Check if two arrays are equal or not

Difficulty Level : Easy
Last Updated : 08 Feb, 2021

Given two given arrays of equal length, the task is to find if given arrays are equal or not. Two arrays are said to be equal if both of them contain the same set of elements, arrangements (or permutation) of elements may be different though.

Note: If there are repetitions, then counts of repeated elements must also be the same for two arrays to be equal.

Examples :

Input  : arr1[] = {1, 2, 5, 4, 0};
         arr2[] = {2, 4, 5, 0, 1}; 
Output : Yes

Input  : arr1[] = {1, 2, 5, 4, 0, 2, 1};
         arr2[] = {2, 4, 5, 0, 1, 1, 2}; 
Output : Yes
 
Input : arr1[] = {1, 7, 1};
        arr2[] = {7, 7, 1};
Output : No

A simple solution is to sort both arrays and then linearly compare elements.

C++

// C++ program to find given two array
// are equal or not
#include <bits/stdc++.h>
using namespace std;
 
// Returns true if arr1[0..n-1] and arr2[0..m-1]
// contain same elements.
bool areEqual(int arr1[], int arr2[], int n, int m)
{
    // If lengths of array are not equal means
    // array are not equal
    if (n != m)
        return false;
 
    // Sort both arrays
    sort(arr1, arr1 + n);
    sort(arr2, arr2 + m);
 
    // Linearly compare elements
    for (int i = 0; i < n; i++)
        if (arr1[i] != arr2[i])
            return false;
 
    // If all elements were same.
    return true;
}
 
// Driver Code
int main()
{
    int arr1[] = { 3, 5, 2, 5, 2 };
    int arr2[] = { 2, 3, 5, 5, 2 };
    int n = sizeof(arr1) / sizeof(int);
    int m = sizeof(arr2) / sizeof(int);
 
    if (areEqual(arr1, arr2, n, m))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program to find given two array
// are equal or not
import java.io.*;
import java.util.*;
 
class GFG {
    // Returns true if arr1[0..n-1] and arr2[0..m-1]
    // contain same elements.
    public static boolean areEqual(int arr1[], int arr2[])
    {
        int n = arr1.length;
        int m = arr2.length;
 
        // If lengths of array are not equal means
        // array are not equal
        if (n != m)
            return false;
 
        // Sort both arrays
        Arrays.sort(arr1);
        Arrays.sort(arr2);
 
        // Linearly compare elements
        for (int i = 0; i < n; i++)
            if (arr1[i] != arr2[i])
                return false;
 
        // If all elements were same.
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr1[] = { 3, 5, 2, 5, 2 };
        int arr2[] = { 2, 3, 5, 5, 2 };
 
        if (areEqual(arr1, arr2))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python3

# Python3 program to find given
# two array are equal or not
 
# Returns true if arr1[0..n-1] and
# arr2[0..m-1] contain same elements.
 
 
def areEqual(arr1, arr2, n, m):
 
    # If lengths of array are not
    # equal means array are not equal
    if (n != m):
        return False
 
    # Sort both arrays
    arr1.sort()
    arr2.sort()
 
    # Linearly compare elements
    for i in range(0, n - 1):
        if (arr1[i] != arr2[i]):
            return False
 
    # If all elements were same.
    return True
 
 
# Driver Code
arr1 = [3, 5, 2, 5, 2]
arr2 = [2, 3, 5, 5, 2]
n = len(arr1)
m = len(arr2)
 
if (areEqual(arr1, arr2, n, m)):
    print("Yes")
else:
    print("No")
 
# This code is contributed
# by Shivi_Aggarwal.

C#

// C# program to find given two array
// are equal or not
using System;
 
class GFG {
 
    // Returns true if arr1[0..n-1] and
    // arr2[0..m-1] contain same elements.
    public static bool areEqual(int[] arr1, int[] arr2)
    {
        int n = arr1.Length;
        int m = arr2.Length;
 
        // If lengths of array are not
        // equal means array are not equal
        if (n != m)
            return false;
 
        // Sort both arrays
        Array.Sort(arr1);
        Array.Sort(arr2);
 
        // Linearly compare elements
        for (int i = 0; i < n; i++)
            if (arr1[i] != arr2[i])
                return false;
 
        // If all elements were same.
        return true;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr1 = { 3, 5, 2, 5, 2 };
        int[] arr2 = { 2, 3, 5, 5, 2 };
 
        if (areEqual(arr1, arr2))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by anuj_67.

