365. Water and Jug Problem

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

Approach #1: Math. [Java]

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;
    }
    
    private int GCD(int a, int b) {
        return b == 0 ? a : GCD(b, a % b);
    }
}

　　

Analysis:

This is a pure Math problem. We need the knowledge of number theory to cover the proof and solution. 

The basic idea is to use the property of Bezout's identity and check is z is a multiple of GCD(x, y)

Quote from wiki:

Bezout's identity (also called Bezout's lemma) is a theorem in the elementary theory of numbers:

let a and b be nonzero integers and let d be their greatest common divisor. Then there exist integers x and y such that ax + by = d.

In addition, the greatest common divisord d is the smallest positive integer that can be written ad ax + by every integer of the form ax + by is a multiple of the freatest common divisor d. 

If a or b is negative this means we are emptying a jug of x or y gallons respectively.

Similarly if a or b is positive this means we are filling a jug of x or y gallons respectively.

x = 4, y = 6, z = 8

GCD(4, 6) = 2

8 is multiple of 2, so that this input is valid and we have:

-1 * 4 + 2 * 6 = 8

In this case, there is a solution obtained by filling the 6 gallon jug twice and emptying the 4 gallon jug once. (Solution. Fill the 6 gallon jug and empty 4 gallons to the 4 gallon jug. Empty the 4 gallon jug. Now empty the remaining two gallons from the 6 gallon jug to the 4 gallon jug. Next refill the 6 gallon jug. This gives 8 gallons in the end.)

Reference:

https://leetcode.com/problems/water-and-jug-problem/discuss/83715/Math-solution-Java-solution