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  • Difficulty Level : Easy
  • Last Updated : 29 Jul, 2022
 

Write a function to check whether two given strings are anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “abcd” and “dabc” are an anagram of each other.

check-whether-two-strings-are-anagram-of-each-other

We strongly recommend that you click here and practice it, before moving on to the solution.

Method 1 (Use Sorting)  

  1. Sort both strings
  2. Compare the sorted strings

Below is the implementation of the above idea:

C++




// C++ program to check whether two strings are anagrams
// of each other
#include <bits/stdc++.h>
using namespace std;
 
/* function to check whether two strings are anagram of
each other */
bool areAnagram(string str1, string str2)
{
    // Get lengths of both strings
    int n1 = str1.length();
    int n2 = str2.length();
 
    // If length of both strings is not same, then they
    // cannot be anagram
    if (n1 != n2)
        return false;
 
    // Sort both the strings
    sort(str1.begin(), str1.end());
    sort(str2.begin(), str2.end());
 
    // Compare sorted strings
    for (int i = 0; i < n1; i++)
        if (str1[i] != str2[i])
            return false;
 
    return true;
}
 
// Driver code
int main()
{
    string str1 = "test";
    string str2 = "ttew";
 
    // Function Call
    if (areAnagram(str1, str2))
        cout << "The two strings are anagram of each other";
    else
        cout << "The two strings are not anagram of each "
                "other";
 
    return 0;
}

Java




// JAVA program to check whether two strings
// are anagrams of each other
import java.io.*;
import java.util.Arrays;
import java.util.Collections;
 
class GFG {
 
    /* function to check whether two strings are
    anagram of each other */
    static boolean areAnagram(char[] str1, char[] str2)
    {
        // Get lengths of both strings
        int n1 = str1.length;
        int n2 = str2.length;
 
        // If length of both strings is not same,
        // then they cannot be anagram
        if (n1 != n2)
            return false;
 
        // Sort both strings
        Arrays.sort(str1);
        Arrays.sort(str2);
 
        // Compare sorted strings
        for (int i = 0; i < n1; i++)
            if (str1[i] != str2[i])
                return false;
 
        return true;
    }
 
    /* Driver Code*/
    public static void main(String args[])
    {
        char str1[] = { 't', 'e', 's', 't' };
        char str2[] = { 't', 't', 'e', 'w' };
       
        // Function Call
        if (areAnagram(str1, str2))
            System.out.println("The two strings are"
                               + " anagram of each other");
        else
            System.out.println("The two strings are not"
                               + " anagram of each other");
    }
}
 
// This code is contributed by Nikita Tiwari.

Python




class Solution:
 
    # Function is to check whether two strings are anagram of each other or not.
    def isAnagram(self, a, b):
 
        if sorted(a) == sorted(b):
            return True
        else:
            return False
 
# {
#  Driver Code Starts
 
if __name__ == '__main__':
    t = int(input())
    for i in range(t):
        a, b = map(str, input().strip().split())
        if(Solution().isAnagram(a, b)):
            print("The two strings are anagram of each other")
        else:
            print("The two strings are not anagram of each other")
# } Driver Code Ends

C#




// C# program to check whether two
// strings are anagrams of each other
using System;
using System.Collections;
class GFG {
 
    /* function to check whether two
strings are anagram of each other */
    public static bool areAnagram(ArrayList str1,
                                  ArrayList str2)
    {
        // Get lengths of both strings
        int n1 = str1.Count;
        int n2 = str2.Count;
 
        // If length of both strings is not
        // same, then they cannot be anagram
        if (n1 != n2) {
            return false;
        }
 
        // Sort both strings
        str1.Sort();
        str2.Sort();
 
