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Count of 0s in an N-level hexagon

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  • Last Updated : 10 Mar, 2022

Given an integer N, the task is to find the count of 0s in an N-level hexagon. 
 

Examples: 
 

Input: N = 2 
Output: 7
Input: N = 3 
Output: 19 
 

 

Approach: For the values of N = 1, 2, 3, … it can be observed that a series will be formed as 1, 7, 19, 37, 61, 91, 127, 169, …. It’s a difference series where differences are in AP as 6, 12, 18, …
Therefore the Nth term of will be 1 + {6 + 12 + 18 +…..(n – 1) terms} 
= 1 + (n – 1) * (2 * 6 + (n – 1 – 1) * 6) / 2 
= 1 + (n – 1) * (12 + (n – 2) * 6) / 2 
= 1 + (n – 1) * (12 + 6n – 12) / 2 
= 1 + (n – 1) * (6n) / 2 
= 1 + (n – 1) * (3n)
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
// 0s in an n-level hexagon
int count(int n)
{
    return 3 * n * (n - 1) + 1;
}
 
// Driver code
int main()
{
    int n = 3;
 
    cout << count(n);
 
    return 0;
}

Java




// Java implementation of the above approach
class GFG
{
     
    // Function to return the count of
    // 0s in an n-level hexagon
    static int count(int n)
    {
        return 3 * n * (n - 1) + 1;
    }
     
    // Driver code
    public static void main(String args[])
    {
        int n = 3;
     
        System.out.println(count(n));
    }
}
 
// This code is contributed by AnkitRai01

Python3




# Python3 implementation of the approach
 
# Function to return the count of
# 0s in an n-level hexagon
def count(n):
    return 3 * n * (n - 1) + 1
 
# Driver code
n = 3
 
print(count(n))
 
# This code is contributed by Mohit Kumar

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the count of
// 0s in an n-level hexagon
static int count(int n)
{
    return 3 * n * (n - 1) + 1;
}
 
// Driver code
static public void Main ()
{
    int n = 3;
     
    Console.Write(count(n));
}
}
 
// This code is contributed by ajit

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the count of
// 0s in an n-level hexagon
function count(n)
{
    return 3 * n * (n - 1) + 1;
}
 
// Driver code
var n = 3;
document.write(count(n));
 
// This code is contributed by rutvik_56.
</script>

Output: 

19

 

Time Complexity: O(1)

Auxiliary Space: O(1)


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