Manacher’s Algorithm – Linear Time Longest Palindromic Substring – Part 3

Last Updated : 24 Mar, 2023

In Manacher’s Algorithm Part 1 and Part 2, we gone through some of the basics, understood LPS length array and how to calculate it efficiently based on four cases. Here we will implement the same.
We have seen that there are no new character comparison needed in case 1 and case 2. In case 3 and case 4, necessary new comparison are needed.
In following figure,

Manacher's Algorithm – Linear Time Longest Palindromic Substring

If at all we need a comparison, we will only compare actual characters, which are at “odd” positions like 1, 3, 5, 7, etc.
Even positions do not represent a character in string, so no comparison will be performed for even positions.
If two characters at different odd positions match, then they will increase LPS length by 2.

There are many ways to implement this depending on how even and odd positions are handled. One way would be to create a new string 1^st where we insert some unique character (say #, $ etc) in all even positions and then run algorithm on that (to avoid different way of even and odd position handling). Other way could be to work on given string itself but here even and odd positions should be handled appropriately.

Here we will start with given string itself. When there is a need of expansion and character comparison required, we will expand in left and right positions one by one. When odd position is found, comparison will be done and LPS Length will be incremented by ONE. When even position is found, no comparison done and LPS Length will be incremented by ONE (So overall, one odd and one even positions on both left and right side will increase LPS Length by TWO).

Implementation:

C++

// A C++ program to implement Manacher’s Algorithm 
#include <bits/stdc++.h> 
using namespace std; 
  
void findLongestPalindromicString(string text) 
{ 
    int N = text.length(); 
    if (N == 0) 
        return; 
  
    // Position count 
    N = 2 * N + 1; 
  
    // LPS Length Array 
    int L[N]; 
    L[0] = 0; 
    L[1] = 1; 
  
    // centerPosition 
    int C = 1; 
  
    // centerRightPosition 
    int R = 2; 
  
    // currentRightPosition 
    int i = 0; 
  
    // currentLeftPosition 
    int iMirror; 
    int expand = -1; 
    int diff = -1; 
    int maxLPSLength = 0; 
    int maxLPSCenterPosition = 0; 
    int start = -1; 
    int end = -1; 
  
    // Uncomment it to print LPS Length array 
    // printf("%d %d ", L[0], L[1]); 
    for (i = 2; i < N; i++) { 
        // Get currentLeftPosition iMirror 
        // for currentRightPosition i 
        iMirror = 2 * C - i; 
  
        // Reset expand - means no 
        // expansion required 
        expand = 0; 
        diff = R - i; 
  
        // If currentRightPosition i is 
        // within centerRightPosition R 
        if (diff >= 0) { 
  
            // Case 1 
            if (L[iMirror] < diff) 
                L[i] = L[iMirror]; 
  
            // Case 2 
            else if (L[iMirror] == diff && R == N - 1) 
                L[i] = L[iMirror]; 
  
            // Case 3 
            else if (L[iMirror] == diff && R < N - 1) { 
                L[i] = L[iMirror]; 
  
                // Expansion required 
                expand = 1; 
            } 
  
            // Case 4 
            else if (L[iMirror] > diff) { 
                L[i] = diff; 
  
                // Expansion required 
                expand = 1; 
            } 
        } 
        else { 
            L[i] = 0; 
  
            // Expansion required 
            expand = 1; 
        } 
  
        if (expand == 1) { 
  
            // Attempt to expand palindrome centered 
            // at currentRightPosition i. Here for odd 
            // positions, we compare characters and 
            // if match then increment LPS Length by ONE 
            // If even position, we just increment LPS 
            // by ONE without any character comparison 
  
            while (((i + L[i]) < N && (i - L[i]) > 0) 
                   && (((i + L[i] + 1) % 2 == 0) 
                       || (text[(i + L[i] + 1) / 2] 
                           == text[(i - L[i] - 1) / 2]))) { 
                L[i]++; 
            } 
        } 
  
