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Sand Timer Flip Counting Problem
• Difficulty Level : Easy
• Last Updated : 11 May, 2021

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``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<

## 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

 ``
Output:
`0 1`

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