Number of coloured 0’s in an N-level hexagon
Last Updated :
10 Mar, 2022
Given an integer N, the task is to find the count of coloured 0s in an N-level hexagon when the 0s are coloured in the following way:
Examples:
Input: N = 2
Output: 5
Input: N = 3
Output: 12
Approach: For the values of N = 1, 2, 3, … it can be observed that a series will be formed as 1, 5, 12, 22, 35, …. It’s a difference series where differences are in AP as 4, 7, 10, 13, ….
Therefore the Nth term of will be 1 + {4 + 7 + 10 +13 +…..(n – 1) terms}
= 1 + (n – 1) * (2 * 4 + (n – 1 – 1) * 3) / 2
= 1 + (n – 1) * (8 + (n – 2) * 3) / 2
= 1 + (n – 1) * (8 + 3n – 6) / 2
= 1 + (n – 1) * (3n + 2) / 2
= n * (3 * n – 1) / 2
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int count( int n)
{
return n * (3 * n - 1) / 2;
}
int main()
{
int n = 3;
cout << count(n);
return 0;
}
|
Java
class GFG
{
static int count( int n)
{
return n * ( 3 * n - 1 ) / 2 ;
}
public static void main(String[] args)
{
int n = 3 ;
System.out.println(count(n));
}
}
|
Python3
def count(n) :
return n * ( 3 * n - 1 ) / / 2 ;
if __name__ = = "__main__" :
n = 3 ;
print (count(n));
|
C#
using System;
class GFG
{
static int count( int n)
{
return n * (3 * n - 1) / 2;
}
public static void Main(String[] args)
{
int n = 3;
Console.WriteLine(count(n));
}
}
|
Javascript
<script>
function count(n)
{
return parseInt(n * (3 * n - 1) / 2);
}
var n = 3;
document.write(count(n));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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