Given two strings S1 and S2, the task is to check if S2 contains an anagram of S1 as its substring.
Examples:
Input: S1 = “ab”, S2 = “bbpobac”
Output: Yes
Explanation: String S2 contains anagram “ba” of S1 (“ba”).Input: S1 = “ab”, S2 = “cbddaoo”
Output: No
Approach: Follow the steps below to solve the problem:
- Initialize two Hashmaps s1hash and s2hash, to store the frequency of alphabets of the two strings.
- If the length of S1 is greater than the length of S2, then print “NO”.
- Iterate over the characters of the string S1 and update s1hash.
- Iterate over the characters of the string S2 using the Sliding Window technique and update the HashMap accordingly.
- For any substring of S2 of length equal to the length of S1, if both the Hashmaps are found to be equal, print “YES”.
- Otherwise, print “NO”.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to check if string s2// contains anagram of the string// s1 as its substringbool checkAnagram(string s1, string s2){ // Stores frequencies of // characters in substrings of s2 vector<int> s2hash(26, 0); // Stores frequencies of // characters in s1 vector<int> s1hash(26, 0); int s1len = s1.size(); int s2len = s2.size(); // If length of s2 exceeds length of s1 if (s1len > s2len) return false; int left = 0, right = 0; // Store frequencies of characters in first // substring of length s1len in string s2 while (right < s1len) { s1hash[s1[right] - 'a'] += 1; s2hash[s2[right] - 'a'] += 1; right++; } right -= 1; // Perform Sliding Window technique while (right < s2len) { // If hashmaps are found to be // identical for any substring if (s1hash == s2hash) return true; right++; if (right != s2len) s2hash[s2[right] - 'a'] += 1; s2hash[s2[left] - 'a'] -= 1; left++; } return false;}// Driver Codeint main(){ string s1 = "ab"; string s2 = "bbpobac"; if (checkAnagram(s1, s2)) cout << "YES\n"; else cout << "No\n"; return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG { // Function to check if string s2 // contains anagram of the string // s1 as its substring public static boolean checkAnagram(String s1, String s2) { // Stores frequencies of // characters in substrings of s2 int s2hash[] = new int[26]; // Stores frequencies of // characters in s1 int s1hash[] = new int[26]; int s1len = s1.length(); int s2len = s2.length(); // If length of s2 exceeds length of s1 if (s1len > s2len) return false; int left = 0, right = 0; // Store frequencies of characters in first // substring of length s1len in string s2 while (right < s1len) { s1hash[s1.charAt(right) - 'a'] += 1; s2hash[s2.charAt(right) - 'a'] += 1; right++; } right -= 1; // Perform Sliding Window technique while (right < s2len) { // If hashmaps are found to be // identical for any substring if (Arrays.equals(s1hash, s2hash)) return true; right++; if (right != s2len) s2hash[s2.charAt(right) - 'a'] += 1; s2hash[s2.charAt(left) - 'a'] -= 1; left++; } return false; } // Driver Code public static void main(String[] args) { String s1 = "ab"; String s2 = "bbpobac"; if (checkAnagram(s1, s2)) System.out.println("YES"); else System.out.println("No"); }}// This code is contributed by kingash. |
Python3
# Python 3 Program to implement# the above approach# Function to check if string s2# contains anagram of the string# s1 as its substringdef checkAnagram(s1, s2): # Stores frequencies of # characters in substrings of s2 s2hash = [0 for i in range(26)] # Stores frequencies of # characters in s1 s1hash = [0 for i in range(26)] s1len = len(s1) s2len = len(s2) # If length of s2 exceeds length of s1 if (s1len > s2len): return False left = 0 right = 0 # Store frequencies of characters in first # substring of length s1len in string s2 while (right < s1len): s1hash[ord(s1[right]) - 97] += 1 s2hash[ord(s2[right]) - 97] += 1 right += 1 right -= 1 # Perform Sliding Window technique while (right < s2len): # If hashmaps are found to be # identical for any substring if (s1hash == s2hash): return True right += 1 if (right != s2len): s2hash[ord(s2[right]) - 97] += 1 s2hash[ord(s2[left]) - 97] -= 1 left += 1 return False# Driver Codeif __name__ == '__main__': s1 = "ab" s2 = "bbpobac" if (checkAnagram(s1, s2)): print("YES") else: print("No") # This code is contributed by ipg2016107 |
C#
// C# Program to implement// the above approachusing System;using System.Collections.Generic;class GFG{ // Function to check if string s2// contains anagram of the string// s1 as its substringstatic bool checkAnagram(string s1, string s2){ // Stores frequencies of // characters in substrings of s2 List<int> s2hash = new List<int>(); for(int i=0;i<26;i++) s2hash.Add(0); // Stores frequencies of // characters in s1 List<int> s1hash = new List<int>(); for(int i=0;i<26;i++) s1hash.Add(0); int s1len = s1.Length; int s2len = s2.Length; // If length of s2 exceeds length of s1 if (s1len > s2len) return false; int left = 0, right = 0; // Store frequencies of characters in first // substring of length s1len in string s2 while (right < s1len) { s1hash[s1[right] - 'a'] += 1; s2hash[s2[right] - 'a'] += 1; right++; } right -= 1; // Perform Sliding Window technique while (right < s2len) { // If hashmaps are found to be // identical for any substring if (s1hash == s2hash) return true; right++; if(right != s2len) s2hash[s2[right] - 'a'] += 1; s2hash[s2[left] - 'a'] -= 1; left++; } return false;}// Driver Codepublic static void Main(){ string s1 = "ab"; string s2 = "bbpobac"; if (checkAnagram(s1, s2)==true) Console.WriteLine("NO"); else Console.WriteLine("YES");}}// This code is contributed by bgangawar59. |
Javascript
<script>// Javascript Program to implement// the above approach// Function to check if string s2// contains anagram of the string// s1 as its substringfunction checkAnagram(s1, s2){ // Stores frequencies of // characters in substrings of s2 var s2hash = Array(26).fill(0); // Stores frequencies of // characters in s1 var s1hash = Array(26).fill(0); var s1len = s1.length; var s2len = s2.length; // If length of s2 exceeds length of s1 if (s1len > s2len) return false; var left = 0, right = 0; // Store frequencies of characters in first // substring of length s1len in string s2 while (right < s1len) { s1hash[s1[right].charCodeAt(0) - 'a'.charCodeAt(0)] += 1; s2hash[s2[right].charCodeAt(0) - 'a'.charCodeAt(0)] += 1; right++; } right -= 1; // Perform Sliding Window technique while (right < s2len) { var ans = true; // If hashmaps are found to be // identical for any substring for(var i =0; i<26; i++) { if(s1hash[i]!=s2hash[i]) { ans = false; } } if(ans) return true; right++; if (right != s2len) s2hash[s2[right].charCodeAt(0) - 'a'.charCodeAt(0)] += 1; s2hash[s2[left].charCodeAt(0) - 'a'.charCodeAt(0)] -= 1; left++; } return false;}// Driver Codevar s1 = "ab";var s2 = "bbpobac";if (checkAnagram(s1, s2)) document.write( "YES");else document.write( "No");// This code is contributed by importantly.</script> |
Output:
YES
Time Complexity: O(26 * len(S2))
Auxiliary Space: O(26)
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