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)