# 0516. Longest Palindromic Subsequence

<https://leetcode.com/problems/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.

## ac1: recursion + memorization

If the two ends of a string are the same, then they must be included in the longest palindrome subsequence. Otherwise, both ends cannot be included in the longest palindrome subsequence.

<https://leetcode.com/problems/longest-palindromic-subsequence/discuss/99111/Evolve-from-brute-force-to-dp>

```
int longestPalindromeSubseq(string s) {
        int n = s.size();
        vector<vector<int>> mem(n,vector<int>(n));
        return longestPalindromeSubseq(0,n-1, s,mem); 
    }
    int longestPalindromeSubseq(int l, int r, string &s, vector<vector<int>>& mem) {
        if(l==r) return 1;
        if(l>r) return 0;
        if(mem[l][r]) return mem[l][r];
        return mem[l][r] = s[l]==s[r] ? 2 + longestPalindromeSubseq(l+1,r-1, s,mem) : 
            max(longestPalindromeSubseq(l+1,r, s,mem),longestPalindromeSubseq(l,r-1, s,mem)); 

    }
```

## ac2: dp

similar: <https://leetcode.com/problems/longest-palindromic-substring>

```java
class Solution {
    public int longestPalindromeSubseq(String s) {
        // edge cases
        if (s == null || s.length() == 0) return 0;

        int n = s.length();
        int[][] dp = new int[n][n];

        for (int i = 0; i < n; i++) {
            for (int j = i; j >= 0; j--) {
                if (s.charAt(j) == s.charAt(i)) {
                    dp[j][i] = j == i || j + 1 == i ? i - j + 1 : 2 + dp[j+1][i-1];
                } else {
                    dp[j][i] = Math.max(dp[j+1][i], dp[j][i-1]);
                }
            }
        }

        return dp[0][n-1];
    }
}
```


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