From 0b6bfd864a52bfd13211611b70942fc4fee526c8 Mon Sep 17 00:00:00 2001
From: Gaurav Saha <72292857+gaurav-312@users.noreply.github.com>
Date: Mon, 3 Oct 2022 00:05:17 +0530
Subject: [PATCH] code for longest palindromic substring

Example 1:

Input: s = "babad"
Output: "bab"
Explanation: "aba" is also a valid answer.

Example 2:

Input: s = "cbbd"
Output: "bb"
---
 .../Longest_Palindromic_Substring.cpp         | 50 +++++++++++++++++++
 1 file changed, 50 insertions(+)
 create mode 100644 algorithms/CPlusPlus/Dynamic-Programming/Longest_Palindromic_Substring.cpp

diff --git a/algorithms/CPlusPlus/Dynamic-Programming/Longest_Palindromic_Substring.cpp b/algorithms/CPlusPlus/Dynamic-Programming/Longest_Palindromic_Substring.cpp
new file mode 100644
index 00000000..5c006b0f
--- /dev/null
+++ b/algorithms/CPlusPlus/Dynamic-Programming/Longest_Palindromic_Substring.cpp
@@ -0,0 +1,50 @@
+class Solution 
+{
+public:
+    string longestPalindrome(string s) 
+    {
+        int n=s.size(); 
+        
+        //dp[i][j] will be false if substring s[i..j] is not palindrome, Else true
+        vector<vector<bool>> dp(n, vector<bool> (n,false));
+        
+        //All substrings of length 1 are palindromes
+        for(int i=0; i<n; i++) dp[i][i]=true;
+        
+        int start=0, lenMax=1;
+        
+        // check for sub-string of length 2.
+        for(int i=0; i<n-1; i++) 
+        {
+            if(s[i]==s[i+1])
+            {
+                dp[i][i+1] = true;
+                start = i;
+                lenMax = 2;
+            }
+        }
+        
+        // Check for sub-string of length greater than 2.
+        for(int k=3; k<=n; k++)  // k is length of substring
+        {
+            for(int i=0; i<n-k+1; i++) // Fix the starting index
+            {
+                int j=i+k-1; // Get the ending index of substring from starting index i and length k
+                
+                // checking for sub-string from ith index to jth index if s[i+1] to s[j-1] is a palindrome
+                if(dp[i+1][j-1]==true && s[i]==s[j])
+                {
+                    dp[i][j] = true;
+                    
+                    if(k>lenMax)
+                    {
+                        lenMax = k;
+                        start=i;
+                    }
+                }
+            }
+        }
+        
+        return s.substr(start, lenMax); //print sub-string from start to lenMax
+    }
+};