LeetCode 3504: Longest Palindrome After Substring Concatenation II Solution

Master LeetCode problem 3504 (Longest Palindrome After Substring Concatenation 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.

3504. Longest Palindrome After Substring Concatenation II

Hard

Two Pointers String Dynamic Programming

Problem Explanation

Explanation

To find the longest palindrome that can be formed by concatenating substrings from strings s and t, we need to consider all possible combinations of substrings from both strings. We can do this by iterating through all possible substrings of s and t, concatenating them, and checking if the resulting string is a palindrome.

For each pair of substrings, we can check if the concatenated string is a palindrome by comparing characters from the start and end of the string. If all characters match, then the concatenated string is a palindrome.

The time complexity of this approach is O(n^4) where n is the length of strings s and t since we are iterating through all possible substrings. The space complexity is O(1) as we are not using any extra space.

Solution Code

class Solution {
    public int longestPalindrome(String s, String t) {
        int maxLen = 0;
        for (int i = 0; i <= s.length(); i++) {
            for (int j = 0; j <= t.length(); j++) {
                for (int len = 0; i + len <= s.length() && j + len <= t.length(); len++) {
                    String subS = s.substring(i, i + len);
                    String subT = t.substring(j, j + len);
                    String concat = subS + subT;
                    if (isPalindrome(concat)) {
                        maxLen = Math.max(maxLen, concat.length());
                    }
                }
            }
        }
        return maxLen;
    }

    private boolean isPalindrome(String str) {
        int left = 0;
        int right = str.length() - 1;
        while (left < right) {
            if (str.charAt(left) != str.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}

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