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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 perimetervoid 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 Codeint main(){     // Perimeter of the rectangle.    int N = 20;     maxRectanglesPossible(N);     return 0;}

## Java

 // Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*; class GFG { // Function to find the maximum number// of distinct rectangles with given perimeterstatic 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 perimeterdef 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 approachusing System;class GFG { // Function to find the maximum number// of distinct rectangles with given perimeterstatic 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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