# 1771. Maximize Palindrome Length From Subsequences

## Description

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.

## Solutions

Solution 1: Dynamic Programming

First, we concatenate strings word1 and word2 to get string $s$. Then we can transform the problem into finding the length of the longest palindromic subsequence in string $s$. However, when calculating the final answer, we need to ensure that at least one character in the palindrome string comes from word1 and another character comes from word2.

We define $f[i][j]$ as the length of the longest palindromic subsequence in the substring of string $s$ with index range $[i, j]$.

If $s[i] = s[j]$, then $s[i]$ and $s[j]$ must be in the longest palindromic subsequence, at this time $f[i][j] = f[i + 1][j - 1] + 2$. At this point, we also need to judge whether $s[i]$ and $s[j]$ come from word1 and word2. If so, we update the maximum value of the answer to $ans=\max(ans, f[i][j])$.

If $s[i] \neq s[j]$, then $s[i]$ and $s[j]$ will definitely not appear in the longest palindromic subsequence at the same time, at this time $f[i][j] = max(f[i + 1][j], f[i][j - 1])$.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of string $s$.

• class Solution {
public int longestPalindrome(String word1, String word2) {
String s = word1 + word2;
int n = s.length();
int[][] f = new int[n][n];
for (int i = 0; i < n; ++i) {
f[i][i] = 1;
}
int ans = 0;
for (int i = n - 2; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
if (s.charAt(i) == s.charAt(j)) {
f[i][j] = f[i + 1][j - 1] + 2;
if (i < word1.length() && j >= word1.length()) {
ans = Math.max(ans, f[i][j]);
}
} else {
f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
}
}
}
return ans;
}
}

• class Solution {
public:
int longestPalindrome(string word1, string word2) {
string s = word1 + word2;
int n = s.size();
int f[n][n];
memset(f, 0, sizeof f);
for (int i = 0; i < n; ++i) f[i][i] = 1;
int ans = 0;
for (int i = n - 2; ~i; --i) {
for (int j = i + 1; j < n; ++j) {
if (s[i] == s[j]) {
f[i][j] = f[i + 1][j - 1] + 2;
if (i < word1.size() && j >= word1.size()) {
ans = max(ans, f[i][j]);
}
} else {
f[i][j] = max(f[i + 1][j], f[i][j - 1]);
}
}
}
return ans;
}
};

• class Solution:
def longestPalindrome(self, word1: str, word2: str) -> int:
s = word1 + word2
n = len(s)
f = [[0] * n for _ in range(n)]
for i in range(n):
f[i][i] = 1
ans = 0
for i in range(n - 2, -1, -1):
for j in range(i + 1, n):
if s[i] == s[j]:
f[i][j] = f[i + 1][j - 1] + 2
if i < len(word1) <= j:
ans = max(ans, f[i][j])
else:
f[i][j] = max(f[i + 1][j], f[i][j - 1])
return ans


• func longestPalindrome(word1 string, word2 string) (ans int) {
s := word1 + word2
n := len(s)
f := make([][]int, n)
for i := range f {
f[i] = make([]int, n)
f[i][i] = 1
}
for i := n - 2; i >= 0; i-- {
for j := i + 1; j < n; j++ {
if s[i] == s[j] {
f[i][j] = f[i+1][j-1] + 2
if i < len(word1) && j >= len(word1) && ans < f[i][j] {
ans = f[i][j]
}
} else {
f[i][j] = max(f[i+1][j], f[i][j-1])
}
}
}
return ans
}

• function longestPalindrome(word1: string, word2: string): number {
const s = word1 + word2;
const n = s.length;
const f: number[][] = Array.from({ length: n }, () => Array.from({ length: n }, () => 0));
for (let i = 0; i < n; ++i) {
f[i][i] = 1;
}
let ans = 0;
for (let i = n - 2; ~i; --i) {
for (let j = i + 1; j < n; ++j) {
if (s[i] === s[j]) {
f[i][j] = f[i + 1][j - 1] + 2;
if (i < word1.length && j >= word1.length) {
ans = Math.max(ans, f[i][j]);
}
} else {
f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
}
}
}
return ans;
}


• impl Solution {
pub fn longest_palindrome(word1: String, word2: String) -> i32 {
let s: Vec<char> = format!("{}{}", word1, word2).chars().collect();
let n = s.len();
let mut f = vec![vec![0; n]; n];
for i in 0..n {
f[i][i] = 1;
}
let mut ans = 0;
for i in (0..n - 1).rev() {
for j in i + 1..n {
if s[i] == s[j] {
f[i][j] = f[i + 1][j - 1] + 2;
if i < word1.len() && j >= word1.len() {
ans = ans.max(f[i][j]);
}
} else {
f[i][j] = f[i + 1][j].max(f[i][j - 1]);
}
}
}
ans
}
}