022-07-17T07:49:41.658786+00:00 | 2022-07-17T07:49:41.658833+00:00 | 217 | false | ```\n\nclass Solution {\npublic:\n int kInversePairs(int n, int k) {\n int m=1000000007;\n vector<vector<int>> dp(n+1,vector<int>(k+1,0));\n for(int i=1;i<=n;i++) dp[i][0]=1;\n for(int i=1;i<=n;i++) {\n for(int j=1;j<=k;j++) {\n dp[i][j]=((long long)dp[i-1][j]-(j... | 2 | 0 | ['Dynamic Programming'] | 1 |
k-inverse-pairs-array | C++ 6-lines solution 98.08%/100% | 9ms/5.7MB | O(nk)/O(k) | DP+state compression | c-6-lines-solution-9808100-9ms57mb-onkok-mqe8 | \nint kInversePairs(int n, int k) {\n long dp[2][k+1],mod=1e9+7;\n memset(dp,0,sizeof(dp));\n for(int i=1;i<=n;++i)\n for(int j= | SunGod1223 | NORMAL | 2022-07-17T07:28:55.553366+00:00 | 2022-07-17T07:28:55.553414+00:00 | 108 | false | ```\nint kInversePairs(int n, int k) {\n long dp[2][k+1],mod=1e9+7;\n memset(dp,0,sizeof(dp));\n for(int i=1;i<=n;++i)\n for(int j=dp[i&1][0]=1,c=i*(i-1)/2;j<=min(k,c);++j)\n dp[i&1][j]=(dp[i&1][j-1]+dp[(i-1)&1][j]-(j>i?dp[(i-1)&1][j-i]:j==i))%mod;\n return (dp[n&1]... | 2 | 0 | ['Dynamic Programming'] | 0 |
k-inverse-pairs-array | 4 java DP solutions | 4-java-dp-solutions-by-legit_123-4ylx | Method - 1 : DP , memoization , TLE\n\n\n// T(n,k) = T(n-1,k-1) + T(n-1,k-2) + ..... + T(n-1,k-n+1)\n// TC : O(n*k*min(n,k))\n// SC : O(n*k)\n\nclass Solution { | legit_123 | NORMAL | 2022-07-17T07:24:41.784380+00:00 | 2022-07-17T07:24:41.784407+00:00 | 284 | false | Method - 1 : DP , memoization , TLE\n\n```\n// T(n,k) = T(n-1,k-1) + T(n-1,k-2) + ..... + T(n-1,k-n+1)\n// TC : O(n*k*min(n,k))\n// SC : O(n*k)\n\nclass Solution {\n public int kInversePairs(int n, int k) {\n Long dp[][] = new Long[n+1][k+1];\n long mod = (long)1e9+7;\n long ans = fun(n,k,mod,dp... | 2 | 0 | ['Dynamic Programming', 'Memoization', 'Java'] | 0 |
k-inverse-pairs-array | C++ || Evolution from Memoization to Bottom UP DP || Easy to Understand | c-evolution-from-memoization-to-bottom-u-ko5y | \n// Recursion + Memoization //\n// Time Complexity : O(N*K*min(N,K)) //\n// Verdict : TLE //\n\nclass Solution {\npublic:\n \n typedef long long ll;\n | anu_2021 | NORMAL | 2022-07-17T07:17:03.623939+00:00 | 2022-07-17T07:19:25.205905+00:00 | 371 | false | ```\n// Recursion + Memoization //\n// Time Complexity : O(N*K*min(N,K)) //\n// Verdict : TLE //\n\nclass Solution {\npublic:\n \n typedef long long ll;\n \n const ll M = 1e9 + 7;\n \n ll mod(ll a){\n return ((a%M)+M)%M;\n }\n \n ll mul(ll a,ll b){\n return mod(mod(a)*mod(b));\n... | 2 | 0 | ['Dynamic Programming', 'Memoization', 'C', 'Prefix Sum'] | 0 |
k-inverse-pairs-array | c solution | c-solution-by-erhfu-ycvd | /\n\u53EF\u4EE5\u60F3\u6210\u63DB\u4F4D\u7F6E\u7684\u611F\u89BA \u82E5\u6709n\u500B\u6578\u5B57 \u5047\u8A2D\u4ED6\u5011\u4E00\u958B\u59CB\u662F\u5F9E\u5C0F\u63 | Erhfu | NORMAL | 2022-07-17T07:13:51.455287+00:00 | 2022-07-18T03:47:37.135975+00:00 | 88 | false | /*\n\u53EF\u4EE5\u60F3\u6210\u63DB\u4F4D\u7F6E\u7684\u611F\u89BA \u82E5\u6709n\u500B\u6578\u5B57 \u5047\u8A2D\u4ED6\u5011\u4E00\u958B\u59CB\u662F\u5F9E\u5C0F\u6392\u5230\u5927 \u6B64\u6642 k = 0\n\u6392\u9664\u6389\u6700\u5C0F\u7684\u7B2C\u4E00\u500B\u6578\u5B57(\u56E0\u70BA\u4ED6\u4E0D\u80FD\u66F4\u5F80\u524D\u9762\u7... | 2 | 0 | ['C'] | 0 |
