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

Credits:
Special thanks to @vinod23 for adding this problem and creating all test cases.

/**
* Math problem
* 能测的体积是两个瓶子最大公约数的倍数
* 比较麻烦的是0 的情况！！
*/

/**
* Math problem
* 能测的体积是两个瓶子最大公约数的倍数
*  比较麻烦的是0 的情况！！
* 可用数论的知识 Bézout's identity 求最大公约数
*/

public class Solution {
public boolean canMeasureWater(int x, int y, int z) {
if(x+y < z) return false;
if(x == 0 && y == 0 && z == 0) return true;
if(x == 0 || y == 0 && z != 0) return false;
return z % gcd1(x,y) == 0;
}

// Bézout's identity
//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
public int gcd1(int x, int y){
while(y != 0){
int temp = y;
y = x % y;
x = temp;
}
return x;
}

public int gcd2(int x, int y){
int res = 1;
int end = Math.min(x, y);
if(end == 1) return 1;
for(int i = 2; i <= end; i++){
if(x % i == 0 && y % i == 0)
res = i;
}
return res;
}
}