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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1701_1800.s1745_palindrome_partitioning_iv;

// #Hard #String #Dynamic_Programming #2022_04_29_Time_10_ms_(100.00%)_Space_44_MB_(95.27%)

/**
 * 1745 - Palindrome Partitioning IV\.
 *
 * Hard
 *
 * Given a string `s`, return `true` _if it is possible to split the string_ `s` _into three **non-empty** palindromic substrings. Otherwise, return_ `false`.
 *
 * A string is said to be palindrome if it the same string when reversed.
 *
 * **Example 1:**
 *
 * **Input:** s = "abcbdd"
 *
 * **Output:** true
 *
 * **Explanation:** "abcbdd" = "a" + "bcb" + "dd", and all three substrings are palindromes.
 *
 * **Example 2:**
 *
 * **Input:** s = "bcbddxy"
 *
 * **Output:** false
 *
 * **Explanation:** s cannot be split into 3 palindromes.
 *
 * **Constraints:**
 *
 * *   `3 <= s.length <= 2000`
 * *   `s` consists only of lowercase English letters.
**/
public class Solution {
    public boolean checkPartitioning(String s) {
        final int len = s.length();
        char[] ch = s.toCharArray();
        int[] dp = new int[len + 1];
        dp[0] = 0x01;
        for (int i = 0; i < len; i++) {
            for (int l : new int[] {i - 1, i}) {
                int r = i;
                while (l >= 0 && r < len && ch[l] == ch[r]) {
                    dp[r + 1] |= dp[l] << 1;
                    l--;
                    r++;
                }
            }
        }
        return (dp[len] & 0x08) == 0x08;
    }
}