# Palindrome Partitioning II Leetcode Solution 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.

One 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.

## 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: 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 <= 2000`
• `s` consists of lowercase English letters only.

### Palindrome Partitioning II Leetcode Solutions in Python

```class Solution:
def minCut(self, s: str) -> int:
n = len(s)
cut =  * 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 Solutionsin CPP

```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[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 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[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.