Total number of triangles formed when there are H horizontal and V vertical lines

Given a triangle ABC. H horizontal lines from side AB to AC (as shown in fig.) and V vertical lines from vertex A to side BC are drawn, the task is to find the total no. of triangles formed.

Examples:

Input: H = 2, V = 2
Output: 18

As we see in the image above, total triangles formed are 18.



Input: H = 3, V = 4
Output: 60

Approach: As we see in the images below, we can derive a general formula for above problem:

  1. If there are only h horizontal lines then total triangles are (h + 1).
  2. If there are only v vertical lines then total triangles are (v + 1) * (v + 2) / 2..
  3. So, total triangles are Triangles formed by horizontal lines * Triangles formed by vertical lines i.e. (h + 1) * (( v + 1) * (v + 2) / 2).

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;
#define LLI long long int
  
// Function to return total triangles
LLI totalTriangles(LLI h, LLI v)
{
    // Only possible triangle is
    // the given triangle
    if (h == 0 && v == 0)
        return 1;
  
    // If only vertical lines are present
    if (h == 0)
        return ((v + 1) * (v + 2) / 2);
  
    // If only horizontal lines are present
    if (v == 0)
        return (h + 1);
  
    // Return total triangles
    LLI Total = (h + 1) * ((v + 1) * (v + 2) / 2);
  
    return Total;
}
  
// Driver code
int main()
{
    int h = 2, v = 2;
    cout << totalTriangles(h, v);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG {
  
    // Function to return total triangles
    public static int totalTriangles(int h, int v)
    {
        // Only possible triangle is
        // the given triangle
        if (h == 0 && v == 0)
            return 1;
  
        // If only vertical lines are present
        if (h == 0)
            return ((v + 1) * (v + 2) / 2);
  
        // If only horizontal lines are present
        if (v == 0)
            return (h + 1);
  
        // Return total triangles
        int total = (h + 1) * ((v + 1) * (v + 2) / 2);
  
        return total;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int h = 2, v = 2;
        System.out.print(totalTriangles(h, v));
    }
}

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C#

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// C# implementation of the approach 
using System;
  
class GFG 
  
    // Function to return total triangles 
    public static int totalTriangles(int h, int v) 
    
        // Only possible triangle is 
        // the given triangle 
        if (h == 0 && v == 0) 
            return 1; 
  
        // If only vertical lines are present 
        if (h == 0) 
            return ((v + 1) * (v + 2) / 2); 
  
        // If only horizontal lines are present 
        if (v == 0) 
            return (h + 1); 
  
        // Return total triangles 
        int total = (h + 1) * ((v + 1) * (v + 2) / 2); 
  
        return total; 
    
  
    // Driver code 
    public static void Main() 
    
        int h = 2, v = 2; 
        Console.Write(totalTriangles(h, v)); 
    
  
// This code is contributed by Ryuga

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Python3

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# Python3 implementation of the approach
  
# Function to return total triangles
def totalTriangles(h, v):
      
    # Only possible triangle is 
    # the given triangle
    if (h == 0 and v == 0):
        return 1
  
    # If only vertical lines are present
    if (h == 0):
        return ((v + 1) * (v + 2) / 2)
  
    # If only horizontal lines are present
    if (v == 0):
        return (h + 1)
  
    # Return total triangles
    total = (h + 1) * ((v + 1) * (v + 2) / 2)
  
    return total
  
# Driver code
h = 2
v = 2
print(int(totalTriangles(h, v)))

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PHP

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<?php
// PHP implementation of the above approach
  
// Function to return total triangles 
function totalTriangles($h, $v
    // Only possible triangle is 
    // the given triangle 
    if ($h == 0 && $v == 0) 
        return 1; 
  
    // If only vertical lines are present 
    if ($h == 0) 
        return (($v + 1) * ($v + 2) / 2); 
  
    // If only horizontal lines are present 
    if ($v == 0) 
        return ($h + 1); 
  
    // Return total triangles 
    $Total = ($h + 1) * (($v + 1) * 
                         ($v + 2) / 2); 
  
    return $Total
  
// Driver code 
$h = 2;
$v = 2; 
echo totalTriangles($h, $v); 
  
// This code is contributed by Arnab Kundu
?>

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Output:

18

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



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