        // Compare sorted strings
        for (int i = 0; i < n1; i++) {
            if (str1[i] != str2[i]) {
                return false;
            }
        }
 
        return true;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        // create and initialize new ArrayList
        ArrayList str1 = new ArrayList();
        str1.Add('t');
        str1.Add('e');
        str1.Add('s');
        str1.Add('t');
        // create and initialize new ArrayList
        ArrayList str2 = new ArrayList();
        str2.Add('t');
        str2.Add('t');
        str2.Add('e');
        str2.Add('w');
 
        // Function call
        if (areAnagram(str1, str2)) {
            Console.WriteLine("The two strings are"
                              + " anagram of each other");
        }
        else {
            Console.WriteLine("The two strings are not"
                              + " anagram of each other");
        }
    }
}
 
// This code is contributed by Shrikant13

Javascript




<script>
 
// JavaScript program to check whether two strings
// are anagrams of each other
 
    /* function to check whether two strings are
    anagram of each other */
    function areAnagram(str1,str2)
    {
        // Get lengths of both strings
        let n1 = str1.length;
        let n2 = str2.length;
   
        // If length of both strings is not same,
        // then they cannot be anagram
        if (n1 != n2)
            return false;
   
        // Sort both strings
        str1.sort();
        str2.sort()
   
        // Compare sorted strings
        for (let i = 0; i < n1; i++)
            if (str1[i] != str2[i])
                return false;
   
        return true;
    }
     
    /* Driver Code*/
    let str1=['t', 'e', 's', 't' ];
    let str2=['t', 't', 'e', 'w' ];
     
    // Function Call
        if (areAnagram(str1, str2))
            document.write("The two strings are"
                               + " anagram of each other<br>");
        else
            document.write("The two strings are not"
                               + " anagram of each other<br>");
 
 
 
 
 
// This code is contributed by rag2127
 
</script>

Output

The two strings are not anagram of each other

Time Complexity: O(nLogn)

Method 2 (Count characters) 
This method assumes that the set of possible characters in both strings is small. In the following implementation, it is assumed that the characters are stored using 8 bit and there can be 256 possible characters. 

  1. Create count arrays of size 256 for both strings. Initialize all values in count arrays as 0.
  2. Iterate through every character of both strings and increment the count of character in the corresponding count arrays.
  3. Compare count arrays. If both count arrays are same, then return true.

Below is the implementation of the above idea:

C++




// C++ program to check if two strings
// are anagrams of each other
#include <bits/stdc++.h>
using namespace std;
#define NO_OF_CHARS 256
 
/* function to check whether two strings are anagram of
each other */
bool areAnagram(char* str1, char* str2)
{
    // Create 2 count arrays and initialize all values as 0
    int count1[NO_OF_CHARS] = { 0 };
    int count2[NO_OF_CHARS] = { 0 };
    int i;
 
    // For each character in input strings, increment count
    // in the corresponding count array
    for (i = 0; str1[i] && str2[i]; i++) {
        count1[str1[i]]++;
        count2[str2[i]]++;
    }
 
    // If both strings are of different length. Removing
    // this condition will make the program fail for strings
    // like "aaca" and "aca"
    if (str1[i] || str2[i])
        return false;
 
    // Compare count arrays
    for (i = 0; i < NO_OF_CHARS; i++)
        if (count1[i] != count2[i])
            return false;
 
    return true;
}
 
/* Driver code*/
int main()
{
    char str1[] = "geeksforgeeks";
    char str2[] = "forgeeksgeeks";
   
    // Function Call
    if (areAnagram(str1, str2))
        cout << "The two strings are anagram of each other";
    else
        cout << "The two strings are not anagram of each "
                "other";
 
    return 0;
}
 
// This is code is contributed by rathbhupendra

C




// C program to check if two strings
// are anagrams of each other
#include <stdio.h>
#define NO_OF_CHARS 256
 
/* function to check whether two strings are anagram of
   each other */
bool areAnagram(char* str1, char* str2)
{
    // Create 2 count arrays and initialize all values as 0
    int count1[NO_OF_CHARS] = { 0 };
    int count2[NO_OF_CHARS] = { 0 };
    int i;
 
