LeetCode 1771: Maximize Palindrome Length From Subsequences Solution
Master LeetCode problem 1771 (Maximize Palindrome Length From Subsequences), 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.
1771. Maximize Palindrome Length From Subsequences
Problem Explanation
Explanation
To maximize the palindrome length, we need to find the longest palindromic subsequence that can be constructed by concatenating subsequences of word1 and word2. We can approach this problem using dynamic programming.
- Initialize a 2D array
dpof sizen x m, wherenis the length ofword1andmis the length ofword2. - Traverse through the
word1andword2characters and fill thedparray based on the following conditions:- If the characters match, increment the diagonal cell by 2.
- Otherwise, take the maximum of the cells above and to the left.
- The final answer will be stored in
dp[n-1][m-1].
Time Complexity: O(n*m) where n is the length of word1 and m is the length of word2.
Space Complexity: O(n*m) for the dp array.
Solution Code
class Solution {
public int longestPalindrome(String word1, String word2) {
int n = word1.length(), m = word2.length();
int[][] dp = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (word1.charAt(i) == word2.charAt(j)) {
dp[i][j] = (i > 0 && j > 0) ? dp[i-1][j-1] + 2 : 2;
} else {
dp[i][j] = Math.max((i > 0) ? dp[i-1][j] : 0, (j > 0) ? dp[i][j-1] : 0);
}
}
}
return (dp[n-1][m-1] > 1) ? dp[n-1][m-1] : 0;
}
}Try It Yourself
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