Manacher's Algorithm - Linear Time Longest Palindromic Substring - Part 4
Last Updated :
23 Jul, 2025
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. In Part 3, we implemented the same.
Here we will review the four cases again and try to see it differently and implement the same.
All four cases depends on LPS length value at currentLeftPosition (L[iMirror]) and value of (centerRightPosition - currentRightPosition), i.e. (R - i). These two information are known before which helps us to reuse previous available information and avoid unnecessary character comparison.
If we look at all four cases, we will see that we 1st set minimum of L[iMirror] and R-i to L[i] and then we try to expand the palindrome in whichever case it can expand.
Above observation may look more intuitive, easier to understand and implement, given that one understands LPS length array, position, index, symmetry property etc.
Implementation:
C++
// A C program to implement Manacher’s Algorithm
#include <stdio.h>
#include <string.h>
char text[100];
int min(int a, int b)
{
int res = a;
if(b < a)
res = b;
return res;
}
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 maxLPSLength = 0;
int maxLPSCenterPosition = 0;
int start = -1;
int end = -1;
int diff = -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;
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
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("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 program to implement Manacher's Algorithm
import java.util.*;
class GFG
{
static void findLongestPalindromicString(String text)
{
int N = text.length();
if (N == 0)
return;
N = 2 * N + 1; // Position count
int[] L = new int[N + 1]; // 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 maxLPSLength = 0;
int maxLPSCenterPosition = 0;
int start = -1;
int end = -1;
int diff = -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;
L[i] = 0;
diff = R - i;
// If currentRightPosition i is within
// centerRightPosition R
if (diff > 0)
L[i] = Math.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
while (((i + L[i]) + 1 < 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]++;
}
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;
System.out.printf("LPS of string is %s : ", 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 text = "babcbabcbaccba";
findLongestPalindromicString(text);
text = "abaaba";
findLongestPalindromicString(text);
text = "abababa";
findLongestPalindromicString(text);
text = "abcbabcbabcba";
findLongestPalindromicString(text);
text = "forgeeksskeegfor";
findLongestPalindromicString(text);
text = "caba";
findLongestPalindromicString(text);
text = "abacdfgdcaba";
findLongestPalindromicString(text);
text = "abacdfgdcabba";
findLongestPalindromicString(text);
text = "abacdedcaba";
findLongestPalindromicString(text);
}
}
// This code is contributed by
// sanjeev2552
# 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# program to implement Manacher's Algorithm
using System;
class GFG
{
static void findLongestPalindromicString(String text)
{
int N = text.Length;
if (N == 0)
return;
N = 2 * N + 1; // Position count
int[] L = new int[N + 1]; // 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 maxLPSLength = 0;
int maxLPSCenterPosition = 0;
int start = -1;
int end = -1;
int diff = -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;
L[i] = 0;
diff = R - i;
// If currentRightPosition i is within
// centerRightPosition R
if (diff > 0)
L[i] = Math.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
while (((i + L[i]) + 1 < 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]);
}
start = (maxLPSCenterPosition - maxLPSLength) / 2;
end = start + maxLPSLength - 1;
Console.Write("LPS of string is {0} : ", text);
for (i = start; i <= end; i++)
Console.Write(text[i]);
Console.WriteLine();
}
// Driver Code
public static void Main(String[] args)
{
String text = "babcbabcbaccba";
findLongestPalindromicString(text);
text = "abaaba";
findLongestPalindromicString(text);
text = "abababa";
findLongestPalindromicString(text);
text = "abcbabcbabcba";
findLongestPalindromicString(text);
text = "forgeeksskeegfor";
findLongestPalindromicString(text);
text = "caba";
findLongestPalindromicString(text);
text = "abacdfgdcaba";
findLongestPalindromicString(text);
text = "abacdfgdcabba";
findLongestPalindromicString(text);
text = "abacdedcaba";
findLongestPalindromicString(text);
}
}
// This code is contributed by PrinciRaj1992
<script>
// Javascript program to implement Manacher's Algorithm
function findLongestPalindromicString(text)
{
let N = text.length;
if (N == 0)
return;
N = 2 * N + 1; // Position count
let L = new Array(N + 1); // LPS Length Array
L[0] = 0;
L[1] = 1;
let C = 1; // centerPosition
let R = 2; // centerRightPosition
let i = 0; // currentRightPosition
let iMirror; // currentLeftPosition
let maxLPSLength = 0;
let maxLPSCenterPosition = 0;
let start = -1;
let end = -1;
let diff = -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;
L[i] = 0;
diff = R - i;
// If currentRightPosition i is within
// centerRightPosition R
if (diff > 0)
L[i] = Math.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
let flr1 = Math.floor((i + L[i] + 1) / 2));
let flr2 = Math.floor((i + L[i] - 1) / 2));
while (((i + L[i]) + 1 < N && (i - L[i]) > 0) &&
(((i + L[i] + 1) % 2 == 0) ||
(text[flr1] ==
text[flr2])))
{
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]);
}
start = (maxLPSCenterPosition - maxLPSLength) / 2;
end = start + maxLPSLength - 1;
document.write("LPS of string is "+text+" : ");
for (i = start; i <= end; i++)
document.write(text[i]);
document.write("<br>");
}
// Driver Code
let text = "babcbabcbaccba";
findLongestPalindromicString(text);
text = "abaaba";
findLongestPalindromicString(text);
text = "abababa";
findLongestPalindromicString(text);
text = "abcbabcbabcba";
findLongestPalindromicString(text);
text = "forgeeksskeegfor";
findLongestPalindromicString(text);
text = "caba";
findLongestPalindromicString(text);
text = "abacdfgdcaba";
findLongestPalindromicString(text);
text = "abacdfgdcabba";
findLongestPalindromicString(text);
text = "abacdedcaba";
findLongestPalindromicString(text);
// 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)
Other Approaches:
We have discussed two approaches here. One in Part 3 and other in current article. In both approaches, we worked on given string. Here we had to handle even and odd positions differently while comparing characters for expansion (because even positions do not represent any character in string).
To avoid this different handling of even and odd positions, we need to make even positions also to represent some character (actually all even positions should represent SAME character because they MUST match while character comparison). One way to do this is to set some character at all even positions by modifying given string or create a new copy of given string. For example, if input string is "abcb", new string should be "#a#b#c#b#" if we add # as unique character at even positions.
The two approaches discussed already can be modified a bit to work on modified string where different handling of even and odd positions will not be needed.
We may also add two DIFFERENT characters (not yet used anywhere in string at even and odd positions) at start and end of string as sentinels to avoid bound check. With these changes string "abcb" will look like "^#a#b#c#b#$" where ^ and $ are sentinels.
This implementation may look cleaner with the cost of more memory.
We are not implementing these here as it's a simple change in given implementations.
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem