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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0101_0200.s0132_palindrome_partitioning_ii;

// #Hard #String #Dynamic_Programming #2022_06_24_Time_2_ms_(99.89%)_Space_42.1_MB_(81.48%)

import java.util.Arrays;

/**
 * 132 - Palindrome Partitioning II\.
 *
 * Hard
 *
 * 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 lower-case English letters only.
**/
public class Solution {
    public int minCut(String s) {
        int n = s.length();
        char[] t = s.toCharArray();
        int[] dp = new int[n + 1];
        Arrays.fill(dp, Integer.MAX_VALUE);
        dp[0] = -1;
        int i = 0;
        while (i < n) {
            expandAround(t, i, i, dp);
            expandAround(t, i, i + 1, dp);
            i++;
        }
        return dp[n];
    }

    private void expandAround(char[] t, int i, int j, int[] dp) {
        while (i >= 0 && j < t.length && t[i] == t[j]) {
            dp[j + 1] = Math.min(dp[j + 1], dp[i] + 1);
            i--;
            j++;
        }
    }
}