        // Track maxLPSLength 
        if (L[i] > maxLPSLength) { 
            maxLPSLength = L[i]; 
            maxLPSCenterPosition = i; 
        } 
  
        // If palindrome centered at 
        // currentRightPosition i expand 
        // beyond centerRightPosition R, 
        // adjust centerPosition C based 
        // on expanded palindrome. 
        if (i + L[i] > R) { 
            C = i; 
            R = i + L[i]; 
        } 
  
        // Uncomment it to print LPS Length array 
        // System.out.print("%d ", L[i]); 
    } 
  
    start = (maxLPSCenterPosition - maxLPSLength) / 2; 
    end = start + maxLPSLength - 1; 
  
    // System.out.print("start: %d end: %d\n", 
    //                  start, end); 
    cout << "LPS of string is " << text << " : "; 
  
    for (i = start; i <= end; i++) 
        cout << text[i]; 
    cout << endl; 
} 
  
int main() 
{ 
  
    string text1 = "babcbabcbaccba"; 
    findLongestPalindromicString(text1); 
  
    string text2 = "abaaba"; 
    findLongestPalindromicString(text2); 
  
    string text3 = "abababa"; 
    findLongestPalindromicString(text3); 
  
    string text4 = "abcbabcbabcba"; 
    findLongestPalindromicString(text4); 
  
    string text5 = "forgeeksskeegfor"; 
    findLongestPalindromicString(text5); 
  
    string text6 = "caba"; 
    findLongestPalindromicString(text6); 
  
    string text7 = "abacdfgdcaba"; 
    findLongestPalindromicString(text7); 
  
    string text8 = "abacdfgdcabba"; 
    findLongestPalindromicString(text8); 
  
    string text9 = "abacdedcaba"; 
    findLongestPalindromicString(text9); 
    return 0; 
} 
  
// This code is contributed by Ishankhandelwals.

C

// A C program to implement Manacher’s Algorithm  
#include <stdio.h>  
#include <string.h>  
  
char text[100];  
void findLongestPalindromicString()  
{  
    int N = strlen(text);  
    if(N == 0)  
        return;  
    N = 2*N + 1; //Position count  
    int L[N]; //LPS Length Array  
    L[0] = 0;  
    L[1] = 1;  
    int C = 1; //centerPosition  
    int R = 2; //centerRightPosition  
    int i = 0; //currentRightPosition  
    int iMirror; //currentLeftPosition  
    int expand = -1;  
    int diff = -1;  
    int maxLPSLength = 0;  
    int maxLPSCenterPosition = 0;  
    int start = -1;  
    int end = -1;  
      
    //Uncomment it to print LPS Length array  
    //printf("%d %d ", L[0], L[1]);  
    for (i = 2; i < N; i++)  
    {  
        //get currentLeftPosition iMirror for currentRightPosition i  
        iMirror = 2*C-i;  
        //Reset expand - means no expansion required  
        expand = 0;  
        diff = R - i;  
        //If currentRightPosition i is within centerRightPosition R  
        if(diff >= 0)  
        {  
            if(L[iMirror] < diff) // Case 1  
                L[i] = L[iMirror];  
            else if(L[iMirror] == diff && R == N-1) // Case 2  
                L[i] = L[iMirror];  
            else if(L[iMirror] == diff && R < N-1) // Case 3  
            {  
                    L[i] = L[iMirror];  
                    expand = 1; // expansion required  
            }  
            else if(L[iMirror] > diff) // Case 4  
            {  
                L[i] = diff;  
                expand = 1; // expansion required  
            }  
        }  
        else
        {  
            L[i] = 0;  
            expand = 1; // expansion required  
        }  
          
        if (expand == 1)  
        {  
            //Attempt to expand palindrome centered at currentRightPosition i  
            //Here for odd positions, we compare characters and  
            //if match then increment LPS Length by ONE  
            //If even position, we just increment LPS by ONE without  
            //any character comparison  
            while (((i + L[i]) < N && (i - L[i]) > 0) &&  
                ( ((i + L[i] + 1) % 2 == 0) ||  
                (text[(i + L[i] + 1)/2] == text[(i-L[i]-1)/2] )))  
            {  
                L[i]++;  
            }  
        }  
  
        if(L[i] > maxLPSLength) // Track maxLPSLength  
        {  
            maxLPSLength = L[i];  
            maxLPSCenterPosition = i;  
        }  
  