k-inverse-pairs-array | Easiest and Clean C++ Code || K Inverse pairs array | easiest-and-clean-c-code-k-inverse-pairs-znkw | Upvote If you like the solution\n\nclass Solution {\npublic:\n \n int kInversePairs(int n, int k) {\n int dp[1001][1001] = {}; //declaring the DP t | imrkr | NORMAL | 2022-07-17T06:06:23.532615+00:00 | 2022-07-17T06:06:23.532652+00:00 | 578 | false | **Upvote** If you like the solution\n```\nclass Solution {\npublic:\n \n int kInversePairs(int n, int k) {\n int dp[1001][1001] = {}; //declaring the DP table\n\n dp[0][0] =1; //intializing 0,0 element 1\n \n for(int i=1 ; i<=n; ++i){ \n long long int cs = 0; // to maintain... | 2 | 0 | ['Dynamic Programming', 'C', 'C++'] | 1 |
k-inverse-pairs-array | Easy solution||Python | easy-solutionpython-by-danurag-ylqo | class Solution:\n def kInversePairs(self, n: int, k: int) -> int:\n M = 1000000007\n dp = [[0](k+1) for _ in range(2)]\n dp[0][0] = 1\n | Danurag | NORMAL | 2022-07-17T04:34:20.359787+00:00 | 2022-07-17T04:34:20.359817+00:00 | 96 | false | class Solution:\n def kInversePairs(self, n: int, k: int) -> int:\n M = 1000000007\n dp = [[0]*(k+1) for _ in range(2)]\n dp[0][0] = 1\n for i in range(1, n+1):\n dp[i%2] = [0]*(k+1)\n dp[i%2][0] = 1\n for j in range(1, k+1):\n dp[i%2][j] = ... | 2 | 0 | ['Dynamic Programming', 'Python'] | 0 |
validate-binary-search-tree | Learn one iterative inorder traversal, apply it to multiple tree questions (Java Solution) | learn-one-iterative-inorder-traversal-ap-o766 | I will show you all how to tackle various tree questions using iterative inorder traversal. First one is the standard iterative inorder traversal using stack. H | issac3 | NORMAL | 2016-05-22T03:38:31+00:00 | 2018-10-26T19:20:07.049810+00:00 | 298,197 | false | I will show you all how to tackle various tree questions using iterative inorder traversal. First one is the standard iterative inorder traversal using stack. Hope everyone agrees with this solution. \n\nQuestion : [Binary Tree Inorder Traversal][1]\n\n public List<Integer> inorderTraversal(TreeNode root) {\n ... | 3,858 | 7 | ['Tree', 'Java'] | 205 |
validate-binary-search-tree | My simple Java solution in 3 lines | my-simple-java-solution-in-3-lines-by-sr-7el7 | public class Solution {\n public boolean isValidBST(TreeNode root) {\n return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);\n }\n | sruzic | NORMAL | 2015-01-12T11:29:13+00:00 | 2018-10-23T06:13:34.561013+00:00 | 205,180 | false | public class Solution {\n public boolean isValidBST(TreeNode root) {\n return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);\n }\n \n public boolean isValidBST(TreeNode root, long minVal, long maxVal) {\n if (root == null) return true;\n if (root.val >... | 1,069 | 18 | ['Recursion', 'Java'] | 97 |
validate-binary-search-tree | C++ in-order traversal, and please do not rely on buggy INT_MAX, INT_MIN solutions any more | c-in-order-traversal-and-please-do-not-r-51ns | class Solution {\n public:\n bool isValidBST(TreeNode* root) {\n TreeNode* prev = NULL;\n return validate(root, prev);\n | wtf0 | NORMAL | 2014-11-01T22:27:41+00:00 | 2018-10-03T10:34:48.475819+00:00 | 124,132 | false | class Solution {\n public:\n bool isValidBST(TreeNode* root) {\n TreeNode* prev = NULL;\n return validate(root, prev);\n }\n bool validate(TreeNode* node, TreeNode* &prev) {\n if (node == NULL) return true;\n if (!validate(node->left, prev)) return... | 625 | 21 | [] | 91 |
validate-binary-search-tree | C++ simple recursive solution | c-simple-recursive-solution-by-jaewoo-jou3 | bool isValidBST(TreeNode* root) {\n return isValidBST(root, NULL, NULL);\n }\n \n bool isValidBST(TreeNode* root, TreeNode* minNode, TreeNode* m | jaewoo | NORMAL | 2015-07-12T21:25:36+00:00 | 2018-10-24T20:13:09.440314+00:00 | 80,066 | false | bool isValidBST(TreeNode* root) {\n return isValidBST(root, NULL, NULL);\n }\n \n bool isValidBST(TreeNode* root, TreeNode* minNode, TreeNode* maxNode) {\n if(!root) return true;\n if(minNode && root->val <= minNode->val || maxNode && root->val >= maxNode->val)\n return fals... | 561 | 5 | [] | 38 |
