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Minimum Characters to Add at Front for Palindrome

Longest Palindromic Substring

Last Updated : 10 Mar, 2025

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Given a string s, the task is to find the longest substring which is a palindrome. If there are multiple answers, then return the first appearing substring.

Examples:

Input: s = “forgeeksskeegfor”
Output: “geeksskeeg”
Explanation: There are several possible palindromic substrings like “kssk”, “ss”, “eeksskee” etc. But the substring “geeksskeeg” is the longest among all.

Input: s = “Geeks”
Output: “ee”

Input: s = “abc”
Output: “a”

Input: s = “”
Output: “”

Table of Content

[Naive Approach] Generating all sub-strings – O(n^3) time and O(1) space

The idea is to generate all substrings.

For each substring, check if it is palindrome or not.
If substring is Palindrome, then update the result on the basis of longest palindromic substring found till now.

C++

#include <bits/stdc++.h> using namespace std;  // Function to check if a substring  // s[low..high] is a palindrome bool checkPal(string &s, int low, int high) {     while (low < high) {         if (s[low] != s[high])             return false;         low++;         high--;     }     return true; }  // function to find the longest palindrome substring string longestPalindrome(string& s) {        // Get length of input string     int n = s.size();      // All substrings of length 1 are palindromes     int maxLen = 1, start = 0;      // Nested loop to mark start and end index     for (int i = 0; i < n; i++) {         for (int j = i; j < n; j++) {                        // Check if the current substring is              // a palindrome             if (checkPal(s, i, j) && (j - i + 1) > maxLen) {                 start = i;                 maxLen = j - i + 1;             }         }     }      return s.substr(start, maxLen); }  int main() {     string s = "forgeeksskeegfor";     cout << longestPalindrome(s) << endl;     return 0; }

Java

// Java program to find the longest // palindromic substring.  import java.util.*;  class GfG {      // Function to check if a substring      // s[low..high] is a palindrome     static boolean checkPal(String s, int low, int high) {         while (low < high) {             if (s.charAt(low) != s.charAt(high))                 return false;             low++;             high--;         }         return true;     }      // Function to find the longest palindrome substring     static String longestPalindrome(String s) {          // Get length of input string         int n = s.length();          // All substrings of length 1 are palindromes         int maxLen = 1, start = 0;          // Nested loop to mark start and end index         for (int i = 0; i < n; i++) {             for (int j = i; j < n; j++) {                  // Check if the current substring is                  // a palindrome                 if (checkPal(s, i, j) && (j - i + 1) > maxLen) {                     start = i;                     maxLen = j - i + 1;                 }             }         }          return s.substring(start, start + maxLen);     }      public static void main(String[] args) {         String s = "forgeeksskeegfor";         System.out.println(longestPalindrome(s));     } }

Python

# Python program to find the longest # palindromic substring.  # Function to check if a substring  # s[low..high] is a palindrome def checkPal(str, low, high):     while low < high:         if str[low] != str[high]:             return False         low += 1         high -= 1     return True  # Function to find the longest palindrome substring def longestPalindrome(s):          # Get length of input string     n = len(s)      # All substrings of length 1 are palindromes     maxLen = 1     start = 0      # Nested loop to mark start and end index     for i in range(n):         for j in range(i, n):              # Check if the current substring is              # a palindrome             if checkPal(s, i, j) and (j - i + 1) > maxLen:                 start = i                 maxLen = j - i + 1      return s[start:start + maxLen]  if __name__ == "__main__":     s = "forgeeksskeegfor"     print(longestPalindrome(s))

C#

// C# program to find the longest // palindromic substring.  using System;  class GfG {      // Function to check if a substring      // s[low..high] is a palindrome     static bool checkPal(string s, int low, int high) {         while (low < high) {             if (s[low] != s[high])                 return false;             low++;             high--;         }         return true;     }      // Function to find the longest palindrome substring     static string longestPalindrome(string s) {          // Get length of input string         int n = s.Length;          // All substrings of length 1 are palindromes         int maxLen = 1, start = 0;          // Nested loop to mark start and end index         for (int i = 0; i < n; i++) {             for (int j = i; j < n; j++) {                  // Check if the current substring is                  // a palindrome                 if (checkPal(s, i, j) && (j - i + 1) > maxLen) {                     start = i;                     maxLen = j - i + 1;                 }             }         }          return s.Substring(start, maxLen);     }      static void Main(string[] args) {         string s = "forgeeksskeegfor";         Console.WriteLine(longestPalindrome(s));     } }

