Check if two strings are permutation of each other

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

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to check whether two strings are 
// Permutations of each other
#include <bits/stdc++.h>
using namespace std;
  
/* function to check whether two strings are 
   Permutation of each other */
bool arePermutation(string str1, string 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 Permutation
    if (n1 != n2)
      return false;
  
    // Sort both 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 program to test to print printDups*/
int main()
{
    string str1 = "test";
    string str2 = "ttew";
    if (arePermutation(str1, str2))
      printf("Yes");
    else
      printf("No");
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to check whether two strings are 
// Permutations of each other 
import java.util.*;
class GfG {
  
/* function to check whether two strings are 
Permutation of each other */
static boolean arePermutation(String str1, String 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 Permutation 
    if (n1 != n2) 
    return false
    char ch1[] = str1.toCharArray();
    char ch2[] = str2.toCharArray();
  
    // Sort both strings 
    Arrays.sort(ch1); 
    Arrays.sort(ch2); 
  
    // Compare sorted strings 
    for (int i = 0; i < n1; i++) 
    if (ch1[i] != ch2[i]) 
        return false
  
    return true
  
/* Driver program to test to print printDups*/
public static void main(String[] args) 
    String str1 = "test"
    String str2 = "ttew"
    if (arePermutation(str1, str2)) 
    System.out.println("Yes"); 
    else
    System.out.println("No"); 
  
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 program to check whether two 
# strings are Permutations of each other
  
# function to check whether two strings 
# are Permutation of each other */
def arePermutation(str1, str2):
      
    # Get lenghts of both strings
    n1 = len(str1)
    n2 = len(str2)
  
    # If length of both strings is not same,
    # then they cannot be Permutation
    if (n1 != n2):
        return False
  
    # Sort both strings
    a = sorted(str1)
    str1 = " ".join(a)
    b = sorted(str2)
    str2 = " ".join(b)
  
    # Compare sorted strings
    for i in range(0, n1, 1):
        if (str1[i] != str2[i]):
            return False
  
    return True
  
# Driver Code
if __name__ == '__main__':
    str1 = "test"
    str2 = "ttew"
    if (arePermutation(str1, str2)):
        print("Yes")
    else:
        print("No")
  
# This code is contributed by
# Sahil_Shelangia

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to check whether two strings are 
// Permutations of each other
using System;
  
class GfG 
{
  
/* function to check whether two strings are 
Permutation of each other */
static bool arePermutation(String str1, String 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 Permutation 
    if (n1 != n2) 
        return false
    char []ch1 = str1.ToCharArray();
    char []ch2 = str2.ToCharArray();
  
    // Sort both strings 
    Array.Sort(ch1); 
    Array.Sort(ch2); 
  
    // Compare sorted strings 
    for (int i = 0; i < n1; i++) 
        if (ch1[i] != ch2[i]) 
            return false
  
    return true
  
/* Driver code*/
public static void Main(String[] args) 
    String str1 = "test"
    String str2 = "ttew"
    if (arePermutation(str1, str2)) 
        Console.WriteLine("Yes"); 
    else
        Console.WriteLine("No"); 
}
}
  
// This code contributed by Rajput-Ji

chevron_right



Output:

No

Time Complexity: Time complexity of this method depends upon the sorting technique used. In the above implementation, quickSort is used which may be O(n^2) in worst case. If we use a O(nLogn) sorting algorithm like merge sort, then the complexity becomes 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to check whether two strings are 
// Permutations of each other
#include <bits/stdc++.h>
using namespace std;
# define NO_OF_CHARS 256
  
/* function to check whether two strings are
   Permutation of each other */
bool arePermutation(string str1, string 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()
{
    string str1 = "geeksforgeeks";
    string str2 = "forgeeksgeeks";
    if ( arePermutation(str1, str2) )
      printf("Yes");
    else
      printf("No");
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// JAVA program to check if two strings
// are Permutations of each other
import java.io.*;
import java.util.*;
  
class GFG{
      
    static int NO_OF_CHARS = 256;
      
    /* function to check whether two strings
    are Permutation of each other */
    static boolean arePermutation(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 ( arePermutation(str1, str2) )
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
  
// This code is contributed by Nikita Tiwari.

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python program to check if two strings are 
# Permutations of each other
NO_OF_CHARS = 256
  
# Function to check whether two strings are
# Permutation of each other
def arePermutation(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 arePermutation(str1, str2):
    print "Yes"
else:
    print "No"
  
# This code is contributed by Bhavya Jain

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to check if two strings
// are Permutations of each other
using System;
class GFG{
      
    static int NO_OF_CHARS = 256;
      
    /* function to check whether two strings
    are Permutation of each other */
    static bool arePermutation(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(String []args)
    {
        char []str1 = ("geeksforgeeks").ToCharArray();
        char []str2 = ("forgeeksgeeks").ToCharArray();
          
        if ( arePermutation(str1, str2) )
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
  
// This code has been contributed by 29AjayKumar

chevron_right



Output:

Yes

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 Permutation of each other. Thanks to Ace for suggesting this optimization.

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ function to check whether two strings are 
// Permutations of each other
bool arePermutation(string str1, string 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;
}

chevron_right


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)

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



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.