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++ 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
Time Complexity: O(n * log n) //the time complexity for using inbuilt sort function is O(N*logN).
Auxiliary Space: O(1), as constant space is used.