Check if two arrays are permutations of each other

Difficulty Level : Easy
Last Updated : 25 Aug, 2021

Given two unsorted arrays of the same size, write a function that returns true if two arrays are permutations of each other, otherwise false.

Examples:

Input: arr1[] = {2, 1, 3, 5, 4, 3, 2}
       arr2[] = {3, 2, 2, 4, 5, 3, 1}
Output: Yes

Input: arr1[] = {2, 1, 3, 5,}
       arr2[] = {3, 2, 2, 4}
Output: No

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A Simple Solution is to sort both arrays and compare sorted arrays. The time complexity of this solution is O(nLogn)
A Better Solution is to use Hashing.

Create a Hash Map for all the elements of arr1[] such that array elements are keys and their counts are values.
Traverse arr2[] and search for each element of arr2[] in the Hash Map. If an element is found then decrement its count in the hash map. If not found, then return false.
If all elements are found then return true.

Below is the implementation of this approach.

C++

// A C++ program to find one array is permutation of other array
#include <bits/stdc++.h>
using namespace std;
 
// Returns true if arr1[] and arr2[] are permutations of each other
bool arePermutations(int arr1[], int arr2[], int n, int m)
{
     
    // Arrays cannot be permutations of one another unless they are
    // of the same length
    if(n != m)
    {
        return false;
    }
   
    // Creates an empty hashMap hM
    map<int, int> hm;
     
    // Traverse through the first array and add elements to hash map
    for (int i = 0; i < n; i++)
    {
        int x = arr1[i];
        hm[x]++;
    }
     
    // Traverse through second array and check if every element is
    // present in hash map
    for (int i = 0; i < m; i++)
    {
        int x = arr2[i];
       
        // If element is not present in hash map or element
        // is not present less number of times
        if(hm[x] == 0)
        {
            return false;
        }
        hm[x]--;
    }
    return true;
}
 
// Driver function to test above function
int main() {
    int arr1[] = {2, 1, 3, 5, 4, 3, 2};
    int arr2[] = {3, 2, 2, 4, 5, 3, 1};
    int n = sizeof(arr1)/sizeof(arr1[0]);
    int m = sizeof(arr2)/sizeof(arr2[0]);
    if (arePermutations(arr1, arr2, n, m))
        cout << "Arrays are permutations of each other" << endl;
    else
        cout << "Arrays are NOT permutations of each other" << endl;
     
    return 0;
}
 
// This code is contributed by avanitrachhadiya2155

Java

// A Java program to find one array is permutation of other array
import java.util.HashMap;
 
class Permutations
{
    // Returns true if arr1[] and arr2[] are permutations of each other
    static Boolean arePermutations(int arr1[], int arr2[])
    {
        // Arrays cannot be permutations of one another unless they are
        // of the same length
        if (arr1.length != arr2.length)
            return false;
       
        // Creates an empty hashMap hM
        HashMap<Integer, Integer> hM = new HashMap<Integer, Integer>();
 
        // Traverse through the first array and add elements to hash map
        for (int i = 0; i < arr1.length; i++)
        {
            int x = arr1[i];
            if (hM.get(x) == null)
                hM.put(x, 1);
            else
            {
                int k = hM.get(x);
                hM.put(x, k+1);
            }
        }
 
        // Traverse through second array and check if every element is
        // present in hash map
        for (int i = 0; i < arr2.length; i++)
        {
            int x = arr2[i];
 
            // If element is not present in hash map or element
            // is not present less number of times
            if (hM.get(x) == null || hM.get(x) == 0)
                return false;
 
            int k = hM.get(x);
            hM.put(x, k-1);
        }
        return true;
    }
 
    // Driver function to test above function
    public static void main(String arg[])
    {
        int arr1[] = {2, 1, 3, 5, 4, 3, 2};
        int arr2[] = {3, 2, 2, 4, 5, 3, 1};
        if (arePermutations(arr1, arr2))
            System.out.println("Arrays are permutations of each other");
        else
            System.out.println("Arrays are NOT permutations of each other");
    }
}

Python3

# Python3 program to find one
# array is permutation of other array
from collections import defaultdict
 
# Returns true if arr1[] and
# arr2[] are permutations of
# each other
def arePermutations(arr1, arr2):
    # Arrays cannot be permutations of one another
    # unless they are of the same length
    if (len(arr1) != len(arr2)):
        return False
       
