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: 5
Explanation:
string “abcba” is the longest substring that is a palindrome which is of length 5.Input: S = “abcdaa”
Output: 2
Explanation:
string “aa” is the longest substring that is a palindrome which is of length 2.
Naive Approach: The simplest approach to solve the problem is to generate all possible substrings of the given string and print the length of the longest substring which is a palindrome.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to obtain the length of // the longest palindromic substring int longestPalSubstr(string str)
{ // Length of given string
int n = str.size();
// Stores the maximum length
int maxLength = 1, start = 0;
// Iterate over the string
for ( int i = 0;
i < str.length(); i++) {
// Iterate over the string
for ( int j = i;
j < str.length(); j++) {
int flag = 1;
// Check for palindrome
for ( int k = 0;
k < (j - i + 1) / 2; k++)
if (str[i + k]
!= str[j - k])
flag = 0;
// If string [i, j - i + 1]
// is palindromic
if (flag
&& (j - i + 1) > maxLength) {
start = i;
maxLength = j - i + 1;
}
}
}
// Return length of LPS
return maxLength;
} // Driver Code int main()
{ // Given string
string str = "forgeeksskeegfor" ;
// Function Call
cout << longestPalSubstr(str);
return 0;
} |
// Java program for the above approach import java.io.*;
class GFG{
// Function to obtain the length of // the longest palindromic substring static int longestPalSubstr(String str)
{ // Length of given string
int n = str.length();
// Stores the maximum length
int maxLength = 1 , start = 0 ;
// Iterate over the string
for ( int i = 0 ; i < str.length(); i++)
{
// Iterate over the string
for ( int j = i; j < str.length(); j++)
{
int flag = 1 ;
// Check for palindrome
for ( int k = 0 ;
k < (j - i + 1 ) / 2 ; k++)
if (str.charAt(i + k) !=
str.charAt(j - k))
flag = 0 ;
// If string [i, j - i + 1]
// is palindromic
if (flag != 0 &&
(j - i + 1 ) > maxLength)
{
start = i;
maxLength = j - i + 1 ;
}
}
}
// Return length of LPS
return maxLength;
} // Driver Code public static void main (String[] args)
{ // Given string
String str = "forgeeksskeegfor" ;
// Function call
System.out.print(longestPalSubstr(str));
} } // This code is contributed by code_hunt |
# Python3 program for the above approach # Function to obtain the length of # the longest palindromic substring def longestPalSubstr( str ):
# Length of given string
n = len ( str )
# Stores the maximum length
maxLength = 1
start = 0
# Iterate over the string
for i in range ( len ( str )):
# Iterate over the string
for j in range (i, len ( str ), 1 ):
flag = 1
# Check for palindrome
for k in range ((j - i + 1 ) / / 2 ):
if ( str [i + k] ! = str [j - k]):
flag = 0
# If string [i, j - i + 1]
# is palindromic
if (flag ! = 0 and
(j - i + 1 ) > maxLength):
start = i
maxLength = j - i + 1
# Return length of LPS
return maxLength
# Driver Code # Given string str = "forgeeksskeegfor"
# Function call print (longestPalSubstr( str ))
# This code is contributed by code_hunt |
// C# program for the above approach using System;
class GFG{
// Function to obtain the length of // the longest palindromic substring static int longestPalSubstr( string str)
{ // Length of given string
int n = str.Length;
// Stores the maximum length
int maxLength = 1, start = 0;
// Iterate over the string
for ( int i = 0; i < str.Length; i++)
{
// Iterate over the string
for ( int j = i; j < str.Length; j++)
{
int flag = 1;
// Check for palindrome
for ( int k = 0;
k < (j - i + 1) / 2; k++)
if (str[i + k] != str[j - k])
flag = 0;
// If string [i, j - i + 1]
// is palindromic
if (flag != 0 &&
(j - i + 1) > maxLength)
{
start = i;
maxLength = j - i + 1;
}
}
}
// Return length of LPS
return maxLength;
} // Driver Code public static void Main ()
{ // Given string
string str = "forgeeksskeegfor" ;
// Function call
Console.Write(longestPalSubstr(str));
} } // This code is contributed by code_hunt |
<script> // JavaScript program for the above approach // Function to obtain the length of // the longest palindromic substring function longestPalSubstr(str)
{ // Length of given string
var n = str.length;
// Stores the maximum length
var maxLength = 1, start = 0;
// Iterate over the string
for ( var i = 0;
i < str.length; i++) {
// Iterate over the string
for ( var j = i;
j < str.length; j++) {
var flag = 1;
// Check for palindrome
for ( var k = 0;
k < (j - i + 1) / 2; k++)
if (str[i + k]
!= str[j - k])
flag = 0;
// If string [i, j - i + 1]
// is palindromic
if (flag
&& (j - i + 1) > maxLength) {
start = i;
maxLength = j - i + 1;
}
}
}
// Return length of LPS
return maxLength;
} // Driver Code // Given string var str = "forgeeksskeegfor" ;
// Function Call document.write( longestPalSubstr(str)); </script> |
10
Time Complexity: O(N3), where N is the length of the given string.
