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516. Longest Palindromic Subsequence

Description

Given a string s, find the longest palindromic subsequence's length in s.

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

 

Example 1:

Input: s = "bbbab"
Output: 4
Explanation: One possible longest palindromic subsequence is "bbbb".

Example 2:

Input: s = "cbbd"
Output: 2
Explanation: One possible longest palindromic subsequence is "bb".

 

Constraints:

  • 1 <= s.length <= 1000
  • s consists only of lowercase English letters.

Solutions

Solution 1: Dynamic Programming

Thinking

A palindromic subsequence need not be contiguous, so center expansion is not enough, and listing all subsequences is too many.

Let \(f[i][j]\) be the LPS length of \(s[i..j]\). Equal ends add \(2\) to the inner interval; otherwise drop one end. Fill by increasing interval length (\(i\) right to left, \(j\) to the right) so dependencies are ready. \(f[0][n-1]\) is the answer.

We define \(f[i][j]\) as the length of the longest palindromic subsequence from the \(i\)-th character to the \(j\)-th character in string \(s\). Initially, \(f[i][i] = 1\), and the values of other positions are all \(0\).

If \(s[i] = s[j]\), then \(f[i][j] = f[i + 1][j - 1] + 2\); otherwise, \(f[i][j] = \max(f[i + 1][j], f[i][j - 1])\).

Since the value of \(f[i][j]\) is related to \(f[i + 1][j - 1]\), \(f[i + 1][j]\), and \(f[i][j - 1]\), we should enumerate \(i\) from large to small, and enumerate \(j\) from small to large.

The answer is \(f[0][n - 1]\).

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

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

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class Solution {
    public int longestPalindromeSubseq(String s) {
        int n = s.length();
        int[][] f = new int[n][n];
        for (int i = 0; i < n; ++i) {
            f[i][i] = 1;
        }
        for (int i = n - 1; 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;
                } else {
                    f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
                }
            }
        }
        return f[0][n - 1];
    }
}

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

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func longestPalindromeSubseq(s string) int {
    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
            } else {
                f[i][j] = max(f[i+1][j], f[i][j-1])
            }
        }
    }
    return f[0][n-1]
}

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function longestPalindromeSubseq(s: string): number {
    const n = s.length;
    const f: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
    for (let i = 0; i < n; ++i) {
        f[i][i] = 1;
    }
    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;
            } else {
                f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
            }
        }
    }
    return f[0][n - 1];
}

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