365. Water and Jug Problem

Posted 2020-08-21 鱼与海洋

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;
    }
}