Auxiliary Space: O(N)
Dynamic Programming Approach: The above approach can be optimized by storing results of Overlapping Subproblems. The idea is similar to this post. Below are the steps:
- Maintain a boolean table[N][N] that is filled in a bottom-up manner.
- The value of table[i][j] is true if the substring is a palindrome, otherwise false.
- To calculate table[i][j], check the value of table[i + 1][j – 1], if the value is true and str[i] is same as str[j], then update table[i][j] true.
- Otherwise, the value of table[i][j] is update as false.
Below is the illustration for the string “geeks”:
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the length of // the longest palindromic substring int longestPalSubstr(string str)
{ // Length of string str
int n = str.size();
// Stores the dp states
bool table[n][n];
// Initialise table[][] as false
memset (table, 0, sizeof (table));
// All substrings of length 1
// are palindromes
int maxLength = 1;
for ( int i = 0; i < n; ++i)
table[i][i] = true ;
// Check for sub-string of length 2
int start = 0;
for ( int i = 0; i < n - 1; ++i) {
// If adjacent character are same
if (str[i] == str[i + 1]) {
// Update table[i][i + 1]
table[i][i + 1] = true ;
start = i;
maxLength = 2;
}
}
// Check for lengths greater than 2
// k is length of substring
for ( int k = 3; k <= n; ++k) {
// Fix the starting index
for ( int i = 0; i < n - k + 1; ++i) {
// Ending index of substring
// of length k
int j = i + k - 1;
// Check for palindromic
// substring str[i, j]
if (table[i + 1][j - 1]
&& str[i] == str[j]) {
// Mark true
table[i][j] = true ;
// Update the maximum length
if (k > maxLength) {
start = i;
maxLength = k;
}
}
}
}
// Return length of LPS
return maxLength;
} // Driver Code int main()
{ // Given string str
string str = "forgeeksskeegfor" ;
// Function Call
cout << longestPalSubstr(str);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to find the length of // the longest palindromic subString static int longestPalSubstr(String str)
{ // Length of String str
int n = str.length();
// Stores the dp states
boolean [][]table = new boolean [n][n];
// All subStrings of length 1
// are palindromes
int maxLength = 1 ;
for ( int i = 0 ; i < n; ++i)
table[i][i] = true ;
// Check for sub-String of length 2
int start = 0 ;
for ( int i = 0 ; i < n - 1 ; ++i)
{
// If adjacent character are same
if (str.charAt(i) == str.charAt(i + 1 ))
{
// Update table[i][i + 1]
table[i][i + 1 ] = true ;
start = i;
maxLength = 2 ;
}
}
// Check for lengths greater than 2
// k is length of subString
for ( int k = 3 ; k <= n; ++k)
{
// Fix the starting index
for ( int i = 0 ; i < n - k + 1 ; ++i)
{
// Ending index of subString
// of length k
int j = i + k - 1 ;
// Check for palindromic
// subString str[i, j]
if (table[i + 1 ][j - 1 ] &&
str.charAt(i) == str.charAt(j))
{
// Mark true
table[i][j] = true ;
// Update the maximum length
if (k > maxLength)
{
start = i;
maxLength = k;
}
}
}
}
// Return length of LPS
return maxLength;
} // Driver Code public static void main(String[] args)
{ // Given String str
String str = "forgeeksskeegfor" ;
// Function Call
System.out.print(longestPalSubstr(str));
} } // This code is contributed by Amit Katiyar |
// C# program for // the above approach using System;
class GFG{
// Function to find the length of // the longest palindromic subString static int longestPalSubstr(String str)
{ // Length of String str
int n = str.Length;
// Stores the dp states
bool [,]table = new bool [n, n];
// All subStrings of length 1
// are palindromes
int maxLength = 1;