        // If palindrome centered at currentRightPosition i  
        // expand beyond centerRightPosition R,  
        // adjust centerPosition C based on expanded palindrome.  
        if (i + L[i] > R)  
        {  
            C = i;  
            R = i + L[i];  
        }  
        //Uncomment it to print LPS Length array  
        //printf("%d ", L[i]);  
    }  
    //printf("\n");  
    start = (maxLPSCenterPosition - maxLPSLength)/2;  
    end = start + maxLPSLength - 1;  
    //printf("start: %d end: %d\n", start, end);  
    printf("LPS of string is %s : ", text);  
    for(i=start; i<=end; i++)  
        printf("%c", text[i]);  
    printf("\n");  
}  
  
  
int main(int argc, char *argv[])  
{  
  
    strcpy(text, "babcbabcbaccba");  
    findLongestPalindromicString();  
  
    strcpy(text, "abaaba");  
    findLongestPalindromicString();  
  
    strcpy(text, "abababa");  
    findLongestPalindromicString();  
  
    strcpy(text, "abcbabcbabcba");  
    findLongestPalindromicString();  
  
    strcpy(text, "forgeeksskeegfor");  
    findLongestPalindromicString();  
  
    strcpy(text, "caba");  
    findLongestPalindromicString();  
  
    strcpy(text, "abacdfgdcaba");  
    findLongestPalindromicString();  
  
    strcpy(text, "abacdfgdcabba");  
    findLongestPalindromicString();  
  
    strcpy(text, "abacdedcaba");  
    findLongestPalindromicString();  
  
    return 0;  
}  
  
// This code is contributed by Ishan Khandelwal

Java

// A Java program to implement Manacher’s Algorithm 
import java.lang.*; 
  
class GFG{ 
      
public static void findLongestPalindromicString( 
                   String text) 
{ 
    int N = text.length(); 
    if(N == 0) 
        return; 
      
    // Position count 
    N = 2 * N + 1;  
      
    // LPS Length Array 
    int []L = new int [N];  
    L[0] = 0; 
    L[1] = 1; 
      
    // centerPosition  
    int C = 1; 
      
    // centerRightPosition 
    int R = 2;  
      
    // currentRightPosition 
    int i = 0;  
      
    // currentLeftPosition 
    int iMirror;  
    int expand = -1; 
    int diff = -1; 
    int maxLPSLength = 0; 
    int maxLPSCenterPosition = 0; 
    int start = -1; 
    int end = -1; 
      
    // Uncomment it to print LPS Length array 
    // printf("%d %d ", L[0], L[1]); 
    for (i = 2; i < N; i++)  
    { 
        // Get currentLeftPosition iMirror 
        // for currentRightPosition i 
        iMirror = 2 * C - i; 
          
        // Reset expand - means no  
        // expansion required 
        expand = 0; 
        diff = R - i; 
          
        // If currentRightPosition i is  
        // within centerRightPosition R 
        if(diff >= 0)  
        { 
              
            // Case 1 
            if(L[iMirror] < diff)  
                L[i] = L[iMirror]; 
                  
            // Case 2 
            else if(L[iMirror] == diff && R == N - 1)  
                L[i] = L[iMirror]; 
              
            // Case 3 
            else if(L[iMirror] == diff && R < N - 1)  
            { 
                    L[i] = L[iMirror]; 
                      
                     // Expansion required 
                    expand = 1; 
            } 
              
            // Case 4 
            else if(L[iMirror] > diff)  
            { 
                L[i] = diff; 
                  
