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

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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[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
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;
}

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Java

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// 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

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Python

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# 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

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C#

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// 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[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))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
  
// This code is contributed by Rajput-Ji

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Output:

Yes


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