Palindrome Partitioning

The Palindrome Partitioning problem involves partitioning a given string such that every substring of the partition is a palindrome. We aim to find the minimum number of cuts needed for a palindrome partitioning of the string.

Problem Statement

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.

Intuition

To solve this problem, we use dynamic programming. We maintain an array dp where dp[i] represents the minimum number of cuts needed for a palindrome partitioning of the substring s[0:i+1].

Dynamic Programming Approach

We use a 2D boolean array isPalindrome to keep track of whether a substring s[i:j+1] is a palindrome. Then we build the dp array using the isPalindrome array.

Pseudocode for Palindrome Partitioning using DP

Initialize:

dp = [0] * len(s)
for i in range(len(s)):
    min_cut = i
    for j in range(i + 1):
        if s[j:i+1] is a palindrome:
            min_cut = 0 if j == 0 else min(min_cut, dp[j-1] + 1)
    dp[i] = min_cut
return dp[-1]

Example Output:

Given the string:

• s = "aab"

The minimum cuts needed for a palindrome partitioning is 1.

Output Explanation:

We can partition the string as "aa" | "b", which requires 1 cut.

Implementing Palindrome Partitioning using DP

Python Implementation

class Solution:
    def minCut(self, s: str) -> int:
        n = len(s)
        dp = [0] * n
        is_palindrome = [[False] * n for _ in range(n)]
        
        for i in range(n):
            min_cut = i
            for j in range(i + 1):
                if s[j] == s[i] and (i - j <= 2 or is_palindrome[j + 1][i - 1]):
                    is_palindrome[j][i] = True
                    min_cut = 0 if j == 0 else min(min_cut, dp[j - 1] + 1)
            dp[i] = min_cut
        
        return dp[-1]

s = "aab"
solution = Solution()
print(solution.minCut(s))  # Output: 1

Java Implementation

class Solution {
    public int minCut(String s) {
        int n = s.length();
        int[] dp = new int[n];
        boolean[][] isPalindrome = new boolean[n][n];

        for (int i = 0; i < n; i++) {
            int minCut = i;
            for (int j = 0; j <= i; j++) {
                if (s.charAt(j) == s.charAt(i) && (i - j <= 2 || isPalindrome[j + 1][i - 1])) {
                    isPalindrome[j][i] = true;
                    minCut = j == 0 ? 0 : Math.min(minCut, dp[j - 1] + 1);
                }
            }
            dp[i] = minCut;
        }

        return dp[n - 1];
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        String s = "aab";
        System.out.println(solution.minCut(s));  // Output: 1
    }
}

C++ Implementation

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

class Solution {
public:
    int minCut(string s) {
        int n = s.length();
        vector<int> dp(n, 0);
        vector<vector<bool>> isPalindrome(n, vector<bool>(n, false));

        for (int i = 0; i < n; i++) {
            int minCut = i;
            for (int j = 0; j <= i; j++) {
                if (s[j] == s[i] && (i - j <= 2 || isPalindrome[j + 1][i - 1])) {
                    isPalindrome[j][i] = true;
                    minCut = j == 0 ? 0 : min(minCut, dp[j - 1] + 1);
                }
            }
            dp[i] = minCut;
        }

        return dp[n - 1];
    }
};

int main() {
    Solution solution;
    string s = "aab";
    cout << solution.minCut(s) << endl;  // Output: 1
    return 0;
}

Complexity Analysis

• Time Complexity: $O(n^2)$, where n is the length of the string.
• Space Complexity: $O(n^2)$, for storing the isPalindrome array.

Conclusion

The Palindrome Partitioning problem can be efficiently solved using dynamic programming. The provided implementations in Python, Java, and C++ demonstrate how to find the minimum cuts needed for a palindrome partitioning of a given string.