                // Expansion required 
                expand = 1;  
            } 
        } 
        else
        { 
            L[i] = 0; 
              
            // Expansion required 
            expand = 1;  
        } 
          
        if (expand == 1) 
        { 
              
            // Attempt to expand palindrome centered 
            // at currentRightPosition i. Here for odd 
            // positions, we compare characters and  
            // if match then increment LPS Length by ONE 
            // If even position, we just increment LPS 
            // by ONE without any character comparison 
            try
            { 
                while (((i + L[i]) < N &&  
                        (i - L[i]) > 0) && 
                      (((i + L[i] + 1) % 2 == 0) || 
                       (text.charAt((i + L[i] + 1) / 2) ==  
                        text.charAt((i - L[i] - 1) / 2)))) 
                { 
                    L[i]++; 
                } 
            } 
            catch (Exception e) 
            { 
                assert true; 
            } 
        } 
          
        // Track maxLPSLength 
        if(L[i] > maxLPSLength)  
        { 
            maxLPSLength = L[i]; 
            maxLPSCenterPosition = i; 
        } 
  
        // If palindrome centered at  
        // currentRightPosition i expand 
        // beyond centerRightPosition R, 
        // adjust centerPosition C based 
        // on expanded palindrome. 
        if (i + L[i] > R)  
        { 
            C = i; 
            R = i + L[i]; 
        } 
          
        //Uncomment it to print LPS Length array 
        //System.out.print("%d ", L[i]); 
    } 
      
    start = (maxLPSCenterPosition -  
             maxLPSLength) / 2; 
    end = start + maxLPSLength - 1; 
      
    //System.out.print("start: %d end: %d\n",  
    //                  start, end); 
    System.out.print("LPS of string is " +  
                      text + " : "); 
      
    for(i = start; i <= end; i++) 
        System.out.print(text.charAt(i)); 
    System.out.println(); 
} 
  
// Driver code 
public static void main(String []args) 
{ 
    String text1="babcbabcbaccba"; 
    findLongestPalindromicString(text1); 
  
    String text2="abaaba"; 
    findLongestPalindromicString(text2); 
  
    String text3= "abababa"; 
    findLongestPalindromicString(text3); 
  
    String text4="abcbabcbabcba"; 
    findLongestPalindromicString(text4); 
  
    String text5="forgeeksskeegfor"; 
    findLongestPalindromicString(text5); 
  
    String text6="caba"; 
    findLongestPalindromicString(text6); 
  
    String text7="abacdfgdcaba"; 
    findLongestPalindromicString(text7); 
  
    String text8="abacdfgdcabba"; 
    findLongestPalindromicString(text8); 
  
    String text9="abacdedcaba"; 
    findLongestPalindromicString(text9); 
} 
} 
  
// This code is contributed by SoumikMondal 

Python3

# Python program to implement Manacher's Algorithm 
   
def findLongestPalindromicString(text): 
    N = len(text) 
    if N == 0: 
        return
    N = 2*N+1    # Position count 
    L = [0] * N 
    L[0] = 0
    L[1] = 1
    C = 1     # centerPosition 
    R = 2     # centerRightPosition 
    i = 0    # currentRightPosition 
    iMirror = 0     # currentLeftPosition 
    maxLPSLength = 0
    maxLPSCenterPosition = 0
    start = -1
    end = -1
    diff = -1
   
    # Uncomment it to print LPS Length array 
    # printf("%d %d ", L[0], L[1]); 
    for i in range(2,N): 
       
        # get currentLeftPosition iMirror for currentRightPosition i 
        iMirror = 2*C-i 
        L[i] = 0
        diff = R - i 
        # If currentRightPosition i is within centerRightPosition R 
        if diff > 0: 
            L[i] = min(L[iMirror], diff) 
   
