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: ”forneveropenskeegfor”,
Output: “neveropenskeeg”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 shash_table = new HashTable()
// Iterate through the string sfor i = 0 to s.length
for j = i to s.lengthhash_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 palindromeif is_palindrome(value)
// Return the longest palindrome
return value
Below is the Implementation of the above approach:
C++
// C++ program to find the longest// palindromic substring using hashing#include <bits/stdc++.h>using namespace std;// Function to find the longest// palindromic substringstring longestPalindromicSubstring(string S){ // Initialize a hash table H with size // n and fill it with 0 int n = S.length(); int hash_table[128]; memset(hash_table, 0, sizeof(hash_table)); bool isPalindrome = true; // Iterate through each character in the // string S and store its index in the // hash table H for (int i = 0; i < n; i++) hash_table[S[i] - 'a'] = i; // Initialize start and end indices of // the longest palindromic substring int start = 0, end = 0; // Iterate through each character // in the string S for (int i = 0; i < n; i++) { // Check if its corresponding index // in the hash table is greater than 0 if (hash_table[S[i] - 'a'] > 0) { // Calculate the length of the palindrome int len = hash_table[S[i] - 'a'] - i; // Check if the substring between // the current character and the // corresponding index in the // hash table is a palindrome isPalindrome = true; for (int j = 0; j < len; j++) { if (S[i + j] != S[hash_table[S[i] - 'a'] - j]) { isPalindrome = false; break; } } // If the substring between the // current character and the // corresponding index in the // hash table is a palindrome, then // update the start and end indices // of the longest palindromic // substring if (isPalindrome && len > end - start) { start = i; end = hash_table[S[i] - 'a']; } } } // Return the longest palindromic // substring if (isPalindrome) { return S.substr(start, end - start + 1); } else { return "not possible"; }}// Driver codeint main(){ string S = "babad"; string S2 = "forneveropenskeegfor"; string S3 = "pawan"; string S4 = "a"; // Function Call cout << longestPalindromicSubstring(S) << endl; cout << longestPalindromicSubstring(S2) << endl; cout << longestPalindromicSubstring(S3) << endl; cout << longestPalindromicSubstring(S4) << endl; return 0;} |
Java
// Java program to find the longest// palindromic substring using hashingimport java.io.*;import java.util.*;class GFG { // Function to find the longest // palindromic substring public static String longestPalindromicSubstring(String S) { // Initialize a hash table H with size // n and fill it with 0 int n = S.length(); int[] hashTable = new int[128]; Arrays.fill(hashTable, 0); boolean isPalindrome = true; // Iterate through each character in the // string S and store its index in the // hash table H for (int i = 0; i < n; i++) hashTable[(int)S.charAt(i)] = i; // Initialize start and end indices of // the longest palindromic substring int start = 0, end = 0; // Iterate through each character // in the string S for (int i = 0; i < n; i++) { // Check if its corresponding index // in the hash table is greater than 0 if (hashTable[(int)S.charAt(i)] > 0) { // Calculate the length of the palindrome int len = hashTable[(int)S.charAt(i)] - i; // Check if the substring between // the current character and the // corresponding index in the // hash table is a palindrome 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 the substring between the // current character and the // corresponding index in the // hash table is a palindrome, then // update the start and end indices // of the longest palindromic // substring if (isPalindrome && len > end - start) { start = i; end = hashTable[(int)S.charAt(i)]; } } } // Return the longest palindromic // substring if (isPalindrome) { return S.substring(start, end + 1); } else { return "not possible"; } } public static void main(String[] args) { String S = "babad"; String S2 = "forneveropenskeegfor"; String S3 = "pawan"; String S4 = "a"; // Function Call System.out.println(longestPalindromicSubstring(S)); System.out.println(longestPalindromicSubstring(S2)); System.out.println(longestPalindromicSubstring(S3)); System.out.println(longestPalindromicSubstring(S4)); }}// This code is contributed by lokesh. |
Python3
