Check if a string contains an anagram of another string as its substring
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. |
Output:
YES
Time Complexity: O(26 * len(S2))
Auxiliary Space: O(26)
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