    // For each character in input strings, increment count
    // in the corresponding count array
    for (i = 0; str1[i] && str2[i]; i++) {
        count1[str1[i]]++;
        count2[str2[i]]++;
    }
 
    // If both strings are of different length. Removing
    // this condition will make the program fail for strings
    // like "aaca" and "aca"
    if (str1[i] || str2[i])
        return false;
 
    // Compare count arrays
    for (i = 0; i < NO_OF_CHARS; i++)
        if (count1[i] != count2[i])
            return false;
 
    return true;
}
 
/* Driver code*/
int main()
{
    char str1[] = "geeksforgeeks";
    char str2[] = "forgeeksgeeks";
   
    // Function Call
    if (areAnagram(str1, str2))
        printf("The two strings are anagram of each other");
    else
        printf("The two strings are not anagram of each "
               "other");
 
    return 0;
}

Java




// JAVA program to check if two strings
// are anagrams of each other
import java.io.*;
import java.util.*;
 
class GFG {
 
    static int NO_OF_CHARS = 256;
 
    /* function to check whether two strings
    are anagram of each other */
    static boolean areAnagram(char str1[], char str2[])
    {
        // Create 2 count arrays and initialize
        // all values as 0
        int count1[] = new int[NO_OF_CHARS];
        Arrays.fill(count1, 0);
        int count2[] = new int[NO_OF_CHARS];
        Arrays.fill(count2, 0);
        int i;
 
        // For each character in input strings,
        // increment count in the corresponding
        // count array
        for (i = 0; i < str1.length && i < str2.length;
             i++) {
            count1[str1[i]]++;
            count2[str2[i]]++;
        }
 
        // If both strings are of different length.
        // Removing this condition will make the program
        // fail for strings like "aaca" and "aca"
        if (str1.length != str2.length)
            return false;
 
        // Compare count arrays
        for (i = 0; i < NO_OF_CHARS; i++)
            if (count1[i] != count2[i])
                return false;
 
        return true;
    }
 
    /* Driver code*/
    public static void main(String args[])
    {
        char str1[] = ("geeksforgeeks").toCharArray();
        char str2[] = ("forgeeksgeeks").toCharArray();
 
        // Function call
        if (areAnagram(str1, str2))
            System.out.println("The two strings are"
                               + "anagram of each other");
        else
            System.out.println("The two strings are not"
                               + " anagram of each other");
    }
}
 
// This code is contributed by Nikita Tiwari.

Python




# Python program to check if two strings are anagrams of
# each other
NO_OF_CHARS = 256
 
# Function to check whether two strings are anagram of
# each other
 
 
def areAnagram(str1, str2):
 
    # Create two count arrays and initialize all values as 0
    count1 = [0] * NO_OF_CHARS
    count2 = [0] * NO_OF_CHARS
 
    # For each character in input strings, increment count
    # in the corresponding count array
    for i in str1:
        count1[ord(i)] += 1
 
    for i in str2:
        count2[ord(i)] += 1
 
    # If both strings are of different length. Removing this
    # condition will make the program fail for strings like
    # "aaca" and "aca"
    if len(str1) != len(str2):
        return 0
 
    # Compare count arrays
    for i in xrange(NO_OF_CHARS):
        if count1[i] != count2[i]:
            return 0
 
    return 1
 
 
# Driver code
str1 = "geeksforgeeks"
str2 = "forgeeksgeeks"
 
# Function call
if areAnagram(str1, str2):
    print "The two strings are anagram of each other"
else:
    print "The two strings are not anagram of each other"
 
# This code is contributed by Bhavya Jain

C#




// C# program to check if two strings
// are anagrams of each other
 
using System;
 
public class GFG {
 
    static int NO_OF_CHARS = 256;
 
    /* function to check whether two strings
    are anagram of each other */
    static bool areAnagram(char[] str1, char[] str2)
    {
        // Create 2 count arrays and initialize
        // all values as 0
        int[] count1 = new int[NO_OF_CHARS];
        int[] count2 = new int[NO_OF_CHARS];
        int i;
 
