> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/dynamic-programming/untitled-5.md).

# Longest Palindromic Subsequence

Given a string s, find the longest palindromic subsequence's length in s. You may assume that the maximum length of s is 1000.

**Example 1:**\
Input:

```
"bbbab"
```

Output:

```
4
```

One possible longest palindromic subsequence is "bbbb".

**Example 2:**\
Input:

```
"cbbd"
```

Output:

```
2
```

One possible longest palindromic subsequence is "bb".

**Constraints:**

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

```java
class Solution {
    public int longestPalindromeSubseq(String s) {
        // dp[i][j] -> length of longest palindromic subsequence between index i and j
        int dp[][] = new int[s.length()][s.length()];
        // now the base case will be that when i=j then longest sequence will be that
        // character itself
        // Therefore ans=1
        for (int i = 0; i < s.length(); i++)
            dp[i][i] = 1;
        // Remember pointer j will always be ahead of pointer i
        // Look at dp[i][j] calls like putting 2 pointers at string
        // Now to use our base cases we need this flow of loop
        for (int i = s.length() - 1; i >= 0; i--) {
            for (int j = i + 1; j < s.length(); j++) {
                if (s.charAt(i) == s.charAt(j)) {
                    dp[i][j] = dp[i + 1][j - 1] + 2;
                } else {
                    dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[0][s.length() - 1];
    }
}
```