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