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++
// 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
// 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[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 |
Javascript
<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> |
Output:
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.