        // For each character in input strings,
        // increment count in the corresponding
        // count array
        for (i = 0; i < str1.Length && i < str2.Length;
             i++) {
            count1[str1[i]]++;
            count2[str2[i]]++;
        }
 
        // If both strings are of different length.
        // Removing this condition will make the program
        // fail for strings like "aaca" and "aca"
        if (str1.Length != str2.Length)
            return false;
 
        // Compare count arrays
        for (i = 0; i < NO_OF_CHARS; i++)
            if (count1[i] != count2[i])
                return false;
 
        return true;
    }
 
    /* Driver code*/
    public static void Main()
    {
        char[] str1 = ("geeksforgeeks").ToCharArray();
        char[] str2 = ("forgeeksgeeks").ToCharArray();
 
        // Function Call
        if (areAnagram(str1, str2))
            Console.WriteLine("The two strings are"
                              + "anagram of each other");
        else
            Console.WriteLine("The two strings are not"
                              + " anagram of each other");
    }
}
 
// This code contributed by 29AjayKumar

Javascript




<script>
// JAVAscript program to check if two strings
// are anagrams of each other
 
let NO_OF_CHARS = 256;
 
/* function to check whether two strings
    are anagram of each other */
function areAnagram(str1, str2)
{
 
    // Create 2 count arrays and initialize
        // all values as 0
        let count1 = new Array(NO_OF_CHARS);
        let count2 = new Array(NO_OF_CHARS);
        for(let i = 0; i < NO_OF_CHARS; i++)
        {
            count1[i] = 0;
            count2[i] = 0;
        }
         
        let i;
  
        // For each character in input strings,
        // increment count in the corresponding
        // count array
        for (i = 0; i < str1.length && i < str2.length;
             i++) {
            count1[str1[i].charCodeAt(0)]++;
            count2[str1[i].charCodeAt(0)]++;
        }
  
        // If both strings are of different length.
        // Removing this condition will make the program
        // fail for strings like "aaca" and "aca"
        if (str1.length != str2.length)
            return false;
  
        // Compare count arrays
        for (i = 0; i < NO_OF_CHARS; i++)
            if (count1[i] != count2[i])
                return false;
  
        return true;
}
 
/* Driver code*/
let str1 = ("geeksforgeeks").split("");
        let str2 = ("forgeeksgeeks").split("");
  
        // Function call
        if (areAnagram(str1, str2))
            document.write("The two strings are"
                               + "anagram of each other<br>");
        else
            document.write("The two strings are not"
                               + " anagram of each other<br>");
 
 
// This code is contributed by avanitrachhadiya2155
</script>

Output

The two strings are anagram of each other

Time Complexity : O(n)

Space Complexity : O(NO_OF_CHAR) = O(256) = O(1) (constant space use)

Method 3 (count characters using one array) 
The above implementation can be further to use only one count array instead of two. We can increment the value in count array for characters in str1 and decrement for characters in str2. Finally, if all count values are 0, then the two strings are anagram of each other. Thanks to Ace for suggesting this optimization.

C++




// C++ program to check if two strings
// are anagrams of each other
#include <bits/stdc++.h>
using namespace std;
#define NO_OF_CHARS 256
 
bool areAnagram(char* str1, char* str2)
{       
    // Create a count array and initialize all values as 0
    int count[NO_OF_CHARS] = { 0 };
    int i;
   
    // For each character in input strings, increment count
    // in the corresponding count array
    for (i = 0; str1[i] && str2[i]; i++) {
        count[str1[i]]++;
        count[str2[i]]--;
    }
   
    // If both strings are of different length. Removing
    // this condition will make the program fail for strings
    // like "aaca" and "aca"
    if (str1[i] || str2[i])
        return false;
 
