# Check if N rectangles of equal area can be formed from (4 * N) integers

• Last Updated : 24 Feb, 2022

Given an integer N and an array arr[] of size 4 * N, the task is to check whether N rectangles of equal area can be formed from this array if each element can be used only once.

Examples:

Input: arr[] = {1, 8, 2, 1, 2, 4, 4, 8}, N = 2
Output: Yes
Two rectangles with sides (1, 8, 1, 8) and (2, 4, 2, 4) can be formed.
Both of these rectangles have the same area.

Input: arr[] = {1, 3, 3, 5, 5, 7, 1, 6}, N = 2
Output: No

Approach:

• Four sides are needed to form a rectangle.
• Given 4 * N integers, utmost N rectangles can be formed using numbers only once.
• The task is to check if the areas of all the rectangles are same. To check this, the array is first sorted.
• The sides are considered as the first two elements and the last two elements.
• Area is calculated and checked if it has the same area as the initially calculated area.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to check whether we can make n``// rectangles of equal area``bool` `checkRectangles(``int``* arr, ``int` `n)``{``    ``bool` `ans = ``true``;` `    ``// Sort the array``    ``sort(arr, arr + 4 * n);` `    ``// Find the area of any one rectangle``    ``int` `area = arr * arr[4 * n - 1];` `    ``// Check whether we have two equal sides``    ``// for each rectangle and that area of``    ``// each rectangle formed is the same``    ``for` `(``int` `i = 0; i < 2 * n; i = i + 2) {``        ``if` `(arr[i] != arr[i + 1]``            ``|| arr[4 * n - i - 1] != arr[4 * n - i - 2]``            ``|| arr[i] * arr[4 * n - i - 1] != area) {` `            ``// Update the answer to false``            ``// if any condition fails``            ``ans = ``false``;``            ``break``;``        ``}``    ``}` `    ``// If possible``    ``if` `(ans)``        ``return` `true``;` `    ``return` `false``;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 8, 2, 1, 2, 4, 4, 8 };``    ``int` `n = 2;` `    ``if` `(checkRectangles(arr, n))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Function to check whether we can make n``// rectangles of equal area``static` `boolean` `checkRectangles(``int``[] arr, ``int` `n)``{``    ``boolean` `ans = ``true``;` `    ``// Sort the array``    ``Arrays.sort(arr);` `    ``// Find the area of any one rectangle``    ``int` `area = arr[``0``] * arr[``4` `* n - ``1``];` `    ``// Check whether we have two equal sides``    ``// for each rectangle and that area of``    ``// each rectangle formed is the same``    ``for` `(``int` `i = ``0``; i < ``2` `* n; i = i + ``2``)``    ``{``        ``if` `(arr[i] != arr[i + ``1``] ||``            ``arr[``4` `* n - i - ``1``] != arr[``4` `* n - i - ``2``] ||``            ``arr[i] * arr[``4` `* n - i - ``1``] != area)``        ``{` `            ``// Update the answer to false``            ``// if any condition fails``            ``ans = ``false``;``            ``break``;``        ``}``    ``}` `    ``// If possible``    ``if` `(ans)``        ``return` `true``;` `    ``return` `false``;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``1``, ``8``, ``2``, ``1``, ``2``, ``4``, ``4``, ``8` `};``    ``int` `n = ``2``;` `    ``if` `(checkRectangles(arr, n))``        ``System.out.print(``"Yes"``);``    ``else``        ``System.out.print(``"No"``);``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python implementation of the approach` `# Function to check whether we can make n``# rectangles of equal area``def` `checkRectangles(arr, n):``    ``ans ``=` `True` `    ``# Sort the array``    ``arr.sort()` `    ``# Find the area of any one rectangle``    ``area ``=` `arr[``0``] ``*` `arr[``4` `*` `n ``-` `1``]` `    ``# Check whether we have two equal sides``    ``# for each rectangle and that area of``    ``# each rectangle formed is the same``    ``for` `i ``in` `range``(``0``, ``2` `*` `n, ``2``):``        ``if` `(arr[i] !``=` `arr[i ``+` `1``]``            ``or` `arr[``4` `*` `n ``-` `i ``-` `1``] !``=` `arr[``4` `*` `n ``-` `i ``-` `2``]``            ``or` `arr[i] ``*` `arr[``4` `*` `n ``-` `i ``-` `1``] !``=` `area):` `            ``# Update the answer to false``            ``# if any condition fails``            ``ans ``=` `False``            ``break` `    ``# If possible``    ``if` `(ans):``        ``return` `True` `    ``return` `False` `# Driver code``arr ``=` `[ ``1``, ``8``, ``2``, ``1``, ``2``, ``4``, ``4``, ``8` `]``n ``=` `2` `if` `(checkRectangles(arr, n)):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)` `# This code is contributed by Sanjit_Prasad`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `// Function to check whether we can make n``// rectangles of equal area``static` `bool` `checkRectangles(``int``[] arr, ``int` `n)``{``    ``bool` `ans = ``true``;` `    ``// Sort the array``    ``Array.Sort(arr);` `    ``// Find the area of any one rectangle``    ``int` `area = arr * arr[4 * n - 1];` `    ``// Check whether we have two equal sides``    ``// for each rectangle and that area of``    ``// each rectangle formed is the same``    ``for` `(``int` `i = 0; i < 2 * n; i = i + 2)``    ``{``        ``if` `(arr[i] != arr[i + 1] ||``            ``arr[4 * n - i - 1] != arr[4 * n - i - 2] ||``            ``arr[i] * arr[4 * n - i - 1] != area)``        ``{` `            ``// Update the answer to false``            ``// if any condition fails``            ``ans = ``false``;``            ``break``;``        ``}``    ``}` `    ``// If possible``    ``if` `(ans)``        ``return` `true``;` `    ``return` `false``;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 1, 8, 2, 1, 2, 4, 4, 8 };``    ``int` `n = 2;` `    ``if` `(checkRectangles(arr, n))``        ``Console.Write(``"Yes"``);``    ``else``        ``Console.Write(``"No"``);``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`Yes`

Time Complexity: O(n * log n)

Auxiliary Space: O(1)

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