LeetCode 1745: Palindrome Partitioning IV Solution

Master LeetCode problem 1745 (Palindrome Partitioning IV), 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.

1745. Palindrome Partitioning IV

Hard

String Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible partitions of the string s and check if each partition results in three palindromic substrings. We can use dynamic programming to check if a given substring is a palindrome quickly. By checking all possible partitions efficiently, we can determine if it is possible to split the string into three palindromic substrings.

Algorithm:

Use dynamic programming to pre-calculate if each substring is a palindrome.
Iterate through all possible partitions of the string s.
Check if each partition results in three palindromic substrings.
Return true if a valid partition is found, otherwise return false.

Time Complexity: O(n^2), where n is the length of the input string s. Space Complexity: O(n^2), for storing the palindrome information.

Solution Code

class Solution {
    public boolean checkPartitioning(String s) {
        int n = s.length();
        boolean[][] isPalindrome = new boolean[n][n];
        
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i; j < n; j++) {
                if (s.charAt(i) == s.charAt(j) && (j - i <= 2 || isPalindrome[i + 1][j - 1])) {
                    isPalindrome[i][j] = true;
                }
            }
        }
        
        for (int i = 1; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                if (isPalindrome[0][i - 1] && isPalindrome[i][j - 1] && isPalindrome[j][n - 1]) {
                    return true;
                }
            }
        }
        
        return false;
    }
}

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