    // See if there is any non-zero value in count array
    for (i = 0; i < NO_OF_CHARS; i++)
        if (count[i])
            return false;
    return true;
}
 
// Driver code
int main()
{
    char str1[] = "geeksforgeeks";
    char str2[] = "forgeeksgeeks";
   
    // Function call
    if (areAnagram(str1, str2))
        cout << "The two strings are anagram of each other";
    else
        cout << "The two strings are not anagram of each "
                "other";
 
    return 0;
}

Java




// Java program to check if two strings
// are anagrams of each other
class GFG{
 
static int NO_OF_CHARS = 256;
 
// function to check if two strings
// are anagrams of each other
static boolean areAnagram(char[] str1,
                          char[] str2)
{
     
    // Create a count array and initialize
    // all values as 0
    int[] count = new int[NO_OF_CHARS];
    int i;
 
    // If both strings are of different
    // length. Removing this condition
    // will make the program fail for
    // strings like "aaca" and "aca"
    if (str1.length != str2.length)
        return false;
   
    // For each character in input strings,
    // increment count in the corresponding
    // count array
    for(i = 0; i < str1.length; i++)
    {
        count[str1[i]]++;
        count[str2[i]]--;
    }
 
    // See if there is any non-zero
    // value in count array
    for(i = 0; i < NO_OF_CHARS; i++)
        if (count[i] != 0)
        {
            return false;
        }
    return true;
}
 
// Driver code
public static void main(String[] args)
{
    char str1[] = "geeksforgeeks".toCharArray();
    char str2[] = "forgeeksgeeks".toCharArray();
 
    // Function call
    if (areAnagram(str1, str2))
        System.out.print("The two strings are " +
                         "anagram of each other");
    else
        System.out.print("The two strings are " +
                         "not anagram of each other");
}
}
 
// This code is contributed by mark_85

Python3




# Python program to check if two strings
# are anagrams of each other
 
NO_OF_CHARS = 256
 
# function to check if two strings
# are anagrams of each other
def areAnagram(str1,str2):
     
    # If both strings are of different
    # length. Removing this condition
    # will make the program fail for
    # strings like "aaca" and "aca"   
    if(len(str1) != len(str2)):
        return False;
       
    # Create a count array and initialize
    # all values as 0
    count=[0 for i in range(NO_OF_CHARS)]
    i=0
     
    # For each character in input strings,
    # increment count in the corresponding
    # count array
    for i in range(len(str1)):
        count[ord(str1[i]) - ord('a')] += 1;
        count[ord(str2[i]) - ord('a')] -= 1;
         
    # See if there is any non-zero
    # value in count array
    for i in range(NO_OF_CHARS):
        if (count[i] != 0):
            return False
         
     
    return True
 
# Driver code
str1="geeksforgeeks"
str2="forgeeksgeeks"
 
# Function call
if (areAnagram(str1, str2)):
    print("The two strings are anagram of each other")
else:
    print("The two strings are not anagram of each other")
     
     
# This code is contributed by patel2127

C#




// C# program to check if two strings
// are anagrams of each other
using System;
 
class GFG{
 
static int NO_OF_CHARS = 256;
 
// function to check if two strings
// are anagrams of each other
static bool areAnagram(char[] str1,
                       char[] str2)
{
   
    // If both strings are of different
    // Length. Removing this condition
    // will make the program fail for
    // strings like "aaca" and "aca"
    if (str1.Length != str2.Length)
        return false;
     
    // Create a count array and initialize
    // all values as 0
    int[] count = new int[NO_OF_CHARS];
    int i;
 
    // For each character in input strings,
    // increment count in the corresponding
    // count array
    for(i = 0; i < str1.Length; i++)
    {
        count[str1[i] - 'a']++;
        count[str2[i] - 'a']--;
    }
 
    // See if there is any non-zero
    // value in count array
    for(i = 0; i < NO_OF_CHARS; i++)
        if (count[i] != 0)
        {
            return false;
        }
         
    return true;
}
 
// Driver code
public static void Main(String []args)
{
    char []str1 = "geeksforgeeks".ToCharArray();
    char []str2 = "forgeeksgeeks".ToCharArray();
 
