Wednesday, October 19, 2016

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

Another math problem. The idea is to find a number z such that z = m * x + n * y, where m, n can be any integer. In the Die hard example,  4 = 3 * (-2) + 5 * 2.  According to Bézout's identity, z must be a multiple of x and y's greatest common divisor. Now the rest of the problem is to find GCD of x and y. Note z must be less than x + y because x and y cannot represent any number that is larger than the sum of their capacity.


public boolean canMeasureWater(int x, int y, int z) {
        int gcd = gcd(x, y);
        if (gcd == 0)
            return z == 0;
        return z <= x + y && z % gcd == 0;
    }
    
    private int gcd(int x, int y) {
        if (x == 0)
            return y;
        return gcd(y % x, x);
    }


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