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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the count of rhombi possiblelong 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 codeint main(){ int h = 2, w = 2; cout << countRhombi(h, w); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG { // Function to return the count of rhombi possiblestatic 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 |
Python 3
# Python 3 implementation of the approach# Function to return the count of # rhombi possibledef 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 codeif __name__ == "__main__": h = 2 w = 2 print(countRhombi(h, w))# This code is contributed by ita_c |
C#
// 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 |
PHP
<?php// PHP implementation of the approach// Function to return the count of // rhombi possiblefunction countRhombi($h, $w){ $ct = 0; // All possible diagonal lengths for ($i = 2; $i <= $h; $i += 2) for ($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 $ct;}// Driver code$h = 2; $w = 2;echo(countRhombi($h, $w));// This code is contributed by Code_Mech?> |
Javascript
<script> // Javascript program to find the frequency of // minimum element in the array // Function to return the count // of rhombi possible function countRhombi(h, w) { let ct = 0; // All possible diagonal lengths for (let i = 2; i <= h; i += 2) for (let 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; } let h = 2, w = 2; document.write(countRhombi(h, w)); </script> |
1
Time Complexity: O(H * W)
Auxiliary Space: O(1)
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