Bézout’s identity is number-theoretic conclusion that should be remembered.

if $$gcd(a_{1},a_{2},\ldots ,a_{n})=d$$

then there are integers $$x_{1},x_{2},\ldots ,x_{n}$$ such that

$$d=a_{1}x_{1}+a_{2}x_{2}+\cdots +a_{n}x_{n}$$

has the following properties:

• $$d$$ is the smallest positive integer of this form
• every number of this form is a multiple of $$d$$

We can apply it with the following problem.

### Description

LeetCode 365

You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs.

If z liters of water is measurable, you must have z liters of water contained within one or both buckets by the end.

Operations allowed:

• Fill any of the jugs completely 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.

Example 1: (From the famous “Die Hard” example)

Input: x = 3, y = 5, z = 4
Output: True


Example 2:

Input: x = 2, y = 6, z = 5
Output: False

### Analysis

Actually, the problem can be transformed a math problem.

Check if there are integers $$m$$ and $$n$$ that satisfies

$$mx+ny=z$$

With Bézout’s identity, we know it is true that $$mx+ny=gcd(x,y)$$. So if there is a integer i that satisfies$$z=gcd(x,y)*i$$, it is possible to measure exactly z litres using these two jugs.

### Solution

class Solution {
public boolean 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;
}
public int gcd(int a,int b)
{
if(b==0) return a;
else return gcd(b,a%b);
}
}


### Reference

[1] https://github.com/EdwardShi92/Leetcode-Solution-Code/blob/master/WaterandJugProblem.java

[2] https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity#Generalizations