Check if two arrays are equal or not

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 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 same for two array 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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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

C/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"); 
    } 
} 

Python 3

# 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 of 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 count of every element in arr2[] matches with count in arr1[].

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

C/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"); 
    } 
} 

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)

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.

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C/C++

Java

Python 3

C#

PHP

C/C++

Java

C#

Recommended Posts: