Maximum count of Equilateral Triangles that can be formed within given Equilateral Triangle

Given two integers N and K where N denotes the unit size of a bigger Equilateral Triangle, the task is to find the number of an equilateral triangle of size K that are present in the bigger triangle of side N.

Examples:

Input: N = 4, K = 3

Output: 3
Explanation:
There are 3 equilateral triangles of 3 unit size which are present in the Bigger equilateral triangle of size 4 units.

Input: N = 4, K = 2
Output: 7
Explanation:
There are 7 equilateral triangles of 2 unit size which are present in the Bigger equilateral triangle of size 4 units. 

Naive Approach: The idea is to iterate over all possible sizes of the bigger equilateral triangle for checking the number of triangles with the required size K and print the total count of triangles.

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

Efficient Approach: To optimize the above approach, observe the following points:

• The number of triangles with a peak in the upward direction of size K present in size N equals to ((N – K +1 ) * (N – K + 2))/2.
• The number of inverted triangles with a peak in the downward direction of size K present in size N equals to ((N – 2K + 1) * (N – 2K + 2))/2.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <iostream>
using namespace std;

// Function to find the number of
// equilateral triangle formed
// within another triangle
int No_of_Triangle(int N, int K)
{
    // Check for the valid condition
    if (N < K)
        return -1;

    else {

        int Tri_up = 0;

        // Number of triangles having
        // upward peak
        Tri_up = ((N - K + 1)
                  * (N - K + 2))
                 / 2;

        int Tri_down = 0;

        // Number of inverted triangles
        Tri_down = ((N - 2 * K + 1)
                    * (N - 2 * K + 2))
                   / 2;

        // Total no. of K sized triangle
        return Tri_up + Tri_down;
    }
}

// Driver Code
int main()
{
    // Given N and K
    int N = 4, K = 2;

    // Function Call
    cout << No_of_Triangle(N, K);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{

// Function to find the number of
// equilateral triangle formed
// within another triangle
static int No_of_Triangle(int N, int K)
{
    // Check for the valid condition
    if (N < K)
        return -1;

    else
    {
        int Tri_up = 0;

        // Number of triangles having
        // upward peak
        Tri_up = ((N - K + 1) * (N - K + 2)) / 2;

        int Tri_down = 0;

        // Number of inverted triangles
        Tri_down = ((N - 2 * K + 1) *
                    (N - 2 * K + 2)) / 2;

        // Total no. of K sized triangle
        return Tri_up + Tri_down;
    }
}

// Driver Code
public static void main(String[] args)
{
    // Given N and K
    int N = 4, K = 2;

    // Function Call
    System.out.print(No_of_Triangle(N, K));
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 program for the above approach

# Function to find the number of
# equilateral triangle formed
# within another triangle
def No_of_Triangle(N, K):

    # Check for the valid condition
    if (N < K):
        return -1;

    else:
        Tri_up = 0;

        # Number of triangles having
        # upward peak
        Tri_up = ((N - K + 1) *
                  (N - K + 2)) // 2;

        Tri_down = 0;

        # Number of inverted triangles
        Tri_down = ((N - 2 * K + 1) *
                    (N - 2 * K + 2)) // 2;

        # Total no. of K sized triangle
        return Tri_up + Tri_down;

# Driver Code
if __name__ == '__main__':
    # Given N and K
    N = 4; K = 2;

    # Function Call
    print(No_of_Triangle(N, K));

# This code is contributed by sapnasingh4991

C#

// C# program for the above approach
using System;
class GFG{

// Function to find the number of
// equilateral triangle formed
// within another triangle
static int No_of_Triangle(int N, int K)
{
    // Check for the valid condition
    if (N < K)
        return -1;

    else
    {
        int Tri_up = 0;

        // Number of triangles having
        // upward peak
        Tri_up = ((N - K + 1) * (N - K + 2)) / 2;

        int Tri_down = 0;

        // Number of inverted triangles
        Tri_down = ((N - 2 * K + 1) *
                    (N - 2 * K + 2)) / 2;

        // Total no. of K sized triangle
        return Tri_up + Tri_down;
    }
}

// Driver Code
public static void Main(String[] args)
{
    // Given N and K
    int N = 4, K = 2;

    // Function Call
    Console.Write(No_of_Triangle(N, K));
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript program for the above approach

// Function to find the number of
// equilateral triangle formed
// within another triangle
function No_of_Triangle(N, K)
{
    // Check for the valid condition
    if (N < K)
        return -1;

    else {

        let Tri_up = 0;

        // Number of triangles having
        // upward peak
        Tri_up = Math.floor(((N - K + 1)
                * (N - K + 2))
                / 2);

        let Tri_down = 0;

        // Number of inverted triangles
        Tri_down = Math.floor(((N - 2 * K + 1)
                    * (N - 2 * K + 2))
                / 2);

        // Total no. of K sized triangle
        return Tri_up + Tri_down;
    }
}

// Driver Code

    // Given N and K
    let N = 4, K = 2;

    // Function Call
    document.write(No_of_Triangle(N, K));


// This code is contributed by Surbhi Tyagi.

</script>

Output:

7

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