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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 ProblemPalindrome 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: 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
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> isPalindrome(n, vector(n, true));

    // dp[i] := min cuts needed for a palindrome partitioning of s[0..i]
    vector 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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