Question

1771. Maximize Palindrome Length From Subsequences

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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 $f[i][j] = max(f[i + 1][j], f[i][j - 1])$.

Finally, we return the answer.

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:
  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

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;
  }
};

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
  }
}