    // Function call
    if (areAnagram(str1, str2))
       Console.Write("The two strings are " +
                     "anagram of each other");
    else
       Console.Write("The two strings are " +
                     "not anagram of each other");
}
}
 
// This code is contributed by shivanisinghss2110

Javascript




<script>
// Javascript program to check if two strings
// are anagrams of each other
let NO_OF_CHARS = 256;
 
// function to check if two strings
// are anagrams of each other
function areAnagram(str1, str2)
{
    // If both strings are of different
    // length. Removing this condition
    // will make the program fail for
    // strings like "aaca" and "aca"
    if (str1.length != str2.length)
        return false;
 
    // Create a count array and initialize
    // all values as 0
    let count = new Array(NO_OF_CHARS);
    for(let i = 0; i < NO_OF_CHARS; i++)
    {
        count[i] = 0;
    }
    let i;
  
    // For each character in input strings,
    // increment count in the corresponding
    // count array
    for(i = 0; i < str1.length; i++)
    {
        count[str1[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
        count[str2[i].charCodeAt(0) - 'a'.charCodeAt(0)]--;
    }
  
    // See if there is any non-zero
    // value in count array
    for(i = 0; i < NO_OF_CHARS; i++)
        if (count[i] != 0)
        {
            return false;
        }
    return true;
}
 
// Driver code
let str1 = "geeksforgeeks".split("");
let str2 = "forgeeksgeeks".split("");
 
// Function call
if (areAnagram(str1, str2))
    document.write("The two strings are " +
                 "anagram of each other");
else
    document.write("The two strings are " +
                 "not anagram of each other");
 
// This code is contributed by unknown2108.
</script>

Output

The two strings are anagram of each other

Time Complexity: O(n)

Time Complexity: O(NO_OF_CHAR) = O(256) = O(1) (constant space use)

Method 4 (Put all characters in HashMap)

In above implementation we are using extra space as we are creating array of 256 characters but we can optimise it using HashMap where we can store character and count of character in HashMap. Idea is to put all characters of one String in HashMap and reducing them as we encounter while looping over other String.

C++




// C++ implementation of the approach
// Function that returns true if a and b
// are anagarams of each other
#include <bits/stdc++.h>
using namespace std;
 
bool isAnagram(string a,string b)
{
 
  // Check if length of both strings is same or not
  if (a.length() != b.length()) {
    return false;
  }
  // Create a HashMap containing Character as Key and
  // Integer as Value. We will be storing character as
  // Key and count of character as Value.
  unordered_map<char,int> Map;
  // Loop over all character of String a and put in
  // HashMap.
  for (int i = 0; i < a.length(); i++) {
    Map[a[i]]++;
  }
  // Now loop over String b
  for (int i = 0; i < b.length(); i++) {
    // Check if current character already exists in
    // HashMap/map
    if (Map.find(b[i]) != Map.end()) {
      // If contains reduce count of that
      // character by 1 to indicate that current
      // character has been already counted as
      // idea here is to check if in last count of
      // all characters in last is zero which
      // means all characters in String a are
      // present in String b.
      Map[b[i]] -= 1;
    } else {
      return false;
    }
  }
 
  // Loop over all keys and check if all keys are 0.
  // If so it means it is anagram.
  for (auto items : Map) {
    if (items.second != 0) {
      return false;
    }
  }
  // Returning True as all keys are zero
  return true;
}
 
// Driver code
int main()
{
  string str1 = "geeksforgeeks";
  string str2 = "forgeeksgeeks";
 
  // Function call
  if (isAnagram(str1, str2))
    cout<<"The two strings are anagram of each other"<<endl;
  else
    cout<<"The two strings are not anagram of each other"<<endl;
}
 
