Manacher’s Algorithm

To find the longest palindromic substring from a string, we can use Manacher’s Algorithm. By selecting each character, we will try to find if there any palindrome using left and right pointer. There is another array to store information, from that information we can easily find how long the palindrome is. For each character, the array will store information. After traversing the whole string, we can find the longest palindromic subsequence from the created array.
The time complexity of this algorithm is O(n).
Input:
String: “levelup”
Output:
Longest palindrome is: level
longestPalindrome(text)
Input − The text to find the longest palindrome
Output − The longest common palindrome in the text
Begin
   n := text size
   if n = 0, then
      return null string
   n := 2n+1

   define array longPal of size n
   longPal[0] := 0 and longPal[1] := 1
   centerIndex := 1
   rightIndex := 2
   right := 0
   maxPalLength := 0
   maxCenterIndex := 0
   start := -1 and end := -1, diff := -1

   for right := 2 to n-1, do
      left := 2*centerIndex – right
      longPal[right] := 0
      diff := rightIndex – right
      if diff > 0, then
         longPal[right] := minimum(longPal[left], diff)
         while (right + longPal[right]) < n AND (right - longPal[right]) > 0 AND
            (right + longPal[right]+1)mod 2 = 0 OR
            text[(right + longPal[right] + 1)/2] = text[(right - longPal[right]-1)/2], do
            increase longPal[right] by 1
      done

      if longPal[right] > maxPalLength, then
         maxPalLength := longPal[right]
         maxCenterIndex := right
      if (right + longPal[right]) > rightIndex, then
         centerIndex := right
         rightIndex := right + longPal[right]
   done
   start := (maxCenterIndex – maxPalLength)/2
   end := start + maxPalLength – 1

   palindrome = substring of text[start..end]
   return palindrome
End
#include<iostream>
using namespace std;

int min(int a, int b) {
   return (a<b)?a:b;
}

string longestPalindrome(string mainString) {
   int n = mainString.size();
   if(n == 0)
      return "";
   n = 2*n + 1; //count the next position
   int longPal[n]; //array to store longest palindrome length
   longPal[0] = 0; longPal[1] = 1;
   int centerIndex = 1;
   int rightIndex = 2;
   int right = 0, left;
   int maxPalLength = 0, maxCenterIndex = 0;
   int start = -1, end = -1, diff = -1;

   for (right = 2; right < n; right++) {
      left  = 2*centerIndex-right; //calculate left position using center and right
      longPal[right] = 0;
      diff = rightIndex - right;

      if(diff > 0)
         longPal[right] = min(longPal[left], diff);
      while ( ((right + longPal[right]) < n && (right - longPal[right]) > 0) &&
         ( ((right + longPal[right] + 1) % 2 == 0) ||
         (mainString[(right + longPal[right] + 1)/2] == mainString[(right - longPal[right] - 1)/2] ))) {
         longPal[right]++;
      }

      if(longPal[right] > maxPalLength) {    //max palindrome length
         maxPalLength = longPal[right];
         axCenterIndex = right;
      }

      if (right + longPal[right] > rightIndex) {
         centerIndex = right;
         rightIndex = right + longPal[right];
      }
   }

   start = (maxCenterIndex - maxPalLength)/2;
   end = start + maxPalLength - 1;
   string palindrome;

   for(int i=start; i<=end; i++)
      palindrome += mainString[i];
      return palindrome;
}

int main(int argc, char *argv[]) {
   string mainString, palindrome;
   cout << "Enter String:";
   cin >> mainString;
   palindrome = longestPalindrome(mainString);
   cout << "Longest palindrome is: " << palindrome << endl;
}
Enter String: levelup
Longest palindrome is: level