Question

1682. Longest Palindromic Subsequence II

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Description

A subsequence of a string s is considered a good palindromic subsequence if:

• It is a subsequence of s.
• It is a palindrome (has the same value if reversed).
• It has an even length.
• No two consecutive characters are equal, except the two middle ones.

For example, if s = "abcabcabb", then "abba" is considered a good palindromic subsequence, while "bcb" (not even length) and "bbbb" (has equal consecutive characters) are not.

Given a string s, return _the length of the longest good palindromic subsequence in _s.

 

Example 1:

Input: s = "bbabab"
Output: 4
Explanation: The longest good palindromic subsequence of s is "baab".

Example 2:

Input: s = "dcbccacdb"
Output: 4
Explanation: The longest good palindromic subsequence of s is "dccd".

 

Constraints:

• 1 <= s.length <= 250
• s consists of lowercase English letters.

Solutions

Solution 1: Memorization Search

We design a function $dfs(i, j, x)$ to represent the length of the longest "good" palindrome subsequence ending with character $x$ in the index range $[i, j]$ of string $s$. The answer is $dfs(0, n - 1, 26)$.

The calculation process of the function $dfs(i, j, x)$ is as follows:

• If $i >= j$, then $dfs(i, j, x) = 0$;
• If $s[i] = s[j]$ and $s[i] \neq x$, then $dfs(i, j, x) = dfs(i + 1, j - 1, s[i]) + 2$;
• If $s[i] \neq s[j]$, then $dfs(i, j, x) = max(dfs(i + 1, j, x), dfs(i, j - 1, x))$.

During the process, we can use memorization search to avoid repeated calculations.

The time complexity is $O(n^2 \times C)$. Where $n$ is the length of the string $s$, and $C$ is the size of the character set. In this problem, $C = 26$.

class Solution:
  def longestPalindromeSubseq(self, s: str) -> int:
    @cache
    def dfs(i, j, x):
      if i >= j:
        return 0
      if s[i] == s[j] and s[i] != x:
        return dfs(i + 1, j - 1, s[i]) + 2
      return max(dfs(i + 1, j, x), dfs(i, j - 1, x))

    ans = dfs(0, len(s) - 1, '')
    dfs.cache_clear()
    return ans

class Solution {
  private int[][][] f;
  private String s;

  public int longestPalindromeSubseq(String s) {
    int n = s.length();
    this.s = s;
    f = new int[n][n][27];
    for (var a : f) {
      for (var b : a) {
        Arrays.fill(b, -1);
      }
    }
    return dfs(0, n - 1, 26);
  }

  private int dfs(int i, int j, int x) {
    if (i >= j) {
      return 0;
    }
    if (f[i][j][x] != -1) {
      return f[i][j][x];
    }
    int ans = 0;
    if (s.charAt(i) == s.charAt(j) && s.charAt(i) - 'a' != x) {
      ans = dfs(i + 1, j - 1, s.charAt(i) - 'a') + 2;
    } else {
      ans = Math.max(dfs(i + 1, j, x), dfs(i, j - 1, x));
    }
    f[i][j][x] = ans;
    return ans;
  }
}

class Solution {
public:
  int f[251][251][27];

  int longestPalindromeSubseq(string s) {
    int n = s.size();
    memset(f, -1, sizeof f);
    function<int(int, int, int)> dfs = [&](int i, int j, int x) -> int {
      if (i >= j) return 0;
      if (f[i][j][x] != -1) return f[i][j][x];
      int ans = 0;
      if (s[i] == s[j] && s[i] - 'a' != x)
        ans = dfs(i + 1, j - 1, s[i] - 'a') + 2;
      else
        ans = max(dfs(i + 1, j, x), dfs(i, j - 1, x));
      f[i][j][x] = ans;
      return ans;
    };
    return dfs(0, n - 1, 26);
  }
};

func longestPalindromeSubseq(s string) int {
  n := len(s)
  f := make([][][]int, n)
  for i := range f {
    f[i] = make([][]int, n)
    for j := range f[i] {
      f[i][j] = make([]int, 27)
      for k := range f[i][j] {
        f[i][j][k] = -1
      }
    }
  }
  var dfs func(i, j, x int) int
  dfs = func(i, j, x int) int {
    if i >= j {
      return 0
    }
    if f[i][j][x] != -1 {
      return f[i][j][x]
    }
    ans := 0
    if s[i] == s[j] && int(s[i]-'a') != x {
      ans = dfs(i+1, j-1, int(s[i]-'a')) + 2
    } else {
      ans = max(dfs(i+1, j, x), dfs(i, j-1, x))
    }
    f[i][j][x] = ans
    return ans
  }
  return dfs(0, n-1, 26)
}