// This code is contributed by shinjanpatra

Java




import java.io.*;
import java.util.*;
 
class GFG {
    public static boolean isAnagram(String a, String b)
    {
        // Check if length of both strings is same or not
        if (a.length() != b.length()) {
            return false;
        }
        // Create a HashMap containing Character as Key and
        // Integer as Value. We will be storing character as
        // Key and count of character as Value.
        HashMap<Character, Integer> map = new HashMap<>();
        // Loop over all character of String a and put in
        // HashMap.
        for (int i = 0; i < a.length(); i++) {
            // Check if HashMap already contain current
            // character or not
            if (map.containsKey(a.charAt(i))) {
                // If contains increase count by 1 for that
                // character
                map.put(a.charAt(i),
                        map.get(a.charAt(i)) + 1);
            }
            else {
                // else put that character in map and set
                // count to 1 as character is encountered
                // first time
                map.put(a.charAt(i), 1);
            }
        }
        // Now loop over String b
        for (int i = 0; i < b.length(); i++) {
            // Check if current character already exists in
            // HashMap/map
            if (map.containsKey(b.charAt(i))) {
                // If contains reduce count of that
                // character by 1 to indicate that current
                // character has been already counted as
                // idea here is to check if in last count of
                // all characters in last is zero which
                // means all characters in String a are
                // present in String b.
                map.put(b.charAt(i),
                        map.get(b.charAt(i)) - 1);
            } else {
                  return false;
            }
        }
        // Extract all keys of HashMap/map
        Set<Character> keys = map.keySet();
        // Loop over all keys and check if all keys are 0.
        // If so it means it is anagram.
        for (Character key : keys) {
            if (map.get(key) != 0) {
                return false;
            }
        }
        // Returning True as all keys are zero
        return true;
    }
    public static void main(String[] args)
    {
        String str1 = "geeksforgeeks";
        String str2 = "forgeeksgeeks";
 
        // Function call
        if (isAnagram(str1, str2))
            System.out.print("The two strings are "
                             + "anagram of each other");
        else
            System.out.print("The two strings are "
                             + "not anagram of each other");
    }
}

Python3




# Python3 implementation of the approach
# Function that returns True if a and b
# are anagarams of each other
def isAnagram(a, b):
     
    # Check if length of both strings is same or not
    if (len(a) != len(b)):
        return False
     
    # Create a HashMap containing Character as Key and
    # Integer as Value. We will be storing character as
    # Key and count of character as Value.
    map = {}
    # Loop over all character of String a and put in
    # HashMap.
    for i in range(len(a)):
        # Check if HashMap already contain current
        # character or not
        if (a[i] in map):
            # If contains increase count by 1 for that
            # character
            map[a[i]] += 1
             
        else:
            # else set that character in map and set
            # count to 1 as character is encountered
            # first time
            map[a[i]] = 1
         
    # Now loop over String b
    for i in range(len(b)):
        # Check if current character already exists in
        # HashMap/map
        if (b[i] in map):
            # If contains reduce count of that
            # character by 1 to indicate that current
            # character has been already counted as
            # idea here is to check if in last count of
            # all characters in last is zero which
            # means all characters in String a are
            # present in String b.
            map[b[i]] -= 1
        else:
            return False
                 
         
    # Extract all keys of HashMap/map
    keys = map.keys()
    # Loop over all keys and check if all keys are 0.
    # If so it means it is anagram.
    for key in keys:
        if (map[key] != 0):
            return  False
     
    # Returning True as all keys are zero
    return True
 
# Driver code
str1 = "geeksforgeeks"
str2 = "forgeeksgeeks"
  
# Function call
if (isAnagram(str1, str2)):
    print("The two strings are anagram of each other")
else:
    print("The two strings are not anagram of each other")
 
 
# This code is contributed by shinjanpatra

C#




using System;
using System.Collections.Generic;
 
public class GFG {
  public static bool isAnagram(String a, String b)
  {
 
    // Check if length of both strings is same or not
    if (a.Length != b.Length) {
      return false;
    }
 
