Sand Timer Flip Counting Problem

Given two integers A and B representing the time taken by two different Sand timers to get empty. The task is to find the count of flips of each timer till the instance at which both the Sand timers get empty at the same time.
Examples:

Input: A = 30, B = 15 
Output: 0 1 
After 15 minutes: 15 0 
Flip timer 2: 15 15 
After another 15 minutes: 0 0
Input: A = 10, B = 8 
Output: 3 4 

Approach: Lowest Common Factor (LCM) of two number will determine the time at which both the Sand timers will get empty together. 
LCM(A, B) = (A * B) / GCD(A, B)
Dividing the LCM by input will give the number of flips for each sand timer respectively. 

C++

//C++14 implementation of the approach
#include<bits/stdc++.h>
using namespace std;

//Recursive function to return
//the gcd of a and b
int gcd(int a, int b){
    //Everything divides 0
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

//Function to print the number of
//flips for both the sand timers
void flip(int a,int b){
    int lcm =(a * b)/gcd(a, b);
    a = lcm/a;
    b = lcm/b;
    cout<<a-1<<" "<<b-1;
}

//Driver code
int main(){

int a = 10;
int b = 5;
flip(a, b);

}

Java

// Java implementation of the approach
class GFG
{

// Recursive function to return
// the gcd of a and b
static int gcd(int a, int b)
{
    // Everything divides 0
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

// Function to print the number of
// flips for both the sand timers
static void flip(int a, int b)
{
    int lcm = (a * b) / gcd(a, b);
    a = lcm / a;
    b = lcm / b;
    System.out.print((a - 1) + " " + (b - 1));
}

// Driver code
public static void main(String[] args)
{
    int a = 10;
    int b = 5;
    flip(a, b);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach

# Recursive function to return
# the gcd of a and b
def gcd(a, b):

    # Everything divides 0
    if (b == 0):
        return a
    return gcd(b, a % b)

# Function to print the number of
# flips for both the sand timers
def flip(a, b):
    lcm = (a * b) // gcd(a, b)
    a = lcm // a
    b = lcm // b
    print(a - 1, b - 1)

# Driver code
a = 10
b = 5
flip(a, b)

# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;

class GFG
{

    // Recursive function to return
    // the gcd of a and b
    static int gcd(int a, int b)
    {
        // Everything divides 0
        if (b == 0)
        return a;

        return gcd(b, a % b);
    }

    // Function to print the number of
    // flips for both the sand timers
    static void flip(int a, int b)
    {
        int lcm = (a * b) / gcd(a, b);
        a = lcm / a;
        b = lcm / b ;
        Console.WriteLine((a - 1) + " " +
                          (b - 1));
    }

    // Driver code
    public static void Main()
    {
        int a = 10;
        int b = 5;
        flip(a, b);
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>
// Javascript implementation of the approach

// Recursive function to return
// the gcd of a and b
function gcd(a, b)
{
    // Everything divides 0
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

// Function to print the number of
// flips for both the sand timers
function flip(a, b){
    let lcm =parseInt((a * b)/gcd(a, b));
    a = parseInt(lcm/a);
    b = parseInt(lcm/b);
    document.write((a-1)+" "+(b-1));
}

// Driver code
let a = 10;
let b = 5;
flip(a, b);

// This code is contributed by subham348.
</script>

Time Complexity: O(min(log a, log b))

Auxiliary Space : O(min(log a, log b))