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++ 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 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 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# 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 |
<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> |
19
Time Complexity: O(1)
Auxiliary Space: O(1)