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Check if a string contains an anagram of another string as its substring

  • Difficulty Level : Expert
  • Last Updated : 03 Jun, 2021

Given two strings S1 and S2, the task is to check if S2 contains an anagram of S1 as its substring.

Examples: 
 

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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:

 

  1. Initialize two Hashmaps s1hash and s2hash, to store the frequency of alphabets of the two strings.
  2. If the length of S1 is greater than the length of S2, then print “NO”.
  3. Iterate over the characters of the string S1 and update s1hash.
  4. Iterate over the characters of the string S2 using the Sliding Window technique and update the HashMap accordingly.
  5. For any substring of S2 of length equal to the length of S1, if both the Hashmaps are found to be equal, print “YES”.
  6. 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 substring
bool 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 Code
int 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 approach
import 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 substring
def 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 Code
if __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 approach
using System;
using System.Collections.Generic;
class GFG{
    
// Function to check if string s2
// contains anagram of the string
// s1 as its substring
static 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 Code
public 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 substring
function 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 Code
var 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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