for ( int i = 0; i < n; ++i)
table[i, i] = true ;
// Check for sub-String
// of length 2
int start = 0;
for ( int i = 0; i < n - 1; ++i)
{
// If adjacent character are same
if (str[i] == str[i + 1])
{
// Update table[i,i + 1]
table[i, i + 1] = true ;
start = i;
maxLength = 2;
}
}
// Check for lengths greater than 2
// k is length of subString
for ( int k = 3; k <= n; ++k)
{
// Fix the starting index
for ( int i = 0; i < n - k + 1; ++i)
{
// Ending index of subString
// of length k
int j = i + k - 1;
// Check for palindromic
// subString str[i, j]
if (table[i + 1, j - 1] &&
str[i] == str[j])
{
// Mark true
table[i, j] = true ;
// Update the maximum length
if (k > maxLength)
{
start = i;
maxLength = k;
}
}
}
}
// Return length of LPS
return maxLength;
} // Driver Code public static void Main(String[] args)
{ // Given String str
String str = "forgeeksskeegfor" ;
// Function Call
Console.Write(longestPalSubstr(str));
} } // This code is contributed by Rajput-Ji |
# Python program for the above approach # Function to find the length of # the longest palindromic subString def longestPalSubstr( str ):
# Length of String str
n = len ( str );
# Stores the dp states
table = [[ False for i in range (n)] for j in range (n)];
# All subStrings of length 1
# are palindromes
maxLength = 1 ;
for i in range (n):
table[i][i] = True ;
# Check for sub-String of length 2
start = 0 ;
for i in range (n - 1 ):
# If adjacent character are same
if ( str [i] = = str [i + 1 ]):
# Update table[i][i + 1]
table[i][i + 1 ] = True ;
start = i;
maxLength = 2 ;
# Check for lengths greater than 2
# k is length of subString
for k in range ( 3 , n + 1 ):
# Fix the starting index
for i in range (n - k + 1 ):
# Ending index of subString
# of length k
j = i + k - 1 ;
# Check for palindromic
# subString str[i, j]
if (table[i + 1 ][j - 1 ] and str [i] = = str [j]):
# Mark True
table[i][j] = True ;
# Update the maximum length
if (k > maxLength):
start = i;
maxLength = k;
# Return length of LPS
return maxLength;
# Driver Code if __name__ = = '__main__' :
# Given String str
str = "forgeeksskeegfor" ;
# Function Call
print (longestPalSubstr( str ));
# This code is contributed by 29AjayKumar |
<script> // javascript program for the above approach // Function to find the length of
// the longest palindromic subString
function longestPalSubstr(str) {
// Length of String str
var n = str.length;
// Stores the dp states
var table = Array(n).fill().map(()=>Array(n).fill( false ));
// All subStrings of length 1
// are palindromes
var maxLength = 1;
for ( var i = 0; i < n; ++i)
table[i][i] = true ;
// Check for sub-String of length 2
var start = 0;
for (i = 0; i < n - 1; ++i) {
// If adjacent character are same
if (str.charAt(i) == str.charAt(i + 1)) {
// Update table[i][i + 1]
table[i][i + 1] = true ;
start = i;
maxLength = 2;
}
}
// Check for lengths greater than 2
// k is length of subString
for (k = 3; k <= n; ++k) {
// Fix the starting index
for (i = 0; i < n - k + 1; ++i) {
// Ending index of subString
// of length k
var j = i + k - 1;
// Check for palindromic
// subString str[i, j]
if (table[i + 1][j - 1] && str.charAt(i) == str.charAt(j)) {
// Mark true
table[i][j] = true ;
// Update the maximum length
if (k > maxLength) {
start = i;
maxLength = k;
}
}
}
}
// Return length of LPS
return maxLength;
}
// Driver Code
// Given String str
var str = "forgeeksskeegfor" ;
// Function Call
document.write(longestPalSubstr(str));
// This code is contributed by umadevi9616 </script> |
10
Time Complexity: O(N2), where N is the length of the given string.