JavaScript

// JavaScript program to find the longest // palindromic substring.  // Function to check if a substring  // s[low..high] is a palindrome function checkPal(s, low, high) {     while (low < high) {         if (s[low] !== s[high])             return false;         low++;         high--;     }     return true; }  // Function to find the longest palindrome substring function longestPalindrome(s) {      // Get length of input string     const n = s.length;      // All substrings of length 1 are palindromes     let maxLen = 1, start = 0;      // Nested loop to mark start and end index     for (let i = 0; i < n; i++) {         for (let j = i; j < n; j++) {              // Check if the current substring is              // a palindrome             if (checkPal(s, i, j) && (j - i + 1) > maxLen) {                 start = i;                 maxLen = j - i + 1;             }         }     }      return s.substring(start, start + maxLen); }  // Driver Code const s = "forgeeksskeegfor"; console.log(longestPalindrome(s));

Output

geeksskeeg

[Better Approach – 1] Using Dynamic Programming – O(n^2) time and O(n^2) space

The idea is to use Dynamic Programming to store the status of smaller substrings and use these results to check if a longer substring forms a palindrome.

The main idea behind the approach is that if we know the status (i.e., palindrome or not) of the substring ranging [i, j], we can find the status of the substring ranging [i-1, j+1] by only matching the character str[i-1] and str[j+1].
If the substring from i to j is not a palindrome, then the substring from i-1 to j+1 will also not be a palindrome. Otherwise, it will be a palindrome only if str[i-1] and str[j+1] are the same.
Base on this fact, we can create a 2D table (say table[][] which stores status of substring str[i . . . j] ), and check for substrings with length from 1 to N. For each length find all the substrings starting from each character i and find if it is a palindrom or not using the above idea. The longest length for which a palindrome formed will be the required asnwer.

Note: Refer to Longest Palindromic Substring using Dynamic Programming for detailed approach and code.

[Better Approach – 2] Using Expansion from center – O(n^2) time and O(1) space

The idea is to traverse each character in the string and treat it as a potential center of a palindrome, trying to expand around it in both directions while checking if the expanded substring remains a palindrome.

For each position, we check for both odd-length palindromes (where the current character is the center) and even-length palindromes (where the current character and the next character together form the center).
As we expand outward from each center, we keep track of the start position and length of the longest palindrome found so far, updating these values whenever we find a longer valid palindrome.

Step-by-step approach:

Use two pointers, low and hi, for the left and right end of the current palindromic substring being found.
Then checks if the characters at s[low] and s[hi] are the same.
- If they are, it expands the substring to the left and right by decrementing low and incrementing hi.
- It continues this process until the characters at s[low] and s[hi] are unequal or until the indices are in bounds.
If the length of the current palindromic substring becomes greater than the maximum length, it updates the maximum length.

C++

// C++ program to find the longest // palindromic substring. #include <bits/stdc++.h> using namespace std;  // Function to find the longest palindrome substring string longestPalindrome(string &s) {          int n = s.length();     if (n == 0) return "";      int start = 0, maxLen = 1;      // Traverse the input string     for (int i = 0; i < n; i++) {          // THIS RUNS TWO TIMES          // for both odd and even length         // palindromes. j = 0 means odd         // and j = 1 means even length         for (int j = 0; j <= 1; j++) {             int low = i;             int high = i + j;               // Expand substring while it is a palindrome             // and in bounds             while (low >= 0 && high < n && s[low] == s[high]) {                 int currLen = high - low + 1;                 if (currLen > maxLen) {                     start = low;                     maxLen = currLen;                 }                 low--;                 high++;             }         }     }      return s.substr(start, maxLen); }  int main() {     string s = "forgeeksskeegfor";     cout << longestPalindrome(s) << endl;     return 0; }

