From ecc1f491c0787b00363b18905042faa2bb73c126 Mon Sep 17 00:00:00 2001
From: its-sai <88180989+its-sai@users.noreply.github.com>
Date: Wed, 19 Oct 2022 21:56:19 +0530
Subject: [PATCH] Create Longest Palindromic Subsequence

I want  to contribute by adding a  very popular strings problem in your amazing repository
---
 .../strings/Longest Palindromic Subsequence   | 51 +++++++++++++++++++
 1 file changed, 51 insertions(+)
 create mode 100644 src/main/java/com/rampatra/strings/Longest Palindromic Subsequence

diff --git a/src/main/java/com/rampatra/strings/Longest Palindromic Subsequence b/src/main/java/com/rampatra/strings/Longest Palindromic Subsequence
new file mode 100644
index 00000000..f02236f8
--- /dev/null
+++ b/src/main/java/com/rampatra/strings/Longest Palindromic Subsequence	
@@ -0,0 +1,51 @@
+       import java.util.*;
+       import java.io.*;
+       
+       class home{
+             public static void main(String args[]){
+             Scanner sc= new Scanner (System.in);
+             int t=sc.nextInt();
+             while(t-->0){
+             String s=sc.next();
+             Solution obj = new Solution();
+             System.out.println(obj.longestPalinSubseq(s);
+             }
+           }
+        }
+         
+class Solution
+{
+    public int longestPalinSubseq(String A)
+    {
+         
+         int n = A.length();
+        
+        StringBuilder st = new StringBuilder();
+        st.append(A);
+        st.reverse();
+        
+        String B = st.toString();
+        
+        // a palindrome reads the same from front and reverse.
+        // hence, the longest palindromic subsequence is
+        // basically the longest common subsequence in
+        // A and reverse(A)
+        int[][] dp = new int[n + 1][n + 1];
+        for (int i = 0; i <= n; ++i)
+        {
+            for (int j = 0; j <= n; ++j)
+            {
+                if (i == 0 || j == 0)
+                    dp[i][j] = 0;
+                else
+                {
+                    if (A.charAt(i - 1) == B.charAt(j - 1))
+                        dp[i][j] = 1 + dp[i - 1][j - 1];
+                    else
+                        dp[i][j] = Math.max (dp[i - 1][j], dp[i][j - 1]);
+                }
+            }
+        }
+        return dp[n][n];
+            }
+}