    # Creates an empty hashMap hM
    hM = defaultdict (int)
 
    # Traverse through the first
    # array and add elements to
    # hash map
    for i in range (len(arr1)):
         
        x = arr1[i]
        hM[x] += 1
         
    # Traverse through second array
    # and check if every element is
    # present in hash map
    for i in range (len(arr2)):
        x = arr2[i]
 
        # If element is not present
        # in hash map or element
        # is not present less number
        # of times
        if x not in hM or hM[x] == 0:
             return False
 
        hM[x] -= 1
        
    return True
 
# Driver code
if __name__ == "__main__":
   
    arr1 = [2, 1, 3, 5, 4, 3, 2]
    arr2 = [3, 2, 2, 4, 5, 3, 1]
    if (arePermutations(arr1, arr2)):
        print("Arrays are permutations of each other")
    else:
        print("Arrays are NOT permutations of each other")
         
# This code is contributed by Chitranayal

C#

// C# program to find one array
// is permutation of other array
using System;
using System.Collections.Generic;
 
public class Permutations {
    // Returns true if arr1[] and arr2[]
    // are permutations of each other
    static Boolean arePermutations(int[] arr1, int[] arr2)
    {
        // Arrays cannot be permutations of one another if
        // they are not the the same length
        if (arr1.Length != arr2.Length)
            return false;
 
        // Creates an empty hashMap hM
        Dictionary<int, int> hM = new Dictionary<int, int>();
 
        // Traverse through the first array
        // and add elements to hash map
        for (int i = 0; i < arr1.Length; i++) {
            int x = arr1[i];
            if (!hM.ContainsKey(x))
                hM.Add(x, 1);
            else {
                int k = hM[x];
                hM.Remove(x);
                hM.Add(x, k + 1);
            }
        }
 
        // Traverse through second array and check if every
        // element is present in hash map (and the same
        // number of times)
        for (int i = 0; i < arr2.Length; i++) {
            int x = arr2[i];
 
            // If element is not present in hash map or
            // element is not present the same number of
            // times
            if (!hM.ContainsKey(x))
                return false; // Not present in the hash map
 
            int k = hM[x];
            if (k == 0)
                return false; // Not present the same number
                              // of times
            hM.Remove(x);
            hM.Add(x, k - 1);
        }
        return true;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr1 = { 2, 1, 3, 5, 4, 3, 2 };
        int[] arr2 = { 3, 2, 2, 4, 5, 3, 1 };
        if (arePermutations(arr1, arr2))
            Console.WriteLine("Arrays are permutations of each other");
        else
            Console.WriteLine("Arrays are NOT permutations of each other");
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

<script>
 
// A Javascript program to find one array
/// is permutation of other array
     
    // Returns true if arr1[] and arr2[]
    // are permutations of each other
    function arePermutations(arr1,arr2)
    {
        // Arrays cannot be permutations of
        // one another unless they are
        // of the same length
        if (arr1.length != arr2.length)
            return false;
        
        // Creates an empty hashMap hM
        let hM = new Map();
  
        // Traverse through the first array
        // and add elements to hash map
        for (let i = 0; i < arr1.length; i++)
        {
            let x = arr1[i];
            if (!hM.has(x))
                hM.set(x, 1);
            else
            {
                let k = hM[x];
                hM.set(x, k+1);
            }
        }
  
        // Traverse through second array and
        // check if every element is
        // present in hash map
        for (let i = 0; i < arr2.length; i++)
        {
            let x = arr2[i];
  
            // If element is not present in
            // hash map or element
            // is not present less number of times
            if (!hM.has(x) || hM[x] == 0)
                return false;
  
            let k = hM[x];
            hM.set(x, k-1);
        }
        return true;
    }
     
    // Driver function to test above function
    let arr1=[2, 1, 3, 5, 4, 3, 2];
    let arr2=[3, 2, 2, 4, 5, 3, 1];
    if (arePermutations(arr1, arr2))
        document.write(
        "Arrays are permutations of each other"
        );
    else
        document.write(
        "Arrays are NOT permutations of each other"
        );
     
    // This code is contributed by rag2127
     
</script>

Output

Arrays are permutations of each other

The time complexity of this method is O(n) under the assumption that we have a hash function inserts and finds elements in O(1) time.
This article is contributed by Ravi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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