120. Palindrome Partitioning II

Precompute which substrings are palindromes, then let dp[i] be the fewest cuts for the prefix ending at i: 0 if the whole prefix is a palindrome, else 1 + the best earlier cut before a palindromic tail.

Two DPs

1. Palindrome table pal[i][j] = is s[i..j] a palindrome? Fill by increasing length using pal[i][j] = (s[i]==s[j]) and pal[i+1][j-1].
2. Min-cut dp[i] = minimum cuts for s[:i+1]. If s[0..i] is itself a palindrome, dp[i] = 0. Otherwise dp[i] = min over j of dp[j-1] + 1 where s[j..i] is a palindrome.

dp[i] = 0 if pal[0][i] else min(dp[j-1]+1 for j in 1..i if pal[j][i])

Worked example

"aab" → cut once into "aa" | "b" → 1.

Complexity

• Time: O(n²).
• Space: O(n²).

class Solution {
    public int minCut(String s) {
        int n = s.length();
        if (n <= 1) return 0;
        boolean[][] pal = new boolean[n][n];
        for (int i = 0; i < n; i++) pal[i][i] = true;
        for (int len = 2; len <= n; len++) {
            for (int i = 0; i + len - 1 < n; i++) {
                int j = i + len - 1;
                if (s.charAt(i) == s.charAt(j) && (len == 2 || pal[i + 1][j - 1])) {
                    pal[i][j] = true;
                }
            }
        }
        int[] dp = new int[n];
        for (int i = 0; i < n; i++) {
            if (pal[0][i]) {
                dp[i] = 0;
            } else {
                int best = Integer.MAX_VALUE;
                for (int j = 1; j <= i; j++) {
                    if (pal[j][i]) best = Math.min(best, dp[j - 1] + 1);
                }
                dp[i] = best;
            }
        }
        return dp[n - 1];
    }
}