PHP

<?php
// PHP program to find given 
// two array are equal or not
 
// Returns true if arr1[0..n-1]
// and arr2[0..m-1] contain same elements.
function areEqual( $arr1, $arr2, $n, $m)
{
    // If lengths of array 
    // are not equal means
    // array are not equal
    if ($n != $m)
        return false;
 
    // Sort both arrays
    sort($arr1);
    sort($arr2);
 
    // Linearly compare elements
    for ( $i = 0; $i < $n; $i++)
        if ($arr1[$i] != $arr2[$i])
            return false;
 
    // If all elements were same.
    return true;
}
 
// Driver Code
$arr1 = array( 3, 5, 2, 5, 2);
$arr2 = array( 2, 3, 5, 5, 2);
$n = count($arr1);
$m = count($arr2);
 
if (areEqual($arr1, $arr2, $n, $m))
    echo "Yes";
else
    echo "No";
 
// This code is contributed by anuj_67.
?>

Output

Yes

Time Complexity: O(n log n)
Auxiliary Space: O(1)

An Efficient Solution to this approach is to use hashing. We store all elements of arr1[] and their counts in a hash table. Then we traverse arr2[] and check if the count of every element in arr2[] matches with the count in arr1[].

Below is the implementation of the above idea. We use unordered_map to store counts.

C++

// C++ program to find given two array
// are equal or not using hashing technique
#include <bits/stdc++.h>
using namespace std;
 
// Returns true if arr1[0..n-1] and arr2[0..m-1]
// contain same elements.
bool areEqual(int arr1[], int arr2[], int n, int m)
{
    // If lengths of arrays are not equal
    if (n != m)
        return false;
 
    // Store arr1[] elements and their counts in
    // hash map
    unordered_map<int, int> mp;
    for (int i = 0; i < n; i++)
        mp[arr1[i]]++;
 
    // Traverse arr2[] elements and check if all
    // elements of arr2[] are present same number
    // of times or not.
    for (int i = 0; i < n; i++) {
        // If there is an element in arr2[], but
        // not in arr1[]
        if (mp.find(arr2[i]) == mp.end())
            return false;
 
        // If an element of arr2[] appears more
        // times than it appears in arr1[]
        if (mp[arr2[i]] == 0)
            return false;
 
        mp[arr2[i]]--;
    }
 
    return true;
}
 
// Driver Code
int main()
{
    int arr1[] = { 3, 5, 2, 5, 2 };
    int arr2[] = { 2, 3, 5, 5, 2 };
    int n = sizeof(arr1) / sizeof(int);
    int m = sizeof(arr2) / sizeof(int);
 
    if (areEqual(arr1, arr2, n, m))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program to find given two array
// are equal or not using hashing technique
import java.util.*;
import java.io.*;
 
class GFG {
    // Returns true if arr1[0..n-1] and arr2[0..m-1]
    // contain same elements.
    public static boolean areEqual(int arr1[], int arr2[])
    {
        int n = arr1.length;
        int m = arr2.length;
 
        // If lengths of arrays are not equal
        if (n != m)
            return false;
 
        // Store arr1[] elements and their counts in
        // hash map
        Map<Integer, Integer> map
            = new HashMap<Integer, Integer>();
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (map.get(arr1[i]) == null)
                map.put(arr1[i], 1);
            else {
                count = map.get(arr1[i]);
                count++;
                map.put(arr1[i], count);
            }
        }
 
