Comprueba si se pueden formar N rectángulos de igual área a partir de (4 * N) números enteros

Dado un número entero N y una array arr[] de tamaño 4 * N , la tarea es verificar si se pueden formar N rectángulos de igual área a partir de esta array si cada elemento se puede usar solo una vez.

Ejemplos: 

Entrada: arr[] = {1, 8, 2, 1, 2, 4, 4, 8}, N = 2 
Salida: Sí 
Dos rectángulos con lados (1, 8, 1, 8) y (2, 4, 2 , 4) se pueden formar. 
Ambos rectángulos tienen la misma área. 

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

Acercarse:  

• Se necesitan cuatro lados para formar un rectángulo.
• Dados 4 * N enteros, los N rectángulos máximos se pueden formar usando números solo una vez.
• La tarea es verificar si las áreas de todos los rectángulos son iguales. Para verificar esto, primero se ordena la array.
• Los lados se consideran como los dos primeros elementos y los dos últimos elementos.
• El área se calcula y se comprueba si tiene la misma área que el área calculada inicialmente.

A continuación se muestra la implementación del enfoque anterior:  

// 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;
}

// 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

# 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# 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

<script>
 
// Javascript implementation of the approach
 
// Function to check whether we can make n
// rectangles of equal area
function checkRectangles(arr, n)
{
    let ans = true;
 
    // Sort the array
    arr.sort();
 
    // Find the area of any one rectangle
    var 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(let 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
var arr = [ 1, 8, 2, 1, 2, 4, 4, 8 ];
var n = 2;
 
if (checkRectangles(arr, n))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by gauravrajput1
 
</script>

Yes

Complejidad de tiempo: O(n * log n)

Espacio Auxiliar: O(1)