Given a string, find the longest substring which is palindrome. For example, if the given string is “forgeeksskeegfor”, the output should be “geeksskeeg”.

We have discussed dynamic programming solution in the previous post. The time complexity of the Dynamic Programming based solution is O(n^2) and it requires O(n^2) extra space. We can find the longest palindrome substring in (n^2) time with O(1) extra space. The idea is to generate all even length and odd length palindromes and keep track of the longest palindrome seen so far.

Step to generate odd length palindrome:

Fix a centre and expand in both directions for longer palindromes.

Step to generate even length palindrome

Fix two centre ( low and high ) and expand in both directions for longer palindromes.

## C/C++

// A O(n^2) time and O(1) space program to find the longest palindromic substring #include <stdio.h> #include <string.h> // A utility function to print a substring str[low..high] void printSubStr(char* str, int low, int high) { for( int i = low; i <= high; ++i ) printf("%c", str[i]); } // This function prints the longest palindrome substring (LPS) // of str[]. It also returns the length of the longest palindrome int longestPalSubstr(char *str) { int maxLength = 1; // The result (length of LPS) int start = 0; int len = strlen(str); int low, high; // One by one consider every character as center point of // even and length palindromes for (int i = 1; i < len; ++i) { // Find the longest even length palindrome with center points // as i-1 and i. low = i - 1; high = i; while (low >= 0 && high < len && str[low] == str[high]) { if (high - low + 1 > maxLength) { start = low; maxLength = high - low + 1; } --low; ++high; } // Find the longest odd length palindrome with center // point as i low = i - 1; high = i + 1; while (low >= 0 && high < len && str[low] == str[high]) { if (high - low + 1 > maxLength) { start = low; maxLength = high - low + 1; } --low; ++high; } } printf("Longest palindrome substring is: "); printSubStr(str, start, start + maxLength - 1); return maxLength; } // Driver program to test above functions int main() { char str[] = "forgeeksskeegfor"; printf("nLength is: %dn", longestPalSubstr( str ) ); return 0; }

## Java

// Java implementation of O(n^2) time and O(1) space method // to find the longest palindromic substring public class LongestPalinSubstring { // A utility function to print a substring str[low..high] static void printSubStr(String str, int low, int high) { System.out.println(str.substring(low, high + 1)); } // This function prints the longest palindrome substring // (LPS) of str[]. It also returns the length of the // longest palindrome static int longestPalSubstr(String str) { int maxLength = 1; // The result (length of LPS) int start = 0; int len = str.length(); int low, high; // One by one consider every character as center // point of even and length palindromes for (int i = 1; i < len; ++i) { // Find the longest even length palindrome with // center points as i-1 and i. low = i - 1; high = i; while (low >= 0 && high < len && str.charAt(low) == str.charAt(high)) { if (high - low + 1 > maxLength) { start = low; maxLength = high - low + 1; } --low; ++high; } // Find the longest odd length palindrome with // center point as i low = i - 1; high = i + 1; while (low >= 0 && high < len && str.charAt(low) == str.charAt(high)) { if (high - low + 1 > maxLength) { start = low; maxLength = high - low + 1; } --low; ++high; } } System.out.print("Longest palindrome substring is: "); printSubStr(str, start, start + maxLength - 1); return maxLength; } // Driver program to test above function public static void main(String[] args) { String str = "forgeeksskeegfor"; System.out.println("Length is: " + longestPalSubstr(str)); } } // This code is contributed by Sumit Ghosh

## Python

# A O(n^2) time and O(1) space program to find the #longest palindromic substring # This function prints the longest palindrome substring (LPS) # of str[]. It also returns the length of the longest palindrome def longestPalSubstr(string): maxLength = 1 start = 0 length = len(string) low = 0 high = 0 # One by one consider every character as center point of # even and length palindromes for i in xrange(1, length): # Find the longest even length palindrome with center # points as i-1 and i. low = i - 1 high = i while low >= 0 and high < length and string[low] == string[high]: if high - low + 1 > maxLength: start = low maxLength = high - low + 1 low -= 1 high += 1 # Find the longest odd length palindrome with center # point as i low = i - 1 high = i + 1 while low >= 0 and high < length and string[low] == string[high]: if high - low + 1 > maxLength: start = low maxLength = high - low + 1 low -= 1 high += 1 print "Longest palindrome substring is:", print string[start:start + maxLength] return maxLength # Driver program to test above functions string = "forgeeksskeegfor" print "Length is: " + str(longestPalSubstr(string)) # This code is contributed by BHAVYA JAIN

Output:

Longest palindrome substring is: geeksskeeg Length is: 10

Time complexity: O ( n^2 ) where n is the length of input string.

Auxiliary Space: O ( 1 )

We will soon be adding more optimized method as separate post.

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