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,

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).
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>
OutputLPS 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.
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Print the longest palindromic prefix of a given string
Given a string str, the task is to find the longest palindromic prefix of the given string. Examples: Input: str = "abaac" Output: aba Explanation: The longest prefix of the given string which is palindromic is "aba". Input: str = "abacabaxyz" Output: abacaba Explanation: The prefixes of the given s
12 min read
Longest Palindromic Substring using Palindromic Tree | Set 3
Given a string, find the longest substring which is a palindrome. For example, if the given string is âforgeeksskeegforâ, the output should be âgeeksskeegâ. Prerequisite : Palindromic Tree | Longest Palindromic Substring Structure of Palindromic Tree : The palindromic Treeâs actual structure is clos
15+ min read
Largest palindromic number by permuting digits
Given a very large integer n in the form of string, the task is to return the largest palindromic number obtainable by permuting the digits of n. If it is not possible to make a palindromic number, then return an empty string.Examples : Input : "313551"Output : "531135"Explanations : 531135 is the l
10 min read
Length of Longest Palindrome Substring
Given a string S of length N, the task is to find the length of the longest palindromic substring from a given string. Examples: Input: S = "abcbab"Output: 5Explanation: string "abcba" is the longest substring that is a palindrome which is of length 5. Input: S = "abcdaa"Output: 2Explanation: string
15+ min read
Longest palindromic string possible after removal of a substring
Given a string str, the task is to find the longest palindromic string that can be obtained from it after removing a substring. Examples: Input: str = "abcdefghiedcba" Output: "abcdeiedcba" Explanation: Removal of substring "fgh" leaves the remaining string palindromic Input: str = "abba" Output: "a
11 min read
Rearrange string to obtain Longest Palindromic Substring
Given string str, the task is to rearrange the given string to obtain the longest palindromic substring. Examples: Input: str = âgeeksforgeeksâOutput: eegksfskgeeorExplanation: eegksfskgee is the longest palindromic substring after rearranging the string.Therefore, the required output is eegksfskgee
9 min read