Count of 0s in an N-level hexagon
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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