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to use Manacher’s Algorithm. By using this algorithm, for each character c, the longest palindromic substring that has c as its center can be found whose length is odd. But the longest palindromic substring can also have an even length which does not have any center. Therefore, some special characters can be added between each character.
For example, if the given string is “abababc” then it will become “$#a#b#a#b#a#b#c#@”. Now, notice that in this case, for each character c, the longest palindromic substring with the center c will have an odd length.
Below are the steps:
- Add the special characters in the given string S as explained above and let its length be N.
- Initialize an array d[], center, and r with 0 where d[i] stores the length of the left part of the palindrome where S[i] is the center, r denotes the rightmost visited boundary and center denotes the current index of character which is the center of this rightmost boundary.
- While traversing the string S, for each index i, if i is smaller than r then its answer has previously been calculated and d[i] can be set equals to answer for the mirror of character at i with the center which can be calculated as (2*center – i).
- Now, check if there are some characters after r such that the palindrome becomes ever longer.
- If (i + d[i]) is greater than r, update r = (i + d[i]) and center as i.
- After finding the longest palindrome for every character c as the center, print the maximum value of (2*d[i] + 1)/2 where 0 ? i < N because d[i] only stores the left part of the palindrome.
Below is the implementation for the above approach:
// C++ program for the above approach: #include <bits/stdc++.h> using namespace std;
// Function that placed '#' intermediately // before and after each character string UpdatedString(string s){ string newString = "#" ;
// Traverse the string
for ( auto ch : s){
newString += ch;
newString += "#" ;
}
// Return the string
return newString;
} // Function that finds the length of // the longest palindromic substring int Manacher(string s){
// Update the string
s = UpdatedString(s);
// Stores the longest proper prefix
// which is also a suffix
int LPS[s.length()] = {};
int C = 0;
int R = 0;
for ( int i = 0 ; i < s.length() ; i++){
int imir = 2 * C - i;
// Find the minimum length of
// the palindrome
if (R > i){
LPS[i] = min(R-i, LPS[imir]);
}
else {
// Find the actual length of
// the palindrome
LPS[i] = 0;
}
// Exception Handling
while ( ((i + 1 + LPS[i]) < s.length()) && ((i - 1 - LPS[i]) >= 0) && s[i + 1 + LPS[i]] == s[i - 1 - LPS[i]]){
LPS[i] += 1;
}
// Update C and R
if (i + LPS[i] > R){
C = i;
R = i + LPS[i];
}
}
int r = 0, c = -1;
for ( int i = 0 ; i < s.length() ; i++){
r = max(r, LPS[i]);
if (r == LPS[i]){
c = i;
}
}
// Return the length r
return r;
} // Driver code int main()
{ // Given string str
string str = "forgeeksskeegfor" ;
// Function Call
cout << Manacher(str) << endl;
} // This code is contributed by subhamgoyal2014. |
// Java code for the above approach import java.util.Arrays;
class GFG {
// Function that placed '#' intermediately
// before and after each character
static String UpdatedString(String s) {
String newString = "#" ;
// Traverse the string
for ( char ch : s.toCharArray()) {
newString += ch;
newString += "#" ;
}
// Return the string
return newString;
}
// Function that finds the length of
// the longest palindromic substring
static int Manacher(String s) {
// Update the string
s = UpdatedString(s);
// Stores the longest proper prefix
// which is also a suffix
int [] LPS = new int [s.length()];
int C = 0 ;
int R = 0 ;
for ( int i = 0 ; i < s.length(); i++) {
int imir = 2 * C - i;
// Find the minimum length of
// the palindrome
if (R > i) {
LPS[i] = Math.min(R - i, LPS[imir]);
} else {
// Find the actual length of
// the palindrome
LPS[i] = 0 ;
}
// Exception Handling
while (((i + 1 + LPS[i]) < s.length()) && ((i - 1 - LPS[i]) >= 0 ) && (s.charAt(i + 1 + LPS[i]) == s.charAt(i - 1 - LPS[i]))) {
LPS[i] += 1 ;
}
// Update C and R
if (i + LPS[i] > R) {
C = i;
