Check if two arrays are equal or not

Given two arrays, arr1 and arr2 of equal length N, the task is to find if the 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 permutations) of elements might/might not be same.
• 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

Check if two arrays are equal or not using Sorting

Follow the steps below to solve the problem using this approach:

• Sort both the arrays
• Then linearly compare elements of both the arrays
• If all are equal then return true, else return false

Below is the implementation of the above approach:

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's 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);

    // 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[])
    {
        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 };

        // Function call
        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):
        if (arr1[i] != arr2[i]):
            return False

    # If all elements were same.
    return True


# Driver Code
if __name__ == "__main__":
    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")

C#

// C# program for the above approach
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's code
    public static void Main()
    {
        int[] arr1 = { 3, 5, 2, 5, 2 };
        int[] arr2 = { 2, 3, 5, 5, 2 };

        // Function call
        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);

// Function call
if (areEqual($arr1, $arr2, $N, $M))
    echo "Yes";
else
    echo "No";

?>

Javascript

<script>

// JavaScript program for the above approach

    // Returns true if arr1[0..n-1] and arr2[0..m-1]
    // contain same elements.
    function areEqual(arr1, arr2)
    {
        let N = arr1.length;
        let M = arr2.length;

        // 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 (let i = 0; i < N; i++)
            if (arr1[i] != arr2[i])
                return false;

        // If all elements were same.
        return true;
    }

// Driver Code
    let arr1 = [3, 5, 2, 5, 2];
    let arr2 = [2, 3, 5, 5, 2];

    if (areEqual(arr1, arr2))
        document.write("Yes");
    else
        document.write("No");

</script>

Output

Yes

Time Complexity: O(N*log(N)) 
Auxiliary Space: O(1)

Check if two arrays are equal or not using Hashing

Store count of all elements of arr1[] in a hash table. Then traverse arr2[] and check if the count of every element in arr2[] matches with the count of elements of arr1[].

Follow the steps mentioned below to implement the approach:

• First check if length of arr1 is not equal to the length of arr2 then return false
• Then traverse over first array and store the count of every element in the hash map
• Then traverse over second array and decrease the count of its elements in the hash map. If that element is not present or the count of that element is 
  zero in the hash map, then return false, else decrease the count of that element
• Return true at the end, as both the arrays are equal by now

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#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;
        // decrease the count of arr2 elements in the
        // unordered map
        mp[arr2[i]]--;
    }

    return true;
}

// Driver's 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);

    // Function call
    if (areEqual(arr1, arr2, N, M))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program for the above approach
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 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's 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");
    }
}

Python3

# Python3 program for the above approach

# Returns true if arr1[0..N-1] and
# arr2[0..M-1] contain same elements.


def is_arr_equal(arr1, arr2):
        # Check if the length of arrays are
    # equal or not: A Easy Logic Check
    if len(arr1) != len(arr2):
        return False

    # Create a dict named count to
    # store counts of each element
    count = {}
    # Store the elements of arr1
    # and their counts in the dictionary
    for i in arr1:
        if i in count:
                # Element already in dict, simply increment its count
            count[i] += 1
        else:
                # Element found for first time, initialize it with value 1.
            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 element
        # is not in count or if any element
        # appears more no. of times than in arr1
        if i not in count or count[i] == 0:
            return False
        else:
                # If element is found, decrement
                # its value in the dictionary
            count[i] -= 1
    # Return true if both arr1 and
    # arr2 are equal
    return True


# Driver Code
if __name__ == "__main__":
    arr1 = [3, 5, 2, 5, 2]
    arr2 = [2, 3, 5, 5, 2]

    if is_arr_equal(arr1, arr2):
        print("Yes")
    else:
        print("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's 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))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

/* This code contributed by PrinciRaj1992 */

Javascript

<script>

//  Javascript program to find given two array
// are equal or not using hashing technique

    // Returns true if arr1[0..N-1] and arr2[0..M-1]
    // contain same elements.
    function areEqual(arr1, arr2)
    {
        let N = arr1.length;
        let M = arr2.length;

        // If lengths of arrays are not equal
        if (N != M)
            return false;

        // Store arr1[] elements and their counts in
        // hash map
        let map
            = new Map();
        let count = 0;
        for (let i = 0; i < N; i++) {
            if (map.get(arr1[i]) == null)
                map.set(arr1[i], 1);
            else {
                count = map.get(arr1[i]);
                count++;
                map.set(arr1[i], count);
            }
        }

        // Traverse arr2[] elements and check if all
        // elements of arr2[] are present same number
        // of times or not.
        for (let i = 0; i < N; i++) {
            // If there is an element in arr2[], but
            // not in arr1[]
            if (!map.has(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.set(arr2[i], count);
        }

        return true;
    }

    // Driver code

    let arr1 = [3, 5, 2, 5, 2];
    let arr2 = [2, 3, 5, 5, 2];

    // Function call
        if (areEqual(arr1, arr2))
            document.write("Yes");
        else
            document.write("No");

</script>

Output

Yes

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

Check if two arrays are equal or not using Map

• Initialise a map say unmap
• Insert all elements of array A into map
• Remove all elements of array B from map
• Check if the size of unmap becomes zero
  • If zero, return true
  • Otherwise, return false

Below is the implementation of the above approach:

C++

// CPP program to find Non-overlapping sum
#include <bits/stdc++.h>
using namespace std;

