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

Namachila November 19, 2025

0 0 4 minutes read

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:

If there are only h horizontal lines then total triangles are (h + 1).
If there are only v vertical lines then total triangles are (v + 1) * (v + 2) / 2..

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++

// 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;
}

Java

// 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));
    }
}

C#

// 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

Python3

# 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)))

PHP

<?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
?>

Javascript

<script>
 
// javascript implementation of the 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
    var total = (h + 1) * ((v + 1) * (v + 2) / 2);
 
    return total;
}
 
// Driver code
var h = 2, v = 2;
document.write(totalTriangles(h, v));
 
// This code contributed by shikhasingrajput
 
</script>

Output:

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

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