        # Attempt to expand palindrome centered at currentRightPosition i 
        # Here for odd positions, we compare characters and 
        # if match then increment LPS Length by ONE 
        # If even position, we just increment LPS by ONE without 
        # any character comparison 
        try: 
            while ((i+L[i]) < N and (i-L[i]) > 0) and \ 
                    (((i+L[i]+1) % 2 == 0) or \ 
                    (text[(i+L[i]+1)//2] == text[(i-L[i]-1)//2])): 
                L[i]+=1
        except Exception as e: 
            pass
   
        if L[i] > maxLPSLength:        # Track maxLPSLength 
            maxLPSLength = L[i] 
            maxLPSCenterPosition = i 
   
        # If palindrome centered at currentRightPosition i 
        # expand beyond centerRightPosition R, 
        # adjust centerPosition C based on expanded palindrome. 
        if i + L[i] > R: 
            C = i 
            R = i + L[i] 
   
    # Uncomment it to print LPS Length array 
    # printf("%d ", L[i]); 
    start = (maxLPSCenterPosition - maxLPSLength) // 2
    end = start + maxLPSLength - 1
    print ("LPS of string is " + text + " : ",text[start:end+1]) 
   
# Driver program 
text1 = "babcbabcbaccba"
findLongestPalindromicString(text1) 
   
text2 = "abaaba"
findLongestPalindromicString(text2) 
   
text3 = "abababa"
findLongestPalindromicString(text3) 
   
text4 = "abcbabcbabcba"
findLongestPalindromicString(text4) 
   
text5 = "forgeeksskeegfor"
findLongestPalindromicString(text5) 
   
text6 = "caba"
findLongestPalindromicString(text6) 
   
text7 = "abacdfgdcaba"
findLongestPalindromicString(text7) 
   
text8 = "abacdfgdcabba"
findLongestPalindromicString(text8) 
   
text9 = "abacdedcaba"
findLongestPalindromicString(text9) 
   
# This code is contributed by BHAVYA JAIN 

C#

// A C# program to implement Manacher’s Algorithm 
using System; 
using System.Diagnostics; 
public class GFG 
{ 
  
public static void findLongestPalindromicString( 
                   String text) 
{ 
    int N = text.Length; 
    if(N == 0) 
        return; 
      
    // Position count 
    N = 2 * N + 1;  
      
    // LPS Length Array 
    int []L = new int [N];  
    L[0] = 0; 
    L[1] = 1; 
      
    // centerPosition  
    int C = 1; 
      
    // centerRightPosition 
    int R = 2;  
      
    // currentRightPosition 
    int i = 0;  
      
    // currentLeftPosition 
    int iMirror;  
    int expand = -1; 
    int diff = -1; 
    int maxLPSLength = 0; 
    int maxLPSCenterPosition = 0; 
    int start = -1; 
    int end = -1; 
      
    // Uncomment it to print LPS Length array 
    // printf("%d %d ", L[0], L[1]); 
    for (i = 2; i < N; i++)  
    { 
        // Get currentLeftPosition iMirror 
        // for currentRightPosition i 
        iMirror = 2 * C - i; 
          
        // Reset expand - means no  
        // expansion required 
        expand = 0; 
        diff = R - i; 
          
        // If currentRightPosition i is  
        // within centerRightPosition R 
        if(diff >= 0)  
        { 
              
            // Case 1 
            if(L[iMirror] < diff)  
                L[i] = L[iMirror]; 
                  
            // Case 2 
            else if(L[iMirror] == diff && R == N - 1)  
                L[i] = L[iMirror]; 
              
            // Case 3 
            else if(L[iMirror] == diff && R < N - 1)  
            { 
                    L[i] = L[iMirror]; 
                      
                     // Expansion required 
                    expand = 1; 
            } 
              
            // Case 4 
            else if(L[iMirror] > diff)  
            { 
                L[i] = diff; 
                  
                // Expansion required 
                expand = 1;  
            } 
        } 
        else
        { 
            L[i] = 0; 
              
