Longest Palindromic Substring
Last Updated : 10 Mar, 2025
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: ""
[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 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 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# 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 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)); [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 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 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# 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 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)); [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:
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