        // Traverse arr2[] elements and check if all
        // elements of arr2[] are present same number
        // of times or not.
        for (int i = 0; i < n; i++) {
            // If there is an element in arr2[], but
            // not in arr1[]
            if (!map.containsKey(arr2[i]))
                return false;
 
            // If an element of arr2[] appears more
            // times than it appears in arr1[]
            if (map.get(arr2[i]) == 0)
                return false;
 
            count = map.get(arr2[i]);
            --count;
            map.put(arr2[i], count);
        }
 
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr1[] = { 3, 5, 2, 5, 2 };
        int arr2[] = { 2, 3, 5, 5, 2 };
 
        if (areEqual(arr1, arr2))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python3

# Python3 program to find if given
# two arrays are equal or not
# using dictionary
from collections import defaultdict
 
# Returns true if arr1[0..n-1] and
# arr2[0..m-1] contain same elements.
 
 
def areEqual(arr1, arr2, n, m):
 
    # If lengths of array are not
    # equal means array are not equal
    if (n != m):
        return False
 
    # Create a defaultdict count to
    # store counts
    count = defaultdict(int)
 
    # Store the elements of arr1
    # and their counts in the dictionary
    for i in arr1:
        count[i] += 1
 
    # Traverse through arr2 and compare
    # the elements and its count with
    # the elements of arr1
    for i in arr2:
 
        # Return false if the elemnent
        # is not in arr2 or if any element
        # appears more no. of times than in arr1
        if (count[i] == 0):
            return False
 
        # If element is found, decrement
        # its value in the dictionary
        else:
            count[i] -= 1
 
    # Return true if both arr1 and
    # arr2 are equal
    return True
 
 
# Driver Code
arr1 = [3, 5, 2, 5, 2]
arr2 = [2, 3, 5, 5, 2]
 
n = len(arr1)
m = len(arr2)
 
if areEqual(arr1, arr2, n, m):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Karthik_Aravind

C#

// C# program to find given two array
// are equal or not using hashing technique
using System;
using System.Collections.Generic;
 
class GFG {
    // Returns true if arr1[0..n-1] and arr2[0..m-1]
    // contain same elements.
    public static bool areEqual(int[] arr1, int[] arr2)
    {
        int n = arr1.Length;
        int m = arr2.Length;
 
        // If lengths of arrays are not equal
        if (n != m)
            return false;
 
        // Store arr1[] elements and their counts in
        // hash map
        Dictionary<int, int> map
            = new Dictionary<int, int>();
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (!map.ContainsKey(arr1[i]))
                map.Add(arr1[i], 1);
            else {
                count = map[arr1[i]];
                count++;
                map.Remove(arr1[i]);
                map.Add(arr1[i], count);
            }
        }
 
        // Traverse arr2[] elements and check if all
        // elements of arr2[] are present same number
        // of times or not.
        for (int i = 0; i < n; i++) {
            // If there is an element in arr2[], but
            // not in arr1[]
            if (!map.ContainsKey(arr2[i]))
                return false;
 
            // If an element of arr2[] appears more
            // times than it appears in arr1[]
            if (map[arr2[i]] == 0)
                return false;
 
            count = map[arr2[i]];
            --count;
 
            if (!map.ContainsKey(arr2[i]))
                map.Add(arr2[i], count);
        }
        return true;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr1 = { 3, 5, 2, 5, 2 };
        int[] arr2 = { 2, 3, 5, 5, 2 };
 
        if (areEqual(arr1, arr2))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
/* This code contributed by PrinciRaj1992 */

Output

Yes

Time Complexity: O(n)
Auxiliary Space: O(n)

An Alternate Solution without comparing each element of the arrays and without using unordered_map (by using XOR). This approach will work only if each element exist only once in an array. For example : array a : { 3 , 3 } and array b : { 5 , 5 }, xor_of_array_a(say b1) = 0 and xor_of_array_b = 0 (say b2) and b1^b2 = 0, but array a and array b are not equal.