            // Expansion required 
            expand = 1;  
        } 
          
        if (expand == 1) 
        { 
              
            // Attempt to expand palindrome centered 
            // at currentRightPosition i. Here for odd 
            // positions, we compare characters and  
            // if match then increment LPS Length by ONE 
            // If even position, we just increment LPS 
            // by ONE without any character comparison 
            try
            { 
                while (((i + L[i]) < N &&  
                        (i - L[i]) > 0) && 
                      (((i + L[i] + 1) % 2 == 0) || 
                       (text[(i + L[i] + 1) / 2] ==  
                        text[(i - L[i] - 1) / 2]))) 
                { 
                    L[i]++; 
                } 
            } 
            catch (Exception) 
            { 
                Debug.Assert(true); 
            } 
        } 
          
        // Track maxLPSLength 
        if(L[i] > maxLPSLength)  
        { 
            maxLPSLength = L[i]; 
            maxLPSCenterPosition = i; 
        } 
  
        // If palindrome centered at  
        // currentRightPosition i expand 
        // beyond centerRightPosition R, 
        // adjust centerPosition C based 
        // on expanded palindrome. 
        if (i + L[i] > R)  
        { 
            C = i; 
            R = i + L[i]; 
        } 
          
        //Uncomment it to print LPS Length array 
        //System.out.print("%d ", L[i]); 
    } 
      
    start = (maxLPSCenterPosition -  
             maxLPSLength) / 2; 
    end = start + maxLPSLength - 1; 
      
    //System.out.print("start: %d end: %d\n",  
    //                  start, end); 
    Console.Write("LPS of string is " +  
                      text + " : "); 
      
    for(i = start; i <= end; i++) 
        Console.Write(text[i]); 
    Console.WriteLine(); 
} 
  
// Driver code 
static public void Main () 
{ 
  
    String text1 = "babcbabcbaccba"; 
    findLongestPalindromicString(text1); 
  
    String text2 = "abaaba"; 
    findLongestPalindromicString(text2); 
  
    String text3 = "abababa"; 
    findLongestPalindromicString(text3); 
  
    String text4 = "abcbabcbabcba"; 
    findLongestPalindromicString(text4); 
  
    String text5 = "forgeeksskeegfor"; 
    findLongestPalindromicString(text5); 
  
    String text6 = "caba"; 
    findLongestPalindromicString(text6); 
  
    String text7 = "abacdfgdcaba"; 
    findLongestPalindromicString(text7); 
  
    String text8 = "abacdfgdcabba"; 
    findLongestPalindromicString(text8); 
  
    String text9 = "abacdedcaba"; 
    findLongestPalindromicString(text9); 
} 
} 
  
// This code is contributed by Dharanendra L V. 

Javascript

<script> 
// A Javascript program to implement Manacher’s Algorithm 
  
function findLongestPalindromicString(text) 
{ 
    let N = text.length; 
    if(N == 0) 
        return; 
       
    // Position count 
    N = 2 * N + 1; 
       
    // LPS Length Array 
    let L = new Array(N); 
    L[0] = 0; 
    L[1] = 1; 
       
    // centerPosition 
    let C = 1; 
       
    // centerRightPosition 
    let R = 2; 
       
    // currentRightPosition 
    let i = 0; 
       
    // currentLeftPosition 
    let iMirror; 
    let expand = -1; 
    let diff = -1; 
    let maxLPSLength = 0; 
    let maxLPSCenterPosition = 0; 
    let start = -1; 
    let end = -1; 
       
    // Uncomment it to print LPS Length array 
    // printf("%d %d ", L[0], L[1]); 
    for (i = 2; i < N; i++) 
    { 
        // Get currentLeftPosition iMirror 
        // for currentRightPosition i 
        iMirror = 2 * C - i; 
           
        // Reset expand - means no 
        // expansion required 
        expand = 0; 
        diff = R - i; 
           
        // If currentRightPosition i is 
        // within centerRightPosition R 
        if(diff >= 0) 
        { 
               
            // Case 1 
            if(L[iMirror] < diff) 
                L[i] = L[iMirror]; 
                   
