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Check if two strings are permutation of each other
  • Difficulty Level : Easy
  • Last Updated : 03 Feb, 2021

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.
 

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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
// 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;
}


Java




// 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");
 
}
}


Python3




# 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


C#




// 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


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++




// 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;
}


Java




// 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.


Python




# 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


C#




// 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


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.
 

CPP




// 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;
}


Java




// Java function to check whether two strings are 
// Permutations of each other
static boolean arePermutation(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; 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] != 0)
            return false;
     return true;
}
 
// This code is contributed by divyesh072019.


Python3




# Python3 function to check whether two strings are
# Permutations of each other
def arePermutation(str1, str2):
 
    # 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
    while(str1[i] and str2[i]):
     
        count[str1[i]] += 1
        count[str2[i]] -= 1
 
    # If both strings are of different length.
    # Removing this condition  will make the
    # program fail for strings like "aaca" and
    # "aca"
    if (str1[i] or str2[i]):
      return False;
 
    # See if there is any non-zero value in
    # count array
    for i in range(NO_OF_CHARS):
        if (count[i]):
            return False;
    return True;
 
# This code is contributed by pratham76.


C#




// C# function to check whether two strings are 
// Permutations of each other
static bool arePermutation(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; 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] != 0)
            return false;
     return true;
}
 
// This code is contributed by divyeshrabadiya07.


If the possible set of characters contains only English alphabets, then we can reduce the size of arrays to 58 and use str[i] – ‘A’ as an index for count arrays because ASCII value of ‘A’ is 65 , ‘B’ is 66, ….. , Z is 90 and ‘a’ is 97 , ‘b’ is 98 , …… , ‘z’ is 122. 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.
 

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