Hello Programmers, In this post you will know how to solve the Palindrome Partitioning II Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.Palindrome Partitioning II Leetcode SolutionsOne more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.ProblemGiven a string s, partition s such that every substring of the partition is a palindrome.Return the minimum cuts needed for a palindrome partitioning of s.Example 1:Input: s = "aab" Output: 1 Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut. Example 2:Input: s = "a" Output: 0 Example 3:Input: s = "ab" Output: 1 Constraints:1 <= s.length <= 2000s consists of lowercase English letters only.Palindrome Partitioning II Leetcode Solutions in Pythonclass Solution: def minCut(self, s: str) -> int: n = len(s) cut = [0] * n dp = [[False] * n for _ in range(n)] for i in range(n): mini = i for j in range(i + 1): if s[j] == s[i] and (j + 1 > i - 1 or dp[j + 1][i - 1]): dp[j][i] = True mini = 0 if j == 0 else min(mini, cut[j - 1] + 1) cut[i] = mini return cut[n - 1]Palindrome Partitioning II Leetcode Solutions in CPPclass Solution { public: int minCut(string s) { const int n = s.length(); // isPalindrome[i][j] := true if s[i..j] is a palindrome vector<vector<bool>> isPalindrome(n, vector<bool>(n, true)); // dp[i] := min cuts needed for a palindrome partitioning of s[0..i] vector<int> dp(n, n); for (int l = 2; l <= n; ++l) for (int i = 0, j = l - 1; j < n; ++i, ++j) isPalindrome[i][j] = s[i] == s[j] && isPalindrome[i + 1][j - 1]; for (int i = 0; i < n; ++i) { if (isPalindrome[0][i]) { dp[i] = 0; continue; } // Try all possible partitions for (int j = 0; j < i; ++j) if (isPalindrome[j + 1][i]) dp[i] = min(dp[i], dp[j] + 1); } return dp.back(); } };Palindrome Partitioning II Leetcode Solutions in Javaclass Solution { public int minCut(String s) { final int n = s.length(); // isPalindrome[i][j] := true if s[i..j] is a palindrome boolean[][] isPalindrome = new boolean[n][n]; for (boolean[] row : isPalindrome) Arrays.fill(row, true); // dp[i] := min cuts needed for a palindrome partitioning of s[0..i] int[] dp = new int[n]; Arrays.fill(dp, n); for (int l = 2; l <= n; ++l) for (int i = 0, j = l - 1; j < n; ++i, ++j) isPalindrome[i][j] = s.charAt(i) == s.charAt(j) && isPalindrome[i + 1][j - 1]; for (int i = 0; i < n; ++i) { if (isPalindrome[0][i]) { dp[i] = 0; continue; } // Try all possible partitions for (int j = 0; j < i; ++j) if (isPalindrome[j + 1][i]) dp[i] = Math.min(dp[i], dp[j] + 1); } return dp[n - 1]; } }Note: This problem Palindrome Partitioning II is generated by Leetcode but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.Next: Balanced Binary Tree Leetcode Solution Post navigationWord Break Leetcode Solution Clone Graph Leetcode Solution