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In this post, you will find the solution for the **Palindrome Partitioning II** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Palindrome Partitioning II– LeetCode Problem

Palindrome Partitioning II– LeetCode Problem

**Problem:**

Given 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:1Explanation: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 <= 2000`

`s`

consists of lowercase English letters only.

Palindrome Partitioning II– LeetCode Solutions

Palindrome Partitioning II Solution in C++:

class 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 Solution in Java:

class 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]; } }

Palindrome Partitioning II Solution in Python:

class 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]

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