0365 - Water and Jug Problem (Medium)

Problem Link

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

Problem Statement

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 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 <= 10^6

Approach 1: Bézout's identity

It's obvious that it is impossible to measure if the target capacity $z$ is greater than the total capacity of jug $x$ and jug $y$. Otherwise, we can express it as a linear combination of $x$ and $y$ and check if $z$ is a linear combination of $x$ and $y$. In order to do so, $z$ has to be a multiple of the gcd of $x$ and $y$.

Written by @wingkwong

class Solution {
public:
    bool canMeasureWater(int x, int y, int z) {
        // impossible case
        if (x + y < z) return false;
        // check if z is a multiple of GCD(x, y) 
        return z % gcd(x, y) == 0;
    }
};