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 bint 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 timersvoid flip(int a,int b){ int lcm =(a * b)/gcd(a, b); a = lcm/a; b = lcm/b; cout<<a-1<<" "<<b-1;}//Driver codeint main(){int a = 10;int b = 5;flip(a, b);} |
Java
// Java implementation of the approachclass GFG {// Recursive function to return// the gcd of a and bstatic 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 timersstatic 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 codepublic 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 timersdef flip(a, b): lcm = (a * b) // gcd(a, b) a = lcm // a b = lcm // b print(a - 1, b - 1)# Driver codea = 10b = 5flip(a, b)# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing 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 bfunction 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 timersfunction 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 codelet 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))
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