Longest Palindromic Substring

Author: sachuverma

Diff for: DSA Crack Sheet/README.md

@@ -62,6 +62,7 @@
 - [Check if strings are rotations of each other or not](https://practice.geeksforgeeks.org/problems/check-if-strings-are-rotations-of-each-other-or-not-1587115620/1# "view question") - [Cpp Solution](./solutions/Check%20if%20strings%20are%20rotations%20of%20each%20other%20or%20not.cpp)
 - [Program to Check if a string is a valid shuffle of two distinct strings](https://www.programiz.com/java-programming/examples/check-valid-shuffle-of-strings "view topic")
 - [Count and Say](https://leetcode.com/problems/count-and-say/ "view question") - [Cpp Solution](./solutions/Count%20and%20Say.cpp)
+- [Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/ "view question") - [Cpp Solution](./solutions/Longest%20Palindromic%20Substring.cpp)
 - []( "view question") - [Cpp Solution](./solutions/.cpp)
 
 ### Searching & Sorting

Diff for: DSA Crack Sheet/solutions/Longest Palindromic Substring.cpp

@@ -0,0 +1,65 @@
+/*
+Longest Palindromic Substring
+=============================
+
+Given a string s, return the longest palindromic substring in s.
+
+Example 1:
+Input: s = "babad"
+Output: "bab"
+Note: "aba" is also a valid answer.
+
+Example 2:
+Input: s = "cbbd"
+Output: "bb"
+
+Example 3:
+Input: s = "a"
+Output: "a"
+
+Example 4:
+Input: s = "ac"
+Output: "a"
+
+Constraints:
+1 <= s.length <= 1000
+s consist of only digits and English letters (lower-case and/or upper-case),
+*/
+
+string longestPalindrome(string s)
+{
+  int n = s.size(), max_len = 1;
+  string ans = s.substr(0, 1);
+  vector<vector<int>> dp(n, vector<int>(n, 0));
+
+  for (int i = 0; i < n; ++i)
+    dp[i][i] = 1;
+  for (int i = 0; i + 1 < n; ++i)
+  {
+    if (s[i] == s[i + 1])
+    {
+      max_len = 2;
+      ans = s.substr(i, 2);
+      dp[i][i + 1] = 1;
+    }
+  }
+
+  for (int gap = 2; gap < n; ++gap)
+  {
+    for (int i = 0; i + gap < n; ++i)
+    {
+      int j = i + gap;
+      if (s[i] == s[j] && dp[i + 1][j - 1])
+      {
+        if (max_len < gap + 1)
+        {
+          max_len = gap + 1;
+          ans = s.substr(i, gap + 1);
+        }
+        dp[i][j] = 1;
+      }
+    }
+  }
+
+  return ans;
+}