validate-binary-search-tree | Solution | solution-by-deleted_user-p0fu | C++ []\nclass Solution {\n\nbool isPossible(TreeNode* root, long long l, long long r){\n if(root == nullptr) return true;\n if(root->val < r and root->va | deleted_user | NORMAL | 2023-02-10T20:19:22.498418+00:00 | 2023-03-09T11:31:16.352384+00:00 | 72,834 | false | ```C++ []\nclass Solution {\n\nbool isPossible(TreeNode* root, long long l, long long r){\n if(root == nullptr) return true;\n if(root->val < r and root->val > l)\n return isPossible(root->left, l, root->val) and \n isPossible(root->right, root->val, r);\n else return fal... | 500 | 0 | ['C++', 'Java', 'Python3'] | 9 |
validate-binary-search-tree | 【Video】Check range of each node. | video-check-range-of-each-node-by-niits-eofa | Intuition\nCheck range of each node.\n\n# Solution Video\n\nhttps://youtu.be/PITUkhnuWHQ\n\n### \u2B50\uFE0F\u2B50\uFE0F Don\'t forget to subscribe to my channe | niits | NORMAL | 2024-08-11T17:22:26.852691+00:00 | 2024-08-11T17:22:26.852721+00:00 | 30,685 | false | # Intuition\nCheck range of each node.\n\n# Solution Video\n\nhttps://youtu.be/PITUkhnuWHQ\n\n### \u2B50\uFE0F\u2B50\uFE0F Don\'t forget to subscribe to my channel! \u2B50\uFE0F\u2B50\uFE0F\n\n**\u25A0 Subscribe URL**\nhttp://www.youtube.com/channel/UC9RMNwYTL3SXCP6ShLWVFww?sub_confirmation=1\n\nSubscribers: 7,384\nTha... | 491 | 0 | ['Tree', 'Binary Search Tree', 'C++', 'Java', 'Python3', 'JavaScript'] | 7 |
validate-binary-search-tree | General Tree Traversal Problems, interview Prep | general-tree-traversal-problems-intervie-onom | Please feel free to suggest similar problems in the comment section.\n\nThere are three types of Tree traversal: Inorder, Preorder and Postorder\n\n Inorder: le | pakilamak | NORMAL | 2020-08-11T20:55:49.724744+00:00 | 2020-08-28T22:37:32.276457+00:00 | 37,963 | false | Please feel free to suggest similar problems in the comment section.\n\nThere are three types of Tree traversal: Inorder, Preorder and Postorder\n\n* **Inorder:** left, root, right\n* **Preorder:** root, left child, right child **OR** root, right child, left child\t\t\t\n* **Postorder:** left child, right child, root ... | 461 | 0 | ['Tree', 'Python'] | 10 |
validate-binary-search-tree | Clean Python Solution | clean-python-solution-by-wz2326-8cd1 | Use recursion. Pass down two parameters: lessThan (which means that all nodes in the the current subtree must be smaller than this value) and largerThan (all mu | wz2326 | NORMAL | 2015-10-14T18:05:23+00:00 | 2018-10-23T20:43:02.837672+00:00 | 62,907 | false | Use recursion. Pass down two parameters: `lessThan` (which means that all nodes in the the current subtree must be smaller than this value) and `largerThan` (all must be larger than it). Compare root of the current subtree with these two values. Then, recursively check the left and right subtree of the current one. Tak... | 224 | 1 | [] | 29 |
validate-binary-search-tree | Another passed Java solution | another-passed-java-solution-by-jeantime-r56y | public class Solution {\n public boolean isValidBST(TreeNode root) {\n return helper(root, null, null);\n }\n \n boolean | jeantimex | NORMAL | 2015-06-28T16:49:18+00:00 | 2018-10-16T19:22:10.299647+00:00 | 36,243 | false | public class Solution {\n public boolean isValidBST(TreeNode root) {\n return helper(root, null, null);\n }\n \n boolean helper(TreeNode root, Integer min, Integer max) {\n if (root == null)\n return true;\n \n if ((min != null &... | 194 | 1 | ['Java'] | 21 |
