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Sand Timer Flip Counting Problem
• Difficulty Level : Easy
• Last Updated : 05 Sep, 2019

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

Output:

0 1

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