Longest Palindromic Substring using hashing in O(nlogn)
Last Updated :
26 Dec, 2023
Given a string S, The task is to find the longest substring which is a palindrome using hashing in O(N log N) time.
Input: S: ”forgeeksskeegfor”,
Output: “geeksskeeg”
Input: S: ”Geeks”,
Output: “ee”
Hashing to Solve the Problem:
The hashing approach to solving the longest palindromic substring problem uses a hash table to store the characters of the string and then uses the hashing function to quickly compare two substrings for equality. The basic idea is to hash each substring of the original string, and then compare the hash values to determine if the substring is a palindrome. The hashing function takes O(N log N) time, where n is the length of the string since it needs to hash each substring of the string. This makes the hashing approach an optimal solution to the longest palindromic substring problem in O(N log N) time.
Approach: There are various approaches to solving the longest palindromic substring problem. Here we will discuss a solution using hashing in O(N log N) time.
The idea behind this approach is to use hashing to store the indices of the string characters. This approach is efficient as it avoids the need to traverse the entire string or check for the characters one by one.
Follow the steps to solve the problem:
- Initialize a hash table hash_table with size N and fill it with 0.
- Iterate through each character in the string S and store its index in the hash table hash_table.
- Iterate through each character in the string S and check if its corresponding index in the hash table is greater than 0. If it is, then check if the substring between the current character and the corresponding index in the hash table is a palindrome.
- If the substring between the current character and the corresponding index in the hash table is a palindrome, then store the length of the substring in a variable max_length and update the start and end indices of the longest palindromic substring.
- Repeat steps 3 and 4 for all characters in the string S.
- Return the longest palindromic substring.
Pseudo-code:
The pseudo-code for the hashing approach to solving the longest palindromic substring problem is as follows:
// Create a hash table to store characters
// of the string s
hash_table = new HashTable()
// Iterate through the string s
for i = 0 to s.length
for j = i to s.length
hash_value = hash_function(s.substring(i, j))
hash_table.add(hash_value)
// Iterate through the hash table for value in hash_table
// Check if the value is a palindrome
if is_palindrome(value)
// Return the longest palindrome
return value
Below is the Implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
string longestPalindromicSubstring(string S)
{
int n = S.length();
int hash_table[128];
memset (hash_table, 0, sizeof (hash_table));
bool isPalindrome = true ;
for ( int i = 0; i < n; i++)
hash_table[S[i] - 'a' ] = i;
int start = 0, end = 0;
for ( int i = 0; i < n; i++) {
if (hash_table[S[i] - 'a' ] > 0) {
int len = hash_table[S[i] - 'a' ] - i;
isPalindrome = true ;
for ( int j = 0; j < len; j++) {
if (S[i + j]
!= S[hash_table[S[i] - 'a' ] - j]) {
isPalindrome = false ;
break ;
}
}
if (isPalindrome && len > end - start) {
start = i;
end = hash_table[S[i] - 'a' ];
}
}
}
if (isPalindrome) {
return S.substr(start, end - start + 1);
}
else {
return "not possible" ;
}
}
int main()
{
string S = "babad" ;
string S2 = "forgeeksskeegfor" ;
string S3 = "pawan" ;
string S4 = "a" ;
cout << longestPalindromicSubstring(S) << endl;
cout << longestPalindromicSubstring(S2) << endl;
cout << longestPalindromicSubstring(S3) << endl;
cout << longestPalindromicSubstring(S4) << endl;
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
public static String
longestPalindromicSubstring(String S)
{
int n = S.length();
int [] hashTable = new int [ 128 ];
Arrays.fill(hashTable, 0 );
boolean isPalindrome = true ;
for ( int i = 0 ; i < n; i++)
hashTable[( int )S.charAt(i)] = i;
int start = 0 , end = 0 ;
for ( int i = 0 ; i < n; i++) {
if (hashTable[( int )S.charAt(i)] > 0 ) {
int len = hashTable[( int )S.charAt(i)] - i;
isPalindrome = true ;
for ( int j = 0 ; j < len; j++) {
if (S.charAt(i + j)
!= S.charAt(