validate-binary-search-tree | C++ THE SIMPLEST!!! O(n) Easy Solution | c-the-simplest-on-easy-solution-by-yehud-qaxh | \nclass Solution {\npublic:\n void inOrder(TreeNode* root) {\n if (!root)\n return;\n inOrder(root->left);\n tree.push_back(r | yehudisk | NORMAL | 2020-12-16T15:10:59.788487+00:00 | 2020-12-16T15:10:59.788529+00:00 | 19,659 | false | ```\nclass Solution {\npublic:\n void inOrder(TreeNode* root) {\n if (!root)\n return;\n inOrder(root->left);\n tree.push_back(root->val);\n inOrder(root->right);\n }\n \n bool isValidBST(TreeNode* root) {\n if (!root)\n return true;\n \n ... | 164 | 2 | ['C'] | 26 |
validate-binary-search-tree | Python/JS/Java/Go/C++ O(n) by DFS and rule [w/ Visualization] 有中文解析文章 | pythonjsjavagoc-on-by-dfs-and-rule-w-vis-150l | O( n ) sol. by DFS and rule.\n\n\u672C\u984C\u5C0D\u61C9\u7684\u4E2D\u6587\u89E3\u984C\u6587\u7AE0\n\n\u7528Python\u5BE6\u73FEBinary Search Tree\u4E8C\u5143\u64 | brianchiang_tw | NORMAL | 2020-12-16T08:07:08.127921+00:00 | 2024-09-02T13:56:35.495395+00:00 | 24,131 | false | O( n ) sol. by DFS and rule.\n\n[\u672C\u984C\u5C0D\u61C9\u7684\u4E2D\u6587\u89E3\u984C\u6587\u7AE0](https://vocus.cc/article/650988e9fd897800019a383f)\n\n[\u7528Python\u5BE6\u73FEBinary Search Tree\u4E8C\u5143\u641C\u5C0B\u6A39](https://vocus.cc/article/66d4426bfd897800012df5fc)\n\n---\n\n# Visualization\n![image.png]... | 133 | 0 | ['Tree', 'Depth-First Search', 'Recursion', 'C', 'Python', 'C++', 'Java', 'Go', 'Python3', 'JavaScript'] | 16 |
validate-binary-search-tree | Python version based on inorder traversal | python-version-based-on-inorder-traversa-0mhc | # Definition for a binary tree node\n # class TreeNode:\n # def __init__(self, x):\n # self.val = x\n # self.left = None\n # | google | NORMAL | 2015-03-20T00:50:43+00:00 | 2018-10-19T21:40:46.716073+00:00 | 33,411 | false | # Definition for a binary tree node\n # class TreeNode:\n # def __init__(self, x):\n # self.val = x\n # self.left = None\n # self.right = None\n \n class Solution:\n # @param root, a tree node\n # @return a boolean\n # 7:38\n def isValidB... | 123 | 3 | ['Tree', 'Python'] | 11 |
validate-binary-search-tree | C++ recursive and iterative | c-recursive-and-iterative-by-jianchao-li-fhoe | For the recursive solution, we set a lower bound and a upper bound for the tree. When we recurse on the left subtree, the upper bound becomes the value of its r | jianchao-li | NORMAL | 2019-02-10T02:30:10.579129+00:00 | 2019-02-10T02:30:10.579173+00:00 | 10,141 | false | For the recursive solution, we set a lower bound and a upper bound for the tree. When we recurse on the left subtree, the upper bound becomes the value of its root. When we recurse on the right subtree, the lower bound becomes the value of its root.\n\n```cpp\nclass Solution {\npublic:\n bool isValidBST(TreeNode* ro... | 106 | 2 | [] | 7 |
validate-binary-search-tree | My java inorder iteration solution | my-java-inorder-iteration-solution-by-sc-j42a | the idea is to do a inorder Traversal and keep the value of the\n\n public boolean isValidBST (TreeNode root){\n \t\t Stack stack = new Stack ();\n \ | scott | NORMAL | 2015-01-18T03:41:31+00:00 | 2018-10-14T03:36:22.705740+00:00 | 35,106 | false | the idea is to do a inorder Traversal and keep the value of the\n\n public boolean isValidBST (TreeNode root){\n \t\t Stack<TreeNode> stack = new Stack<TreeNode> ();\n \t\t TreeNode cur = root ;\n \t\t TreeNode pre = null ;\t\t \n \t\t while (!stack.isEmpty() || cur != null) {\t\t\t \n \t\... | 98 | 1 | [] | 14 |
