Check whether two strings are anagram of each other

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 same characters, only the order of characters can be different. For example, “abcd” and “dabc” are anagram of each other. check-whether-two-strings-are-anagram-of-each-other

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Method 1 (Use Sorting)
1) Sort both strings
2) Compare the sorted strings

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"; 
    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 program to test to print printDups*/
    public static void main(String args[]) 
    { 
        char str1[] = { 't', 'e', 's', 't' }; 
        char str2[] = { 't', 't', 'e', 'w' }; 
        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 program to test the above function  
str1 = "test"
str2 = "ttew"
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'); 
  
        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.
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.

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"; 
    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 program to test to print printDups*/
int main() 
{ 
    char str1[] = "geeksforgeeks"; 
    char str2[] = "forgeeksgeeks"; 
    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 program to test to print printDups*/
    public static void main(String args[]) 
    { 
        char str1[] = ("geeksforgeeks").toCharArray(); 
        char str2[] = ("forgeeksgeeks").toCharArray(); 
  
        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 program to test the above functions 
str1 = "geeksforgeeks"
str2 = "forgeeksgeeks"
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 program to test to print printDups*/
    public static void Main() 
    { 
        char[] str1 = ("geeksforgeeks").ToCharArray(); 
        char[] str2 = ("forgeeksgeeks").ToCharArray(); 
  
        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

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.

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; 
} 

If the possible set of characters contains only English alphabets, then we can reduce the size of arrays to 52 and use str[i] – ‘A’ as an index for count arrays. This will further optimize this method.

Time Complexity: O(n)

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