# C++ Program to check if two Arrays are Equal or not

• Difficulty Level : Hard
• Last Updated : 05 Jul, 2022

Given two arrays arr1[] and arr2[] of length N and M respectively, the task is to check if the two arrays are equal or not.

Note: Arrays are said to be equal if and only if both arrays contain the same elements and the frequencies of each element in both arrays are the same.

Examples:

Input: arr1[] = {1, 2, 3, 4, 5}, arr2[] = {5, 4, 3, 2, 1}
Output : Equal
Explanation: As both the arrays contain same elements.

Input: arr1[] = {1, 5, 2, 7, 3, 8}, arr2[] = {8, 2, 3, 5, 7, 1}
Output : Equal

Naive Approach: The basic way to solve the problem is as follows:

Apply sorting on both the arrays and then match each element of one array with the element at same index of the other array.

Follow the steps mentioned below to implement the idea.

• Check if the length of both the arrays are equal or not
• Then sort both the arrays, so that we can compare every equal element.
• Linearly iterate over both the arrays and check if the elements are equal or not,
• If equal then print Equal and if not then print Not equal.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement the approach` `#include ``using` `namespace` `std;` `// Function to check if both arrays are equal``bool` `checkArrays(``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 elements are same``    ``return` `true``;``}` `// Driver Code``int` `main()``{``    ``int` `arr1[] = { 1, 2, 3, 4, 5 };``    ``int` `arr2[] = { 5, 4, 3, 2, 1 };``    ``int` `N = ``sizeof``(arr1) / ``sizeof``(``int``);``    ``int` `M = ``sizeof``(arr2) / ``sizeof``(``int``);` `    ``// Function call``    ``if` `(checkArrays(arr1, arr2, N, M))``        ``cout << ``"Equal"``;``    ``else``        ``cout << ``"Not Equal"``;``    ``return` `0;``}`

Output

`Equal`

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

Efficient Approach: The problem can be solved efficiently using Hashing (unordered_map),

Use hashing to count the frequency of each element of both arrays. Then traverse through the hashtables of the arrays and match the frequencies of the elements.

Follow the below mentioned steps to solve the problem:

• Check if the length of both the arrays are equal or not
• Create an unordered map and store all elements and frequency of elements of arr1[] in the map.
• Traverse over arr2[] and check if count of each and every element in arr2[] matches with the count in arr1[]. This can be done by decrementing the frequency while traversing in arr2[].

Below is the implementation of the above approach:

## C++

 `// C++ program to implement the approach` `#include ``using` `namespace` `std;` `// Function to check if both arrays are equal``bool` `checkArrays(``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 frequencies 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 Code``int` `main()``{``    ``int` `arr1[] = { 1, 2, 3, 4, 5 };``    ``int` `arr2[] = { 4, 3, 1, 5, 2 };``    ``int` `N = ``sizeof``(arr1) / ``sizeof``(``int``);``    ``int` `M = ``sizeof``(arr2) / ``sizeof``(``int``);` `    ``// Function call``    ``if` `(checkArrays(arr1, arr2, N, M))``        ``cout << ``"Equal"``;``    ``else``        ``cout << ``"Not Equal"``;``    ``return` `0;``}`

Output

`Equal`

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

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