validate-binary-search-tree | ✔️Python One-liner 96.6% with detailed explantion | recursion | simple ^_^ | python-one-liner-966-with-detailed-expla-95fz | There are two steps:\n1. set a range (at first we set it to (-infinity, infinity)\n2. see if every node is in their own range\n\nHere\'s the one line code:\n\nc | luluboy168 | NORMAL | 2022-08-11T01:35:46.898746+00:00 | 2022-08-11T02:31:18.969283+00:00 | 10,907 | false | There are two steps:\n1. set a range (at first we set it to (-infinity, infinity)\n2. see if every node is in their own range\n\nHere\'s the one line code:\n```\nclass Solution:\n def isValidBST(self, node: ... | 93 | 2 | ['Python'] | 8 |
validate-binary-search-tree | Python3 100% using easy recursion | python3-100-using-easy-recursion-by-bria-es7r | If you just want a solution scroll down, otherwise here\'s an explanation.\n\nLet\'s start with a simple definition of the nodes of a binary search tree!\nA par | briantvann | NORMAL | 2018-07-07T20:22:04.845569+00:00 | 2018-10-12T22:33:12.485540+00:00 | 11,822 | false | If you just want a solution scroll down, otherwise here\'s an explanation.\n\nLet\'s start with a simple definition of the nodes of a binary search tree!\nA parent node is greater than its left child but smaller than its right.\n```\n 2\n / \\\n 1 3 // valid tree!\n```\n\n```\n 5\n / \\\n 1 4 // not va... | 85 | 1 | [] | 16 |
validate-binary-search-tree | Easy Java 0ms Solution | easy-java-0ms-solution-by-deleted_user-huel | The trick is to do an inorder traversal of the tree and check that the value of each node visited is greater than the value of the previous node.\n\n/**\n * Def | deleted_user | NORMAL | 2019-03-19T20:40:42.985388+00:00 | 2019-03-19T20:40:42.985429+00:00 | 4,446 | false | The trick is to do an inorder traversal of the tree and check that the value of each node visited is greater than the value of the previous node.\n```\n/**\n * Definition for a binary tree node.\n * public class TreeNode {\n * int val;\n * TreeNode left;\n * TreeNode right;\n * TreeNode(int x) { val = x... | 67 | 0 | ['Recursion', 'Java'] | 3 |
validate-binary-search-tree | 1 ms Java Solution using Recursion | 1-ms-java-solution-using-recursion-by-ry-ek1a | public class Solution {\n public boolean isValidBST(TreeNode root) {\n return isValid(root, null, null);\n }\n \n public boolean isValid(Tree | ryanswizzle | NORMAL | 2015-12-06T06:46:06+00:00 | 2018-09-11T03:14:39.277293+00:00 | 15,230 | false | public class Solution {\n public boolean isValidBST(TreeNode root) {\n return isValid(root, null, null);\n }\n \n public boolean isValid(TreeNode root, Integer min, Integer max) {\n if(root == null) return true;\n if(min != null && root.val <= min) return false;\n if(max != n... | 67 | 0 | [] | 4 |
validate-binary-search-tree | C++ recursive solution passing all test cases | c-recursive-solution-passing-all-test-ca-5hiv | Keep in Mind\n * Passing NULL to int will cast NULL to 0 in integer. \n * So, It will give wrong result for [0, null, -1].\n * If someone will pass INT_MIN as d | kpilkk | NORMAL | 2020-12-04T15:06:32.561572+00:00 | 2020-12-04T15:06:32.561601+00:00 | 11,782 | false | ### Keep in Mind\n * Passing `NULL` to `int` will cast `NULL` to `0` in integer. \n * So, It will give wrong result for [0, null, -1].\n * If someone will pass `INT_MIN` as default value, then It\'ll fail for case - [-2147483648] and vice-versa.\n * That\'s why pointers are used here.\n * One can use **TreeNode** add... | 54 | 2 | ['Recursion', 'C', 'C++'] | 11 |
