Water and Jug Problem in C++

Suppose we have two jugs with capacities x and y liters. There is an infinite amount of water supply available to us. Now we need to determine whether it is possible to measure exactly z liters using these two jugs. If z liters of water are measurable, we must have z liters of water contained within one or both buckets by the end.

We can do these few operations −

Fill any of the jugs fully with water.
Empty any of the jugs.
Pour water from one jug into another till the other jug is completely full or the first jug itself is empty.

So if x = 2 and y = 5, and z = 4, then it will return true.

To solve this, we will follow these steps −

if x + y < z, then return false
if x = z or y = z, or x + y = z, then return true
return true z is divisible by gcd of x and y, otherwise false

Example (C++)

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h&g;
using namespace std;
class Solution {
   public:
   bool canMeasureWater(int x, int y, int z) {
      if(x + y < z) return false;
      if(x == z || y == z || x + y == z) return true;
      return z % __gcd(x, y) == 0;
   }
};
main(){
   Solution ob;
   cout << (ob.canMeasureWater(3,5,4));
}