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

• Difficulty Level : Expert
• Last Updated : 14 Jan, 2022

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, 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 1st 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).

C

 // A C program to implement Manacher’s Algorithm#include #include  char text;void findLongestPalindromicString(){    int N = strlen(text);    if(N == 0)        return;    N = 2*N + 1; //Position count    int L[N]; //LPS Length Array    L = 0;    L = 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, L);    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;}

Java

 // A Java program to implement Manacher’s Algorithmimport 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;    L = 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, L);    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 codepublic 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 =  * N    L = 0    L = 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, L);    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 programtext1 = "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 Algorithmusing 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;    L = 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, L);    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 codestatic 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



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

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.