1682. Longest Palindromic Subsequence II 🔒

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, $C = 26$.

Python3JavaC++Go

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)
}

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