validate-binary-search-tree | Simple, Beginner Friendly & Dry Run || Recursive Solution || Time O(n) Space O(n) || Gits ✅✅👈👈 | simple-beginner-friendly-dry-run-recursi-suk2 | Intuition \uD83D\uDC48\n\nIn the question given, the root of a binary tree is provided, and we have to determine whether the given tree is a binary search tree | GiteshSK_12 | NORMAL | 2024-02-15T16:31:36.960888+00:00 | 2024-02-15T16:31:36.960919+00:00 | 10,487 | false | # Intuition \uD83D\uDC48\n\nIn the question given, the root of a binary tree is provided, and we have to determine whether the given tree is a binary search tree or not. To solve this question, we should know what a binary search tree is.\n\n**Understanding BSTs**\n\nA Binary Search Tree (BST) is a binary tree where fo... | 44 | 0 | ['Binary Search Tree', 'C', 'Binary Tree', 'Python', 'C++', 'Java', 'Python3', 'Ruby', 'JavaScript', 'C#'] | 4 |
validate-binary-search-tree | ✅Short || C++ || Java || PYTHON || Explained Solution✔️ || Beginner Friendly ||🔥 🔥 BY MR CODER | short-c-java-python-explained-solution-b-o330 | Please UPVOTE if you LIKE!!\nWatch this video for the better explanation of the code.\nAlso you can SUBSCRIBE \uD83E\uDC83 this channel for the daily leetcode c | mrcoderrm | NORMAL | 2022-08-11T08:24:19.460760+00:00 | 2022-08-11T14:57:47.269970+00:00 | 9,066 | false | **Please UPVOTE if you LIKE!!**\n***Watch this video for the better explanation of the code.***\n**Also you can SUBSCRIBE \uD83E\uDC83 this channel for the daily leetcode challange solution.**\nhttps://www.youtube.com/watch?v=geBeUvcMMwo\n\n**C++**\n```\nclass Solution {\npublic:\n void helper(TreeNode* root, vector... | 41 | 0 | ['C', 'Python', 'Java'] | 4 |
validate-binary-search-tree | javascript : 97% faster simple recursion | javascript-97-faster-simple-recursion-by-8t88 | \nvar isValidBST = function(root, min=null, max=null) {\n if (!root) return true;\n if (min && root.val <= min.val) return false;\n if (max && root.val | shankydoodle | NORMAL | 2020-03-05T18:39:27.095611+00:00 | 2020-03-05T18:39:27.095640+00:00 | 9,540 | false | ```\nvar isValidBST = function(root, min=null, max=null) {\n if (!root) return true;\n if (min && root.val <= min.val) return false;\n if (max && root.val >= max.val) return false;\n return isValidBST(root.left, min, root) && isValidBST(root.right, root, max);\n};\n``` | 39 | 0 | ['Recursion', 'JavaScript'] | 9 |
validate-binary-search-tree | faster than 100% CPP solution : Don't repeat my mistake : "INT_MIN, INT_MAX" | faster-than-100-cpp-solution-dont-repeat-v2q2 | You might be doing the same mistake that I did, using INT_MAX, INT_MIN.\nMy first solution : \n\nclass Solution {\npublic:\n bool isValidBST(TreeNode* root) | rehman0211 | NORMAL | 2021-06-19T10:18:17.313313+00:00 | 2021-06-19T10:19:42.981181+00:00 | 5,946 | false | You might be doing the same mistake that I did, using **INT_MAX**, **INT_MIN**.\nMy first solution : \n```\nclass Solution {\npublic:\n bool isValidBST(TreeNode* root) {\n return fun(root, INT_MAX, INT_MIN);\n }\n \n bool fun(TreeNode* root, int max, int min){\n if(root==NULL){\n re... | 38 | 0 | ['Binary Search Tree', 'C', 'C++'] | 7 |
