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# 0365. Water and Jug Problem

<https://leetcode.com/problems/water-and-jug-problem>

## Description

You are given two jugs with capacities `jug1Capacity` and `jug2Capacity` liters. There is an infinite amount of water supply available. Determine whether it is possible to measure exactly `targetCapacity` liters using these two jugs.

If `targetCapacity` liters of water are measurable, you must have `targetCapacity` liters of water contained **within one or both buckets** by the end.

Operations allowed:

* Fill any of the jugs 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:**

```
**Input:** jug1Capacity = 3, jug2Capacity = 5, targetCapacity = 4
**Output:** true
**Explanation:** The famous [Die Hard](https://www.youtube.com/watch?v=BVtQNK_ZUJg&ab_channel=notnek01) example 
```

**Example 2:**

```
**Input:** jug1Capacity = 2, jug2Capacity = 6, targetCapacity = 5
**Output:** false
```

**Example 3:**

```
**Input:** jug1Capacity = 1, jug2Capacity = 2, targetCapacity = 3
**Output:** true
```

**Constraints:**

* `1 <= jug1Capacity, jug2Capacity, targetCapacity <= 106`

## ac

```java
class Solution {
    public boolean canMeasureWater(int x, int y, int z) {
        return z == 0 || x + y >= z && z % gcd(x, y) == 0;
    }

    public int gcd(int x, int y) {
        return y == 0 ? x : gcd(y, x % y);
    }
}

/*
1) bezout's law, ax + by = c has solution then c == gcd(x, y); 2) so just find gcd, return z == 0 || x + y >= z && z % gcd(x, y) == 0;
*/
```