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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 

7








 

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

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