validate-binary-search-tree | 2 solutions | Easy to understand | Simple | Recursive | Iterative | Python Solution | 2-solutions-easy-to-understand-simple-re-wchq | \n def iterative(self, root):\n if not root: return True\n stack = [(root, -float(\'inf\'), float(\'inf\'))]\n while len(stack):\n | mrmagician | NORMAL | 2020-05-20T23:18:05.605476+00:00 | 2020-05-20T23:18:05.605526+00:00 | 6,442 | false | ```\n def iterative(self, root):\n if not root: return True\n stack = [(root, -float(\'inf\'), float(\'inf\'))]\n while len(stack):\n node, left, right = stack.pop()\n if node.val <= left or node.val >= right: return False\n if node.left: stack.append((node.left,... | 36 | 2 | ['Recursion', 'Iterator', 'Python', 'Python3'] | 8 |
validate-binary-search-tree | [Python] simple dfs, explained | python-simple-dfs-explained-by-dbabichev-2170 | We need to check that some property holds for every node of our tree, so as usual any recursion method should work here. Let us use function dfs(node, low, high | dbabichev | NORMAL | 2020-12-16T08:49:17.341877+00:00 | 2020-12-16T08:49:17.341903+00:00 | 1,276 | false | We need to check that some property holds for every node of our tree, so as usual any recursion method should work here. Let us use function `dfs(node, low, high)`, where:\n1. `node` is node we are currently in\n2. `low` and `high` are bounds we expect to value of this node be in.\n\nNow, let us go to the main algorith... | 34 | 2 | ['Depth-First Search'] | 4 |
validate-binary-search-tree | JAVA | DFS / Recursion | CLEAN | 0 ms ✅ | java-dfs-recursion-clean-0-ms-by-sourin_-50ry | Please Upvote :D\njava []\nclass Solution {\n public boolean isValidBST(TreeNode root) {\n return isValid(root, Long.MIN_VALUE, Long.MAX_VALUE);\n | sourin_bruh | NORMAL | 2022-10-20T07:43:42.341924+00:00 | 2023-02-05T21:38:32.547767+00:00 | 5,925 | false | ### **Please Upvote** :D\n``` java []\nclass Solution {\n public boolean isValidBST(TreeNode root) {\n return isValid(root, Long.MIN_VALUE, Long.MAX_VALUE);\n }\n\n public boolean isValid(TreeNode root, long min, long max) {\n if (root == null) return true;\n if (root.val >= max || root.va... | 33 | 0 | ['Depth-First Search', 'Recursion', 'Java'] | 3 |
validate-binary-search-tree | C++ easy to read in-order traversal solution | c-easy-to-read-in-order-traversal-soluti-grlk | Ensure that every next node of in-order traversal is larger than previous one. Using boolean flag to start with the left most node.\n\n class Solution {\n | paul7 | NORMAL | 2014-12-13T06:20:47+00:00 | 2014-12-13T06:20:47+00:00 | 8,859 | false | Ensure that every next node of in-order traversal is larger than previous one. Using boolean flag to start with the left most node.\n\n class Solution {\n bool first = true;\n int prev = 0;\n public:\n bool isValidBST(TreeNode *root) {\n if(root == NULL) return true;\n \... | 30 | 0 | [] | 2 |
validate-binary-search-tree | My JavaScript solution | my-javascript-solution-by-jeantimex-8m2s | This is a very classic BST problem, we just need to scan every single node in the tree and see if the node's value matches the BST rules, that is all the values | jeantimex | NORMAL | 2017-10-11T15:57:55.123000+00:00 | 2017-10-11T15:57:55.123000+00:00 | 6,215 | false | This is a very classic BST problem, we just need to scan every single node in the tree and see if the node's value matches the BST rules, that is all the values in node's left subtree are less than the value in node, and all the values in node's right subtree are greater than the value in node, if we found a node that ... | 30 | 0 | [] | 6 |
