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Count of 0s in an N-level hexagon
  • Last Updated : 21 Oct, 2019

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

19

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