Count of maximum distinct Rectangles possible with given Perimeter

Given an integer N denoting the perimeter of a rectangle. The task is to find the number of distinct rectangles possible with a given perimeter. 

Examples

Input: N = 10
Output: 4
Explanation: All the rectangles with perimeter 10 are following in the form of (length, breadth):
(1, 4), (4, 1), (2, 3), (3, 2)

Input: N = 8
Output: 3

Approach: This problem can be solved by using the properties of rectangles. Follow the steps below to solve the given problem.

• The perimeter of a rectangle is 2*(length + breadth).
• If N is odd, then there is no rectangle possible. As perimeter can never be odd.
• If N is less than 4 then also, there cannot be any rectangle possible. As the minimum possible length of a side is 1, even if the length of all the sides is 1 then also the perimeter will be 4.
• Now N = 2*(l + b) and (l + b) = N/2.
• So, it is required to find all the pairs whose sum is N/2 which is (N/2) - 1.

Below is the implementation of the above approach.

#include <iostream> using namespace std; // Function to find the maximum number // of distinct rectangles with given perimeter void maxRectanglesPossible(int N) {  // Invalid case  if (N < 4 || N % 2 != 0) {  cout << -1 << "\n";  }  else  // Number of distinct rectangles.  cout << (N / 2) - 1 << "\n"; } // Driver Code int main() {  // Perimeter of the rectangle.  int N = 20;  maxRectanglesPossible(N);  return 0; } 

// Java program for the above approach import java.io.*; import java.lang.*; import java.util.*; class GFG { // Function to find the maximum number // of distinct rectangles with given perimeter static void maxRectanglesPossible(int N) {    // Invalid case  if (N < 4 || N % 2 != 0) {  System.out.println(-1);  }  else  // Number of distinct rectangles.  System.out.println((N / 2) - 1); } // Driver Code  public static void main (String[] args) {  // Perimeter of the rectangle.  int N = 20;  maxRectanglesPossible(N);  } } // This code is contributed by hrithikgarg0388. 

# Function to find the maximum number # of distinct rectangles with given perimeter def maxRectanglesPossible (N): # Invalid case if (N < 4 or N % 2 != 0): print("-1"); else: # Number of distinct rectangles. print(int((N / 2) - 1)); # Driver Code # Perimeter of the rectangle. N = 20; maxRectanglesPossible(N); # This code is contributed by gfgking 

// C# program for the above approach using System; class GFG { // Function to find the maximum number // of distinct rectangles with given perimeter static void maxRectanglesPossible(int N) {    // Invalid case  if (N < 4 || N % 2 != 0) {  Console.WriteLine(-1);  }  else  // Number of distinct rectangles.  Console.WriteLine((N / 2) - 1); } // Driver Code  public static void Main () {  // Perimeter of the rectangle.  int N = 20;  maxRectanglesPossible(N);  } } // This code is contributed by Samim Hossain Mondal. 

 <script>  // Function to find the maximum number  // of distinct rectangles with given perimeter  const maxRectanglesPossible = (N) => {    // Invalid case  if (N < 4 || N % 2 != 0) {  document.write("-1<br/>");  }  else  // Number of distinct rectangles.  document.write(`${(N / 2) - 1}<br/>`);  }  // Driver Code  // Perimeter of the rectangle.  let N = 20;  maxRectanglesPossible(N);  // This code is contributed by rakeshsahni  </script> 

9

Time Complexity: O(1) 
Auxiliary Space: O(1)