LeetCode 1771. Maximize Palindrome Length From Subsequences

DP

You are given two strings, word1 and word2. You want to construct a string in the following manner:

• Choose some non-empty subsequence subsequence1 from word1.

• Choose some non-empty subsequence subsequence2 from word2.

• Concatenate the subsequences: subsequence1 + subsequence2, to make the string.

Return the length of the longest palindrome that can be constructed in the described manner. If no palindromes can be constructed, return 0.

A subsequence of a string s is a string that can be made by deleting some (possibly none) characters from s without changing the order of the remaining characters.

A palindrome is a string that reads the same forward as well as backward.

Example 1:

Input: word1 = "cacb", word2 = "cbba"
Output: 5
Explanation: Choose "ab" from word1 and "cba" from word2 to make "abcba", which is a palindrome.

Example 2:

Input: word1 = "ab", word2 = "ab"
Output: 3
Explanation: Choose "ab" from word1 and "a" from word2 to make "aba", which is a palindrome.

Example 3:

Input: word1 = "aa", word2 = "bb"
Output: 0
Explanation: You cannot construct a palindrome from the described method, so return 0.

Constraints:

• 1 <= word1.length, word2.length <= 1000

• word1 and word2 consist of lowercase English letters.

Solution:

English Version in Youtube

中文版解答Youtube Link

中文版解答Bilibili Link

class Solution {
public:
    int longestPalindrome(string word1, string word2) {
        string s = word1 + word2;
        int n = s.size();
        vector<vector<int>> dp(n, vector<int>(n,0));
        
        int res = 0;
        for (int i = n - 1; i >= 0; i--) {
            dp[i][i] = 1;
            for (int j = i + 1; j < n; j++) {
                if (s[i]==s[j]) {
                    dp[i][j] = dp[i+1][j-1] + 2;
                    if (i < word1.length() && j >= word1.length()) {
                        res = max(res, dp[i][j]);
                    }
                } else {
                    dp[i][j] = max(dp[i+1][j], dp[i][j-1]);
                }
            }
        }
        return res;
    }
};