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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

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