Check whether two strings are anagram of each other

Difficulty Level : Easy
Last Updated : 23 Nov, 2020

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

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Method 1 (Use Sorting)

Sort both strings
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 lenghts 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

# Python program to check whether two strings are
# anagrams of each other
 
# function to check whether two strings are anagram
# of each other
 
 
def areAnagram(str1, str2):
    # Get lengths of both strings
    n1 = len(str1)
    n2 = len(str2)
 
    # If lenght of both strings is not same, then
    # they cannot be anagram
    if n1 != n2:
        return 0
 
    # Sort both strings
    str1 = sorted(str1)
    str2 = sorted(str2)
 
    # Compare sorted strings
    for i in range(0, n1):
        if str1[i] != str2[i]:
            return 0
 
    return 1
 
 
# Driver code
str1 = "test"
str2 = "ttew"
 
# 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 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 lenghts 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 initalize new ArrayList
        ArrayList str1 = new ArrayList();
        str1.Add('t');
        str1.Add('e');
        str1.Add('s');
        str1.Add('t');
        // create and initalize 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

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.

Create count arrays of size 256 for both strings. Initialize all values in count arrays as 0.
Iterate through every character of both strings and increment the count of character in the corresponding count arrays.
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

Output:

The two strings are anagram of each other

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;
 
    // 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']--;
    }
 
    // 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;
 
    // 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

Time Complexity: O(n)

Method 4 (Taking sum)

The problem can be Done in Linear time and constant space.

We initialize a variable say count to 0.
Then we take the sum of all the characters of the first String and then decreasing the value of all the characters from the second String.
If the Count value finally is 0, i.e. before performing any operation then its an anagram, else it is not.

Below is the implementation of the above approach:

C++

// C++ program to check if two strings
// are anagrams of each other
#include <bits/stdc++.h>
using namespace std;
 
bool isAnagram(string c, string d)
{
    if (c.size() != d.size())
        return false;
    int count = 0;
 
    // Take sum of all characters of first String
    for (int i = 0; i < c.size(); i++) {
        count += c[i];
    }
 
    // Subtract the Value of all the characters of second
    // String
    for (int i = 0; i < d.size(); i++) {
        count -= d[i];
    }
 
    // If Count = 0 then they are anagram
    // If count > 0 or count < 0 then they are not anagram
    return (count == 0);
}
 
// Driver code
int main()
{
    char str1[] = "geeksforgeeks";
    char str2[] = "forgeeksgeeks";
   
    // Function call
    if (isAnagram(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 boolean isAnagram(String c, String d)
{
    if (c.length() != d.length())
        return false;
         
    int count = 0;
 
    // Take sum of all characters of
    // first String
    for(int i = 0; i < c.length(); i++)
    {
        count = count + c.charAt(i);
    }
 
    // Subtract the Value of all the characters
    // of second String
    for(int i = 0; i < d.length(); i++)
    {
        count = count - d.charAt(i);
    }
 
    // If Count = 0 then they are anagram
    // If count > 0 or count < 0 then 
    // they are not anagram
    return (count == 0);
}
 
// Driver code
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");
}
}
 
// This code is contributed by mark_85

C#

// C# program to check if two strings
// are anagrams of each other
using System.Collections.Generic; 
using System; 
 
class GFG{
 
static bool isAnagram(string c, string d)
{
    if (c.Length != d.Length)
        return false;
         
    int count = 0;
     
    // Take sum of all characters of
    // first String
    for(int i = 0; i < c.Length; i++)
    {
        count = count + c[i];
    }
 
    // Subtract the Value of all the
    // characters of second String
    for(int i = 0; i < d.Length; i++)
    {
        count = count - d[i];
    }
 
    // If Count = 0 then they are anagram
    // If count > 0 or count < 0 then 
    // they are not anagram
    return (count == 0);
}
 
// Driver code
public static void Main()
{
    string str1 = "geeksforgeeks";
    string str2 = "forgeeksgeeks";
 
    // Function call
    if (isAnagram(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 Stream_Cipher

Output

The two strings are anagram of each other

Time Complexity: O(N)
Auxiliary Space: O(1)

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