hashTable[( int )S.charAt(i)]
- j)) {
isPalindrome = false ;
break ;
}
}
if (isPalindrome && len > end - start) {
start = i;
end = hashTable[( int )S.charAt(i)];
}
}
}
if (isPalindrome) {
return S.substring(start, end + 1 );
}
else {
return "not possible" ;
}
}
public static void main(String[] args)
{
String S = "babad" ;
String S2 = "forgeeksskeegfor" ;
String S3 = "pawan" ;
String S4 = "a" ;
System.out.println(longestPalindromicSubstring(S));
System.out.println(longestPalindromicSubstring(S2));
System.out.println(longestPalindromicSubstring(S3));
System.out.println(longestPalindromicSubstring(S4));
}
}
|
Python3
def longestPalindromicSubstring(S):
n = len (S)
hash_table = [ 0 ] * 128
isPalindrome = True
for i in range (n):
hash_table[ ord (S[i]) - ord ( 'a' )] = i
start,end = 0 , 0
for i in range (n):
if (hash_table[ ord (S[i]) - ord ( 'a' )]):
Len = hash_table[ ord (S[i]) - ord ( 'a' )] - i
isPalindrome = True
for j in range ( Len ):
if (S[i + j]! = S[hash_table[ ord (S[i]) - ord ( 'a' )] - j]):
isPalindrome = False
break
if (isPalindrome and Len >end - start):
start = i
end = hash_table[ ord (S[i]) - ord ( 'a' )]
if (isPalindrome):
return S[start:end + 1 ]
else :
return "not possible"
S = "babad"
S2 = "forgeeksskeegfor"
S3 = "pawan"
S4 = "a"
print (longestPalindromicSubstring(S))
print (longestPalindromicSubstring(S2))
print (longestPalindromicSubstring(S3))
print (longestPalindromicSubstring(S4))
|
C#
using System;
using System.Linq;
using System.Collections;
public class GFG
{
public static String longestPalindromicSubstring(String S)
{
var n = S.Length;
int [] hashTable = new int [128];
System.Array.Fill(hashTable,0);
var isPalindrome = true ;
for ( int i = 0; i < n; i++)
{
hashTable[( int )S[i]] = i;
}
var start = 0;
var end = 0;
for ( int i = 0; i < n; i++)
{
if (hashTable[( int )S[i]] > 0)
{
var len = hashTable[( int )S[i]] - i;
isPalindrome = true ;
for ( int j = 0; j < len; j++)
{
if (S[i + j] != S[hashTable[( int )S[i]] - j])
{
isPalindrome = false ;
break ;
}
}
if (isPalindrome && len > end - start)
{
start = i;
end = hashTable[( int )S[i]];
}
}
}
if (isPalindrome)
{
return S.Substring(start,end + 1-start);
}
else
{
return "not possible" ;
}
}
public static void Main(String[] args)
{
var S = "babad" ;
var S2 = "forgeeksskeegfor" ;
var S3 = "pawan" ;
var S4 = "a" ;
Console.WriteLine(GFG.longestPalindromicSubstring(S));
Console.WriteLine(GFG.longestPalindromicSubstring(S2));
Console.WriteLine(GFG.longestPalindromicSubstring(S3));
Console.WriteLine(GFG.longestPalindromicSubstring(S4));
}
}
|
Javascript
function longestPalindromicSubstring(S)
{
let n = S.length;
let hash_table = new Array(n).fill(0);
let isPalindrome = true ;
for (let i = 0; i < n; i++)
hash_table[S[i].charCodeAt(0) - 'a' .charCodeAt(0)] = i;
let start = 0, end = 0;
for (let i = 0; i < n; i++) {
if (hash_table[S[i].charCodeAt(0) - 'a' .charCodeAt(0)] > 0)
{
let len = hash_table[S[i].charCodeAt(0) - 'a' .charCodeAt(0)] - i;
isPalindrome = true ;
for (let j = 0; j < len; j++) {
if (S[i + j] != S[hash_table[S[i].charCodeAt(0) - 'a' .charCodeAt(0)] - j]) {
isPalindrome = false ;
break ;
}
}
if (isPalindrome && len > end - start) {
start = i;
end = hash_table[S[i].charCodeAt(0) - 'a' .charCodeAt(0)];
}
}
}
if (isPalindrome) {
return S.substring(start, end+1 );
}
else {
return "not possible" ;
}
}
let S = "babad" ;
let S2 = "forgeeksskeegfor" ;
let S3 = "pawan" ;
let S4 = "a" ;
document.write(longestPalindromicSubstring(S) + "<br>" );
document.write(longestPalindromicSubstring(S2)+ "<br>" );
document.write(longestPalindromicSubstring(S3)+ "<br>" );
document.write(longestPalindromicSubstring(S4)+ "<br>" );
|
Output
bab
geeksskeeg
awa
a
Time Complexity: O(n^2), The time complexity of the above algorithm is O(N*logN). Here, N is the length of the string S. The time complexity is due to the fact that for each character in the string S we need to iterate through the entire hash table in order to find its corresponding index.
Auxiliary Space: O(N) The space complexity of the above algorithm is O(N). Here, N is the length of the string S. We need to store the indices of the characters in the string S in a hash table of size n.
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