Count the number of rhombi possible inside a rectangle of given size

Given a rectangle of height H and width W which has the bottom left corner at (0, 0). The task is to count the number of distinct Rhombi that have all points inside or on the border of the rectangle satisfying the following conditions exists:

  • Have non-zero area.
  • Have diagonals parallel to the x and y axes.
  • Have integer coordinates.

Examples:

Input: H = 2, W = 2
Output: 2
There is only one rhombus possible with coordinates (0, 1), (1, 0), (2, 1) and (1, 2).

Input: H = 4, W = 4
Output: 16

Approach: Since the diagonals are parallel to the axis, let’s try fixing the diagonals and creating rhombi on them. For the rhombus to have integer coordinates, the length of the diagonals must be even. Let’s fix the length of the diagonals to i and j, the number of rhombi we can form with these diagonal lengths inside the rectangle would be (H – i + 1) * (W – j + 1). Thus, we iterate over all possible values of i and j and update the count.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of rhombi possible
long long countRhombi(int h, int w)
{
    long long ct = 0;
  
    // All possible diagonal lengths
    for (int i = 2; i <= h; i += 2)
        for (int j = 2; j <= w; j += 2)
  
            // Update rhombi possible with
            // the current diagonal lengths
            ct += (h - i + 1) * (w - j + 1);
  
    // Return the total count
    // of rhombi possible
    return ct;
}
  
// Driver code
int main()
{
    int h = 2, w = 2;
  
    cout << countRhombi(h, w);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.io.*;
  
class GFG 
{
      
// Function to return the count of rhombi possible
static int countRhombi(int h, int w)
{
    int ct = 0;
  
    // All possible diagonal lengths
    for (int i = 2; i <= h; i += 2)
        for (int j = 2; j <= w; j += 2)
  
            // Update rhombi possible with
            // the current diagonal lengths
            ct += (h - i + 1) * (w - j + 1);
  
    // Return the total count
    // of rhombi possible
    return ct;
}
  
    // Driver code
    public static void main (String[] args) 
    {
    int h = 2, w = 2;
    System.out.println (countRhombi(h, w));
    }
}
  
// This code is contributed by jit_t

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Python 3

# Python 3 implementation of the approach

# Function to return the count of
# rhombi possible
def countRhombi(h, w):

ct = 0;

# All possible diagonal lengths
for i in range(2, h + 1, 2):
for j in range(2, w + 1, 2):

# Update rhombi possible with
# the current diagonal lengths
ct += (h – i + 1) * (w – j + 1)

# Return the total count
# of rhombi possible
return ct

# Driver code
if __name__ == “__main__”:

h = 2
w = 2

print(countRhombi(h, w))

# This code is contributed by ita_c

C#

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// C# program to find the frequency of 
// minimum element in the array 
using System;
  
class GFG 
          
    // Function to return the count
    // of rhombi possible 
    static int countRhombi(int h, int w) 
    
        int ct = 0; 
      
        // All possible diagonal lengths 
        for (int i = 2; i <= h; i += 2) 
            for (int j = 2; j <= w; j += 2) 
      
                // Update rhombi possible with 
                // the current diagonal lengths 
                ct += (h - i + 1) * (w - j + 1); 
      
        // Return the total count 
        // of rhombi possible 
        return ct; 
    
      
    // Driver code 
    public static void Main() 
    
        int h = 2, w = 2; 
          
        Console.WriteLine(countRhombi(h, w)); 
    
  
// This code is contributed by Ryuga 

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Output:

1

Time Complexity: O(H * W)



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