            // Case 2 
            else if(L[iMirror] == diff && R == N - 1) 
                L[i] = L[iMirror]; 
               
            // Case 3 
            else if(L[iMirror] == diff && R < N - 1) 
            { 
                    L[i] = L[iMirror]; 
                       
                     // Expansion required 
                    expand = 1; 
            } 
               
            // Case 4 
            else if(L[iMirror] > diff) 
            { 
                L[i] = diff; 
                   
                // Expansion required 
                expand = 1; 
            } 
        } 
        else
        { 
            L[i] = 0; 
               
            // Expansion required 
            expand = 1; 
        } 
           
        if (expand == 1) 
        { 
               
            // Attempt to expand palindrome centered 
            // at currentRightPosition i. Here for odd 
            // positions, we compare characters and 
            // if match then increment LPS Length by ONE 
            // If even position, we just increment LPS 
            // by ONE without any character comparison 
              
               let flr1 = Math.floor((i + L[i] + 1) / 2));            
               let flr2 = Math.floor((i + L[i] - 1) / 2)); 
  
                while (((i + L[i]) < N && 
                        (i - L[i]) > 0) && 
                      (((i + L[i] + 1) % 2 == 0) || 
                       (text[flr] == 
                        text[flr]))) 
                { 
                    L[i]++; 
                } 
              
              
        } 
           
        // Track maxLPSLength 
        if(L[i] > maxLPSLength) 
        { 
            maxLPSLength = L[i]; 
            maxLPSCenterPosition = i; 
        } 
   
        // If palindrome centered at 
        // currentRightPosition i expand 
        // beyond centerRightPosition R, 
        // adjust centerPosition C based 
        // on expanded palindrome. 
        if (i + L[i] > R) 
        { 
            C = i; 
            R = i + L[i]; 
        } 
           
        //Uncomment it to print LPS Length array 
        //System.out.print("%d ", L[i]); 
    } 
       
    start = (maxLPSCenterPosition - 
             maxLPSLength) / 2; 
    end = start + maxLPSLength - 1; 
       
    //System.out.print("start: %d end: %d\n", 
    //                  start, end); 
    document.write("LPS of string is " + 
                      text + " : "); 
       
    for(i = start; i <= end; i++) 
        document.write(text[i]); 
    document.write("<br>"); 
} 
  
// Driver code 
  
let text1="babcbabcbaccba"; 
findLongestPalindromicString(text1); 
  
let text2="abaaba"; 
findLongestPalindromicString(text2); 
  
let text3= "abababa"; 
findLongestPalindromicString(text3); 
  
let text4="abcbabcbabcba"; 
findLongestPalindromicString(text4); 
  
let text5="forgeeksskeegfor"; 
findLongestPalindromicString(text5); 
  
let text6="caba"; 
findLongestPalindromicString(text6); 
  
let text7="abacdfgdcaba"; 
findLongestPalindromicString(text7); 
  
let text8="abacdfgdcabba"; 
findLongestPalindromicString(text8); 
  
let text9="abacdedcaba"; 
findLongestPalindromicString(text9); 
  
  
// This code is contributed by unknown2108 
</script> 

Output

LPS of string is babcbabcbaccba : abcbabcba
LPS of string is abaaba : abaaba
LPS of string is abababa : abababa
LPS of string is abcbabcbabcba : abcbabcbabcba
LPS of string is forgeeksskeegfor : geeksskeeg
LPS of string is caba : aba
LPS of string is abacdfgdcaba : aba
LPS of string is abacdfgdcabba : abba
LPS of string is abacdedcaba : abacdedcaba

Time Complexity: O(N)
Auxiliary Space: O(N)

This is the implementation based on the four cases discussed in Part 2. In Part 4, we have discussed a different way to look at these four cases and few other approaches.

Print the longest palindromic prefix of a given string

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Manacher’s Algorithm – Linear Time Longest Palindromic Substring – Part 3

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