# Count of maximum distinct Rectangles possible with given Perimeter

Last Updated : 09 Feb, 2022

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.

## C++

 `#include ` `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

 `// 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.`

## Python3

 `# 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#

 `// 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.`

## Javascript

 ``

Output

`9`

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

Article Tags :
Practice Tags :