// function for calculating
// Non-overlapping sum of two array
bool isEqual(vector<int> A, vector<int> B)
{

    unordered_map<int, int> unmap;

    // Insert all element of A into map
    for (auto i : A)
        unmap[i]++;

    // Remove all elements of B from map
    for (auto i : B) {
        unmap[i]--;
        if (unmap[i] == 0)
            unmap.erase(i);
    }

    // Check if size of unmap becomes zero
    // If zero, return true
    // Otherwise, return false
    if (unmap.size() == 0)
        return true;
    return false;
}

// Driver code
int main()
{
    vector<int> A = { 5, 4, 9, 2, 3 };
    vector<int> B = { 5, 4, 9, 2, 3 };

    // function call
    cout << isEqual(A, B);
    return 0;
}

// This code is contributed by hkdass001

Java

// Java program to find Non-overlapping sum
import java.util.*;

public class GFG {

  // function for calculating
  // Non-overlapping sum of two array
  static boolean isEqual(int[] A, int[] B)
  {

    HashMap<Integer, Integer> unmap = new HashMap<>();

    // Insert all element of A into map
    for (int i : A)
      unmap.put(i, unmap.getOrDefault(i, 0) + 1);

    // Remove all elements of B from map
    for (int i : B) {
      if (unmap.containsKey(i)) {
        unmap.put(i, unmap.get(i) - 1);
        if (unmap.get(i) == 0)
          unmap.remove(i);
      }
    }

    // Check if size of unmap becomes zero
    // If zero, return true
    // Otherwise, return false
    if (unmap.size() == 0)
      return true;
    return false;
  }

  // Driver code
  public static void main(String[] args)
  {
    int[] A = { 5, 4, 9, 2, 3 };
    int[] B = { 5, 4, 9, 2, 3 };

    // function call
    System.out.println((isEqual(A, B) ? "1" : "0"));
  }
}

// This code is contributed by Karandeep1234

Python3

# Python program to find Non-overlapping sum
from collections import Counter

# function for calculating
# Non-overlapping sum of two array
def isEqual(A, B):
    # Use Counter to count the frequency of elements
    unmap = Counter(A)

    # Remove all elements of B from map
    for i in B:
        unmap[i] -= 1
        if unmap[i] == 0:
            del unmap[i]

    # Check if size of unmap becomes zero
    # If zero, return true
    # Otherwise, return false
    if len(unmap) == 0:
        return True
    return False

# Driver code
A = [5, 4, 9, 2, 3]
B = [5, 4, 9, 2, 3]

# function call
#rint(isEqual(A, B))
print(1 if isEqual(A, B) else 0)

Output

1

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

Check if two arrays are equal or not using Counter Class

We can use the Counter class from the collections module to count the number of occurrences of each element in the arrays and then compare the resulting dictionaries.

Steps:

• Use the Counter class to count the number of occurrences of each element in a and b.
• Use the == operator to compare the resulting Counter objects.
• Return the result of the comparison.
• Import the Counter class from the collections module.

Below is the implementation of the above approach:

Python3

from collections import Counter

def arrays_equal(a, b):
    return Counter(a) == Counter(b)

a = [3, 2, 1, 3, 2, 1]
b = [1, 2, 3, 1, 2, 3]
c = [4, 5, 6, 4, 5, 6]

print(arrays_equal(a, b))  # True
print(arrays_equal(a, c))  # False

# This code is contributed by Susobhan Akhuli

C++

#include <map>
#include <iostream>
#include <algorithm>

bool arrays_equal(std::vector<int> a, std::vector<int> b) {
    std::map<int, int> map_a, map_b;
    for (auto i : a) {
        map_a[i]++;
    }
    for (auto i : b) {
        map_b[i]++;
    }
    return map_a == map_b;
}

int main() {
    std::vector<int> a = {3, 2, 1, 3, 2, 1};
    std::vector<int> b = {1, 2, 3, 1, 2, 3};
    std::vector<int> c = {4, 5, 6, 4, 5, 6};

    std::cout << (arrays_equal(a, b) ? "True" : "False") << std::endl;
    std::cout << (arrays_equal(a, c) ? "True" : "False") << std::endl;
    return 0;
}

Java

import java.util.*;

public class GFG {
    public static boolean arrays_equal(int[] a, int[] b)
    {
        Map<Integer, Integer> aCount = new HashMap<>();
        Map<Integer, Integer> bCount = new HashMap<>();
        for (int i = 0; i < a.length; i++) {
            int count = aCount.getOrDefault(a[i], 0);
            aCount.put(a[i], count + 1);
        }

        for (int i = 0; i < b.length; i++) {
            int count = bCount.getOrDefault(b[i], 0);
            bCount.put(b[i], count + 1);
        }

        return aCount.equals(bCount);
    }

    public static void main(String[] args)
    {
        int[] a = { 3, 2, 1, 3, 2, 1 };
        int[] b = { 1, 2, 3, 1, 2, 3 };
        int[] c = { 4, 5, 6, 4, 5, 6 };
        System.out.println(
            arrays_equal(a, b) ? "True" : "False"); // True
        System.out.println(
            arrays_equal(a, c) ? "True" : "False"); // False
    }
}

// This code is contributed by Susobhan Akhuli

Output

True
False

Time Complexity: O(N) [To count the number of occurrences of each element in the arrays]
Auxiliary Space: O(1)