Java

// Java program to find the longest // palindromic substring.  class GfG {      // Function to find the longest palindrome substring     static String longestPalindrome(String s) {         int n = s.length();         if (n == 0) return "";          int start = 0, maxLen = 1;          // Traverse the input string         for (int i = 0; i < n; i++) {              // THIS RUNS TWO TIMES              // for both odd and even length             // palindromes. j = 0 means odd             // and j = 1 means even length             for (int j = 0; j <= 1; j++) {                 int low = i;                 int high = i + j;                   // Expand substring while it is a palindrome                 // and in bounds                 while (low >= 0 && high < n && s.charAt(low) == s.charAt(high)) {                     int currLen = high - low + 1;                     if (currLen > maxLen) {                         start = low;                         maxLen = currLen;                     }                     low--;                     high++;                 }             }         }          return s.substring(start, start + maxLen);     }      public static void main(String[] args) {         String s = "forgeeksskeegfor";         System.out.println(longestPalindrome(s));     } }

Python

# Python program to find the longest # palindromic substring.  # Function to find the  # longest palindrome substring def longestPalindrome(s):     n = len(s)     if n == 0:         return ""      start, maxLen = 0, 1      # Traverse the input string     for i in range(n):          # THIS RUNS TWO TIMES          # for both odd and even length         # palindromes. j = 0 means odd         # and j = 1 means even length         for j in range(2):             low, high = i, i + j              # Expand substring while it is a palindrome             # and in bounds             while low >= 0 and high < n and s[low] == s[high]:                 currLen = high - low + 1                 if currLen > maxLen:                     start = low                     maxLen = currLen                 low -= 1                 high += 1      return s[start:start + maxLen]  if __name__ == "__main__":     s = "forgeeksskeegfor"     print(longestPalindrome(s))

C#

// C# program to find the longest // palindromic substring.  using System;  class GfG {      // Function to find the longest palindrome substring     static string longestPalindrome(string s) {         int n = s.Length;         if (n == 0) return "";          int start = 0, maxLen = 1;          // Traverse the input string         for (int i = 0; i < n; i++) {              // THIS RUNS TWO TIMES              // for both odd and even length             // palindromes. j = 0 means odd             // and j = 1 means even length             for (int j = 0; j <= 1; j++) {                 int low = i;                 int high = i + j;                   // Expand substring while it is a palindrome                 // and in bounds                 while (low >= 0 && high < n && s[low] == s[high]) {                     int currLen = high - low + 1;                     if (currLen > maxLen) {                         start = low;                         maxLen = currLen;                     }                     low--;                     high++;                 }             }         }          return s.Substring(start, maxLen);     }      static void Main(string[] args) {         string s = "forgeeksskeegfor";         Console.WriteLine(longestPalindrome(s));     } }

JavaScript

// JavaScript program to find the longest // palindromic substring.  // Function to find the longest palindrome substring function longestPalindrome(s) {     const n = s.length;     if (n === 0) return "";      let start = 0, maxLen = 1;      // Traverse the input string     for (let i = 0; i < n; i++) {          // THIS RUNS TWO TIMES          // for both odd and even length         // palindromes. j = 0 means odd         // and j = 1 means even length         for (let j = 0; j <= 1; j++) {             let low = i;             let high = i + j;               // Expand substring while it is a palindrome             // and in bounds             while (low >= 0 && high < n && s[low] === s[high]) {                 const currLen = high - low + 1;                 if (currLen > maxLen) {                     start = low;                     maxLen = currLen;                 }                 low--;                 high++;             }         }     }      return s.substring(start, start + maxLen); }  // Driver Code const s = "forgeeksskeegfor"; console.log(longestPalindrome(s));

Output

geeksskeeg

[Expected Approach] Using Manacher’s Algorithm – O(n) time and O(n) space

We can solve this problem in linear time using Manacher’s Algorithm. Refer the below links for details:

Minimum Characters to Add at Front for Palindrome

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