# Python program to find the longest# palindromic substring using hashing# Function to find the longest# palindromic substringdef longestPalindromicSubstring(S): # Initialize a hash table H with size # n and fill it with 0 n=len(S) hash_table=[0]*128 isPalindrome=True # Iterate through each character in the # string S and store its index in the # hash table H for i in range(n): hash_table[ord(S[i])-ord('a')]=i # Initialize start and end indices of # the longest palindromic substring start,end=0,0 # Iterate through each character # in the string S for i in range(n): # Check if its corresponding index # in the hash table is greater than 0 if(hash_table[ord(S[i])-ord('a')]): # Calculate the length of the palindrome Len=hash_table[ord(S[i])-ord('a')]-i # Check if the substring between # the current character and the # corresponding index in the # hash table is a palindrome isPalindrome=True for j in range(Len): if(S[i+j]!=S[hash_table[ord(S[i])-ord('a')]-j]): isPalindrome=False break # If the substring between the # current character and the # corresponding index in the # hash table is a palindrome, then # update the start and end indices # of the longest palindromic # substring if(isPalindrome and Len>end-start): start=i end=hash_table[ord(S[i])-ord('a')] # Return the longest palindromic # substring if(isPalindrome): return S[start:end+1] else: return "not possible"# Driver codeS = "babad"S2 = "forneveropenskeegfor"S3 = "pawan"S4 = "a"# Function Callprint(longestPalindromicSubstring(S))print(longestPalindromicSubstring(S2))print(longestPalindromicSubstring(S3))print(longestPalindromicSubstring(S4))# This code is contributed by Pushpesh Raj. |
C#
// Include namespace systemusing System;using System.Linq;using System.Collections;public class GFG{ // Function to find the longest // palindromic substring public static String longestPalindromicSubstring(String S) { // Initialize a hash table H with size // n and fill it with 0 var n = S.Length; int[] hashTable = new int[128]; System.Array.Fill(hashTable,0); var isPalindrome = true; // Iterate through each character in the // string S and store its index in the // hash table H for (int i = 0; i < n; i++) { hashTable[(int)S[i]] = i; } // Initialize start and end indices of // the longest palindromic substring var start = 0; var end = 0; // Iterate through each character // in the string S for (int i = 0; i < n; i++) { // Check if its corresponding index // in the hash table is greater than 0 if (hashTable[(int)S[i]] > 0) { // Calculate the length of the palindrome var len = hashTable[(int)S[i]] - i; // Check if the substring between // the current character and the // corresponding index in the // hash table is a palindrome isPalindrome = true; for (int j = 0; j < len; j++) { if (S[i + j] != S[hashTable[(int)S[i]] - j]) { isPalindrome = false; break; } } // If the substring between the // current character and the // corresponding index in the // hash table is a palindrome, then // update the start and end indices // of the longest palindromic // substring if (isPalindrome && len > end - start) { start = i; end = hashTable[(int)S[i]]; } } } // Return the longest palindromic // substring if (isPalindrome) { return S.Substring(start,end + 1-start); } else { return "not possible"; } } public static void Main(String[] args) { var S = "babad"; var S2 = "forneveropenskeegfor"; var S3 = "pawan"; var S4 = "a"; // Function Call Console.WriteLine(GFG.longestPalindromicSubstring(S)); Console.WriteLine(GFG.longestPalindromicSubstring(S2)); Console.WriteLine(GFG.longestPalindromicSubstring(S3)); Console.WriteLine(GFG.longestPalindromicSubstring(S4)); }} |
Javascript
// Javascript program to find the longest// palindromic substring using hashing// Function to find the longest// palindromic substringfunction longestPalindromicSubstring(S){ // Initialize a hash table H with size // n and fill it with 0 let n = S.length; let hash_table = new Array(n).fill(0); let isPalindrome = true; // Iterate through each character in the // string S and store its index in the // hash table H for (let i = 0; i < n; i++) hash_table[S[i].charCodeAt(0) - 'a'.charCodeAt(0)] = i; // Initialize start and end indices of // the longest palindromic substring let start = 0, end = 0; // Iterate through each character // in the string S for (let i = 0; i < n; i++) { // Check if its corresponding index // in the hash table is greater than 0 if (hash_table[S[i].charCodeAt(0) - 'a'.charCodeAt(0)] > 0) { // Calculate the length of the palindrome let len = hash_table[S[i].charCodeAt(0) - 'a'.charCodeAt(0)] - i; // Check if the substring between // the current character and the // corresponding index in the // hash table is a palindrome 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 the substring between the // current character and the // corresponding index in the // hash table is a palindrome, then // update the start and end indices // of the longest palindromic // substring if (isPalindrome && len > end - start) { start = i; end = hash_table[S[i].charCodeAt(0) - 'a'.charCodeAt(0)]; } } } // Return the longest palindromic // substring if (isPalindrome) { return S.substring(start, end+1 ); } else { return "not possible"; }}// Driver code let S = "babad"; let S2 = "forneveropenskeegfor"; let S3 = "pawan"; let S4 = "a"; // Function Call document.write(longestPalindromicSubstring(S) + "<br>"); document.write(longestPalindromicSubstring(S2)+ "<br>"); document.write(longestPalindromicSubstring(S3)+ "<br>"); document.write(longestPalindromicSubstring(S4)+ "<br>"); |
Output
bab neveropenskeeg awa a
Time Complexity: O(nlogn), 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.
Related Articles:
- Introduction to Strings – Data Structure and Algorithm Tutorials
- Introduction to Hashing – Data Structure and Algorithm Tutorials
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