LeetCode 132: Palindrome Partitioning II Solution

Master LeetCode problem 132 (Palindrome Partitioning II), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

132. Palindrome Partitioning II

Hard

String Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We will create a 2D array dp to store whether a substring s[i:j] is a palindrome or not. Then, we will use another 1D array cuts to keep track of the minimum cuts needed up to index i.

The algorithmic steps are as follows:

Initialize the dp array and cuts array.
Iterate over all possible centers for palindrome expansion (odd and even length palindromes) and update the dp array.
Calculate the minimum cuts using the cuts array.

Time Complexity

The time complexity of this solution is O(n^2) where n is the length of the input string s.

Space Complexity

The space complexity of this solution is O(n^2) for the dp array and O(n) for the cuts array.

Solution Code

class Solution {
    public int minCut(String s) {
        int n = s.length();
        boolean[][] dp = new boolean[n][n];
        int[] cuts = new int[n];
        
        for (int i = 0; i < n; i++) {
            int minCuts = i;
            for (int j = 0; j <= i; j++) {
                if (s.charAt(i) == s.charAt(j) && (i - j <= 1 || dp[j + 1][i - 1])) {
                    dp[j][i] = true;
                    minCuts = (j == 0) ? 0 : Math.min(minCuts, cuts[j - 1] + 1);
                }
            }
            cuts[i] = minCuts;
        }
        
        return cuts[n - 1];
    }
}

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