C++14

// C++ program to find given two array
// are equal or not
#include <bits/stdc++.h>
using namespace std;
 
// Returns true if arr1[0..n-1] and arr2[0..m-1]
// contain same elements.
bool areEqual(int arr1[], int arr2[], int n, int m)
{
    // If lengths of array are not equal means
    // array are not equal
    if (n != m)
        return false;
     
    // to store xor of both arrays
    int b1 = arr1[0];
    int b2 = arr2[0];
     
    // find xor of each elements in array
    for (int i = 1; i < n; i++) {
        b1 ^= arr1[i];
    }
 
    for (int i = 1; i < m; i++) {
        b2 ^= arr2[i];
    }
    int all_xor = b1 ^ b2;
     
    // if xor is zero means they are equal (5^5=0)
    if (all_xor == 0)
        return true;
     
    // If all elements were not same, then xor will not be
    // zero
    return false;
}
 
// Driver Code
int main()
{
    int arr1[] = { 3, 6, 7, 5, 2 };
    int arr2[] = { 2, 3, 5, 6, 7 };
    int n = sizeof(arr1) / sizeof(int);
    int m = sizeof(arr2) / sizeof(int);
 
    // Function call
    if (areEqual(arr1, arr2, n, m))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program to find given two array
// are equal or not 
import java.io.*;
import java.util.*;
 
class GFG{
     
// Returns true if arr1[0..n-1] and arr2[0..m-1]
// contain same elements.
public static boolean areEqual(int arr1[],
                               int arr2[])
{
     
    // Length of the two array
    int n = arr1.length;
    int m = arr2.length;
     
    // If lengths of arrays are not equal
    if (n != m)
        return false;
         
    // To store xor of both arrays
    int b1 = arr1[0];
    int b2 = arr2[0];
     
    // Find xor of each elements in array
    for(int i = 1; i < n; i++) 
    {
        b1 ^= arr1[i];
    }
    for(int i = 1; i < m; i++) 
    {
        b2 ^= arr2[i];
    }
    int all_xor = b1 ^ b2;
     
    // If xor is zero means they are 
    // equal (5^5=0)
    if (all_xor == 0)
        return true;
         
    // If all elements were not same,
    // then xor will not be zero
    return false;
}
 
// Driver code
public static void main(String[] args)
{
    int arr1[] = { 3, 5, 2, 5, 2 };
    int arr2[] = { 2, 3, 5, 5, 2 };
     
    // Function call
    if (areEqual(arr1, arr2))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by sayantanbose2001

Python3

# Python3 program to find given
# two array are equal or not
 
# Returns true if arr1[0..n-1] and
# arr2[0..m-1] contain same elements.
 
def areEqual(arr1, arr2, n, m):
 
    # If lengths of array are not
    # equal means array are not equal
    if (n != m):
        return False
    b1 = arr1[0]
    b2 = arr2[0]
     
    # find xor of all elements
    for i in range(1, n - 1):
        b1 ^= arr1[i]
 
    for i in range(1, m - 1):
        b2 ^= arr2[i]
 
    all_xor = b1 ^ b2
     
    # If all elements were same then xor will be zero
    if(all_xor == 0):
        return True
 
    return False
 
 
# Driver Code
arr1 = [3, 5, 2, 5, 2]
arr2 = [2, 3, 5, 5, 2]
n = len(arr1)
m = len(arr2)
 
# Function call
if (areEqual(arr1, arr2, n, m)):
    print("Yes")
else:
    print("No")

Output

Yes

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

This article is contributed by DANISH_RAZA. 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 write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Improved By: vt_m, Shivi_Aggarwal, imrohan, princiraj1992, karthikaravindt88, anupriyanishad

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Related Articles

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

C++14

Java

Python3