    // Create a Dictionary containing char as Key and
    // int as Value. We will be storing character as
    // Key and count of character as Value.
    Dictionary<char, int> map = new Dictionary<char, int>();
 
    // Loop over all character of String a and put in
    // Dictionary.
    for (int i = 0; i < a.Length; i++)
    {
 
      // Check if Dictionary already contain current
      // character or not
      if (map.ContainsKey(a[i]))
      {
 
        // If contains increase count by 1 for that
        // character
        map[a[i]] =
          map[a[i]] + 1;
      }
      else
      {
 
        // else put that character in map and set
        // count to 1 as character is encountered
        // first time
        map.Add(a[i], 1);
      }
    }
 
    // Now loop over String b
    for (int i = 0; i < b.Length; i++)
    {
 
      // Check if current character already exists in
      // Dictionary/map
      if (map.ContainsKey(b[i]))
      {
 
        // If contains reduce count of that
        // character by 1 to indicate that current
        // character has been already counted as
        // idea here is to check if in last count of
        // all characters in last is zero which
        // means all characters in String a are
        // present in String b.
        map[b[i]]=
          map[b[i]] - 1;
      }
      else {
        return false;
      }
    }
 
    // Extract all keys of Dictionary/map
    var keys = map.Keys;
 
    // Loop over all keys and check if all keys are 0.
    // If so it means it is anagram.
    foreach (char key in keys) {
      if (map[key] != 0) {
        return false;
      }
    }
 
    // Returning True as all keys are zero
    return true;
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    String str1 = "geeksforgeeks";
    String str2 = "forgeeksgeeks";
 
    // Function call
    if (isAnagram(str1, str2))
      Console.Write("The two strings are "
                    + "anagram of each other");
    else
      Console.Write("The two strings are "
                    + "not anagram of each other");
  }
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
 
// JavaScript implementation of the approach
// Function that returns true if a and b
// are anagarams of each other
 
function isAnagram(a, b)
{
     
    // Check if length of both strings is same or not
        if (a.length != b.length) {
            return false;
        }
        // Create a HashMap containing Character as Key and
        // Integer as Value. We will be storing character as
        // Key and count of character as Value.
        let map = new Map();
        // Loop over all character of String a and put in
        // HashMap.
        for (let i = 0; i < a.length; i++) {
            // Check if HashMap already contain current
            // character or not
            if (map.has(a[i])) {
                // If contains increase count by 1 for that
                // character
                map.set(a[i],
                        map.get(a[i]) + 1);
            }
            else {
                // else set that character in map and set
                // count to 1 as character is encountered
                // first time
                map.set(a[i], 1);
            }
        }
        // Now loop over String b
        for (let i = 0; i < b.length; i++) {
            // Check if current character already exists in
            // HashMap/map
            if (map.has(b[i])) {
                // If contains reduce count of that
                // character by 1 to indicate that current
                // character has been already counted as
                // idea here is to check if in last count of
                // all characters in last is zero which
                // means all characters in String a are
                // present in String b.
                map.set(b[i],
                        map.get(b[i]) - 1);
            }
            else {
                return false;
            }
        }
        // Extract all keys of HashMap/map
        let keys = map.keys();
        // Loop over all keys and check if all keys are 0.
        // If so it means it is anagram.
        for (let key of keys) {
            if (map.get(key) != 0) {
                return false;
            }
        }
        // Returning True as all keys are zero
        return true;
    }
 
// Driver code
let str1 = "geeksforgeeks";
let str2 = "forgeeksgeeks";
  
// Function call
if (isAnagram(str1, str2))
    document.write("The two strings are anagram of each other");
else
    document.write("The two strings are not anagram of each other");
 
 
// This code is contributed by shinjanpatra
 
</script>

Output

The two strings are anagram of each other

Time Complexity : O(n)

Space Complexity : O(1) (constant space use)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


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