Count of a, b & c after n seconds for given reproduction rate

Given three different alphabets ‘a’, ‘b’ and ‘c’ with a certain rule that after every 2 seconds every ‘a’ changes to a ‘b’, after every 5 seconds every ‘b’ changes to one ‘c’ and after every 12 seconds every ‘c’ changes again into two ‘a’s. 
Starting with one ‘a’, the task is to find the final count of a, b and c after given n seconds.
Examples:

Input: n = 2 
Output: a = 0, b = 1, c = 0 
Initially a = 1, b = 0, c = 0 
At n = 1, nothing will change 
At n = 2, all a will change to b i.e. a = 0, b = 1, c = 0
Input: n = 72 
Output: a = 64, b = 0, c = 0 

Approach: It can be observed that the values of a, b and c will form a pattern after every 60 seconds (which is the LCM of 2, 5 and 12) as follows: 

• At n = 60 -> a = 32¹, b = 0, c = 0
• At n = 120 -> a = 32², b = 0, c = 0
• At n = 180 -> a = 32³, b = 0, c = 0 and so on.

If n is a multiple of 60 then calculate the result from the above observation else calculate the result for the multiple of 60 which is nearest to n say x and then update the result for the seconds from x + 1 to n.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ull unsigned long long

// Function to print the count of
// a, b and c after n seconds
void findCount(int n)
{
    ull a = 1, b = 0, c = 0;

    // Number of multiples of 60 below n
    int x = n / 60;
    a = (ull)pow(32, x);

    // Multiple of 60 nearest to n
    x = 60 * x;

    for (int i = x + 1; i <= n; i++) {

        // Change all a to b
        if (i % 2 == 0) {
            b += a;
            a = 0;
        }

        // Change all b to c
        if (i % 5 == 0) {
            c += b;
            b = 0;
        }

        // Change each c to two a
        if (i % 12 == 0) {
            a += (2 * c);
            c = 0;
        }
    }

    // Print the updated values of a, b and c
    cout << "a = " << a << ", ";
    cout << "b = " << b << ", ";
    cout << "c = " << c;
}

// Driver code
int main()
{
    int n = 72;
    findCount(n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to print the count of
// a, b and c after n seconds
static void findCount(int n)
{
    long a = 1, b = 0, c = 0;

    // Number of multiples of 60 below n
    int x = n / 60;
    a = (long)Math.pow(32, x);

    // Multiple of 60 nearest to n
    x = 60 * x;

    for (int i = x + 1; i <= n; i++)
    {

        // Change all a to b
        if (i % 2 == 0)
        {
            b += a;
            a = 0;
        }

        // Change all b to c
        if (i % 5 == 0)
        {
            c += b;
            b = 0;
        }

        // Change each c to two a
        if (i % 12 == 0)
        {
            a += (2 * c);
            c = 0;
        }
    }

    // Print the updated values of a, b and c
    System.out.println("a = " + a + ", b = " +
                           b + ", c = " + c);
}

// Driver code
public static void main (String[] args)
{
    int n = 72;
    findCount(n);
}
}

// This code is contributed by mits

Python3

# Python3 implementation of the approach

# Function to print the count of
# a, b and c after n seconds
import math
def findCount(n):
    a, b, c = 1, 0, 0;

    # Number of multiples of 60 below n
    x = (int)(n / 60);
    a = int(math.pow(32, x));

    # Multiple of 60 nearest to n
    x = 60 * x;

    for i in range(x + 1, n + 1):

        # Change all a to b
        if (i % 2 == 0):
            b += a;
            a = 0;

        # Change all b to c
        if (i % 5 == 0):
            c += b;
            b = 0;

        # Change each c to two a
        if (i % 12 == 0):
            a += (2 * c);
            c = 0;

    # Print the updated values of a, b and c
    print("a =", a, end = ", ");
    print("b =", b, end = ", ");
    print("c =", c);

# Driver code
if __name__ == '__main__':
    n = 72;
    findCount(n);

# This code is contributed
# by 29AjayKumar

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to print the count of
// a, b and c after n seconds
static void findCount(int n)
{
    long a = 1, b = 0, c = 0;

    // Number of multiples of 60 below n
    int x = n / 60;
    a = (long)Math.Pow(32, x);

    // Multiple of 60 nearest to n
    x = 60 * x;

    for (int i = x + 1; i <= n; i++)
    {

        // Change all a to b
        if (i % 2 == 0)
        {
            b += a;
            a = 0;
        }

        // Change all b to c
        if (i % 5 == 0)
        {
            c += b;
            b = 0;
        }

        // Change each c to two a
        if (i % 12 == 0)
        {
            a += (2 * c);
            c = 0;
        }
    }

    // Print the updated values of a, b and c
    Console.WriteLine("a = " + a + ", b = " + b + ", c = " + c);
}

// Driver code
static void Main()
{
    int n = 72;
    findCount(n);
}
}

// This code is contributed by mits

PHP

<?php
// PHP implementation of the approach

// Function to print the count of
// a, b and c after n seconds
function findCount($n)
{
    $a = 1; $b = 0; $c = 0;

    // Number of multiples of 60 below n
    $x = $n / 60;
    $a = pow(32, $x);

    // Multiple of 60 nearest to n
    $x = 60 * $x;

    for ($i = $x + 1; $i <= $n; $i++)
    {

        // Change all a to b
        if ($i % 2 == 0)
        {
            $b += $a;
            $a = 0;
        }

        // Change all b to c
        if ($i % 5 == 0)
        {
            $c += $b;
            $b = 0;
        }

        // Change each c to two a
        if ($i % 12 == 0)
        {
            $a += (2 * $c);
            $c = 0;
        }
    }

    // Print the updated values of a, b and c
    echo("a = " . $a . ", b = " .
                  $b . ", c = " . $c);
}

// Driver code
$n = 72;
findCount($n);

// This code is contributed
// by Code_Mech.
?>

Javascript

<script>

// JavaScript implementation of the approach
// Function to print the count of
    // a, b and c after n seconds

    function findCount(n) {
        var a = 1, b = 0, c = 0;

        // Number of multiples of 60 below n
        var x = parseInt(n / 60);
        a =  Math.pow(32, x);

        // Multiple of 60 nearest to n
        x = 60 * x;

        for (i = x + 1; i <= n; i++) {

            // Change all a to b
            if (i % 2 == 0) {
                b += a;
                a = 0;
            }

            // Change all b to c
            if (i % 5 == 0) {
                c += b;
                b = 0;
            }

            // Change each c to two a
            if (i % 12 == 0) {
                a += (2 * c);
                c = 0;
            }
        }

        // Print the updated values of a, b and c
        document.write("a = " + a + ", b = " + b + ", c = " + c);
    }

    // Driver code

        var n = 72;
        findCount(n);

// This code contributed by Rajput-Ji

</script>

Output:

a = 64, b = 0, c = 0

Time Complexity: O(n)

Auxiliary Space: O(1)