validate-binary-search-tree | C++ inorder =ascending order | c-inorder-ascending-order-by-jeetaksh-rgde | Intuition\nThe inorder traversal of BST is in ascending order.\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nStore the elements o | Jeetaksh | NORMAL | 2023-01-15T07:46:35.324183+00:00 | 2023-01-15T07:46:35.324237+00:00 | 7,847 | false | # Intuition\nThe inorder traversal of BST is in ascending order.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nStore the elements of inorder traversal of BST in an array. If it doesn\'t follow strictly increasing order, return false.\nReturn true otherwise. \n<!-- Describe your app... | 28 | 0 | ['C++'] | 6 |
validate-binary-search-tree | java simple solution | java-simple-solution-by-rmanish0308-kk4y | \nclass Solution {\n public boolean isValidBST(TreeNode root) {\n return isBST(root,Long.MIN_VALUE,Long.MAX_VALUE);\n }\n boolean isBST(TreeNode | rmanish0308 | NORMAL | 2021-06-17T08:47:35.672077+00:00 | 2021-06-17T08:47:35.672116+00:00 | 2,181 | false | ```\nclass Solution {\n public boolean isValidBST(TreeNode root) {\n return isBST(root,Long.MIN_VALUE,Long.MAX_VALUE);\n }\n boolean isBST(TreeNode root,long min,long max)\n {\n if(root == null)\n return true;\n \n //System.out.println(root.val + " "+min + " "+max);\n ... | 28 | 1 | ['Recursion', 'Java'] | 4 |
validate-binary-search-tree | JavaScript Intuitive Solution Using InOrder Traversal | javascript-intuitive-solution-using-inor-6pf8 | javascript\nvar isValidBST = function(root) {\n \n function inOrder(node) {\n if(!node) return [];\n return [...inOrder(node.left), node.val | control_the_narrative | NORMAL | 2020-05-09T01:20:44.937604+00:00 | 2020-05-09T01:28:11.882756+00:00 | 3,614 | false | ```javascript\nvar isValidBST = function(root) {\n \n function inOrder(node) {\n if(!node) return [];\n return [...inOrder(node.left), node.val, ...inOrder(node.right)]\n }\n \n const sortedArr = inOrder(root);\n \n for(let i = 0; i < sortedArr.length; i++) {\n if(sortedArr[i+1... | 25 | 0 | ['JavaScript'] | 6 |
validate-binary-search-tree | 3 python solutions, clamping window, in-order traversal | 3-python-solutions-clamping-window-in-or-78jh | Solution 1:\n\n\nclass Solution(object):\n def isValidBST(self, root):\n """\n :type root: TreeNode\n :rtype: bool\n """\n | vinhhoang | NORMAL | 2018-02-01T01:29:16.799000+00:00 | 2018-02-01T01:29:16.799000+00:00 | 4,309 | false | *Solution 1:\n\n```\nclass Solution(object):\n def isValidBST(self, root):\n """\n :type root: TreeNode\n :rtype: bool\n """\n if not root:\n return True\n return self.isValidNode(root,float('-inf'), float('inf'))\n \n def isValidNode(self, root, l, r):\n ... | 25 | 0 | ['Sliding Window', 'Python'] | 1 |
validate-binary-search-tree | ✅✨Beats 100% in All Languages, Well Explained | beats-100-in-all-languages-well-explaine-rwxe | Problem:\nGiven the root of a binary tree, determine if it is a valid Binary Search Tree (BST).\n\n\n\n\n\n### Approach:\n1. Perform an inorder traversal of the | donquixote-doflamingo | NORMAL | 2024-10-24T16:15:21.313656+00:00 | 2024-10-24T16:15:21.313691+00:00 | 3,335 | false | ### Problem:\nGiven the root of a binary tree, determine if it is a valid Binary Search Tree (BST).\n\n\n\n\n\n### Approach:\n1. Perform an **inorder traversal** of the tree.\n - During the inorder traver... | 24 | 0 | ['Depth-First Search', 'Binary Search Tree', 'C', 'Python', 'C++', 'Java', 'TypeScript', 'Python3', 'JavaScript'] | 1 |
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