R = i + LPS[i];
}
}
int r = 0 ;
for ( int i = 0 ; i < s.length(); i++) {
r = Math.max(r, LPS[i]);
}
// Return the length r
return r;
}
// Driver code
public static void main(String[] args) {
// Given string str
String str = "forgeeksskeegfor" ;
// Function Call
System.out.println(Manacher(str));
}
} // This code is contributed by lokeshpotta20. |
# Python program for the above approach # Function that placed '#' intermediately # before and after each character def UpdatedString(string):
newString = [ '#' ]
# Traverse the string for char in string:
newString + = [char, '#' ]
# Return the string return ''.join(newString)
# Function that finds the length of # the longest palindromic substring def Manacher(string):
# Update the string
string = UpdatedString(string)
# Stores the longest proper prefix
# which is also a suffix
LPS = [ 0 for _ in range ( len (string))]
C = 0
R = 0
for i in range ( len (string)):
imir = 2 * C - i
# Find the minimum length of
# the palindrome
if R > i:
LPS[i] = min (R - i, LPS[imir])
else :
# Find the actual length of
# the palindrome
LPS[i] = 0
# Exception Handling
try :
while string[i + 1 + LPS[i]] \
= = string[i - 1 - LPS[i]]:
LPS[i] + = 1
except :
pass
# Update C and R
if i + LPS[i] > R:
C = i
R = i + LPS[i]
r, c = max (LPS), LPS.index( max (LPS))
# Return the length r
return r
# Driver code # Given string str str = "forgeeksskeegfor"
# Function Call print (Manacher( str ))
|
// C# program to implement above approach using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{ // Function that placed '#' intermediately
// before and after each character
static string UpdatedString( string s){
string newString = "#" ;
// Traverse the string
foreach ( char ch in s){
newString += ch;
newString += "#" ;
}
// Return the string
return newString;
}
// Function that finds the length of
// the longest palindromic substring
static int Manacher( string s){
// Update the string
s = UpdatedString(s);
// Stores the longest proper prefix
// which is also a suffix
int [] LPS = new int [s.Length];
int C = 0;
int R = 0;
for ( int i = 0 ; i < s.Length ; i++){
int imir = 2 * C - i;
// Find the minimum length of
// the palindrome
if (R > i){
LPS[i] = Math.Min(R-i, LPS[imir]);
}
else {
// Find the actual length of
// the palindrome
LPS[i] = 0;
}
// Exception Handling
while ( ((i + 1 + LPS[i]) < s.Length) &&
((i - 1 - LPS[i]) >= 0) &&
s[i + 1 + LPS[i]] == s[i - 1 - LPS[i]]){
LPS[i] += 1;
}
// Update C and R
if (i + LPS[i] > R){
C = i;
R = i + LPS[i];
}
}
int r = 0;
for ( int i = 0 ; i < s.Length ; i++){
r = Math.Max(r, LPS[i]);
}
// Return the length r
return r;
}
// Driver code
public static void Main( string [] args){
// Given string str
string str = "forgeeksskeegfor" ;
// Function Call
Console.WriteLine(Manacher(str));
}
} // This code is contributed by entertain2022. |
//Javascript code for the above approach // Function that placed '#' intermediately // before and after each character function UpdatedString(string) {
let newString = [ '#' ];
// Traverse the string for (let char of string) {
newString.push(char, '#' );
}
// Return the string return newString.join( '' );
} // Function that finds the length of // the longest palindromic substring function Manacher(string) {
// Update the string
string = UpdatedString(string);
// Stores the longest proper prefix
// which is also a suffix
let LPS = new Array(string.length).fill(0);
let C = 0;
let R = 0;
for (let i = 0; i < string.length; i++) {
let imir = 2 * C - i;
// Find the minimum length of
// the palindrome
if (R > i) {
LPS[i] = Math.min(R-i, LPS[imir]);
} else {
// Find the actual length of
// the palindrome
LPS[i] = 0;
}
// Exception Handling
try {
while (string[i + 1 + LPS[i]] === string[i - 1 - LPS[i]]) {
LPS[i] += 1;
}
} catch (err) {
// pass
}
// Update C and R
if (i + LPS[i] > R) {
C = i;
R = i + LPS[i];
}
}
let r = Math.max(...LPS);
let c = LPS.indexOf(r);
// Return the length r
return r;
} // Driver code // Given string str let str = "forgeeksskeegfor" ;
// Function Call console.log(Manacher(str)); |
10
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(N)