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+---
+id: longest-palindromic-subsequence
+title: Longest Palindromic Subsequence
+sidebar_label: 0516-Longest-Palindromic-Subsequence
+tags:
+- Dynamic Programming
+- String
+description: "Given a string s, find the longest palindromic subsequence's length in s."
+---
+
+## Problem
+
+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.
+
+### Examples
+
+**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.
+
+---
+
+## Approach
+
+To solve this problem, we can use dynamic programming. We will create a 2D array `dp` where `dp[i][j]` represents the length of the longest palindromic subsequence in the substring `s[i:j+1]`.
+
+### Steps:
+
+1. Initialize a 2D array `dp` of size `n x n` where `n` is the length of the string `s`. Set all elements to 0.
+2. For each single character, set `dp[i][i] = 1` because a single character is a palindrome of length 1.
+3. Fill the `dp` array in a bottom-up manner:
+   - For each substring length `l` from 2 to `n`:
+     - For each starting index `i` from 0 to `n-l`:
+       - Set `j = i + l - 1`.
+       - If `s[i] == s[j]`, then `dp[i][j] = dp[i+1][j-1] + 2`.
+       - Otherwise, `dp[i][j] = max(dp[i+1][j], dp[i][j-1])`.
+4. The result will be `dp[0][n-1]`, the length of the longest palindromic subsequence in the entire string.
+
+### Solution
+
+#### Java Solution
+
+```java
+class Solution {
+    public int longestPalindromeSubseq(String s) {
+        int n = s.length();
+        int[][] dp = new int[n][n];
+        
+        for (int i = n - 1; i >= 0; i--) {
+            dp[i][i] = 1;
+            for (int j = i + 1; j < n; 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][n - 1];
+    }
+}
+```
+#### C++ Solution 
+
+```cpp
+class Solution {
+public:
+    int longestPalindromeSubseq(string s) {
+        int n = s.length();
+        vector<vector<int>> dp(n, vector<int>(n, 0));
+        
+        for (int i = n - 1; i >= 0; i--) {
+            dp[i][i] = 1;
+            for (int j = i + 1; j < n; j++) {
+                if (s[i] == s[j]) {
+                    dp[i][j] = dp[i + 1][j - 1] + 2;
+                } else {
+                    dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);
+                }
+            }
+        }
+        
+        return dp[0][n - 1];
+    }
+};
+```
+#### Python Solution
+
+```python
+class Solution:
+    def longestPalindromeSubseq(self, s: str) -> int:
+        n = len(s)
+        dp = [[0] * n for _ in range(n)]
+        
+        for i in range(n - 1, -1, -1):
+            dp[i][i] = 1
+            for j in range(i + 1, n):
+                if s[i] == s[j]:
+                    dp[i][j] = dp[i + 1][j - 1] + 2
+                else:
+                    dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])
+        
+        return dp[0][n - 1]
+```
+### Complexity Analysis
+**Time Complexity:** O(n^2)
+>Reason: We are filling an n x n table, and each cell takes constant time to compute.
+
+**Space Complexity:** O(n^2)
+>Reason: We use a 2D array dp of size n x n to store the intermediate results.
+
+### References
+**LeetCode Problem:** Longest Palindromic Subsequence
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