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# 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.

## 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)

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