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.

Source: leetcode | Difficulty: MEDIUM_HARD
Tags: Graph traversal

Mathematics

Starter code: class Solution:
    def canMeasureWater(self, x: int, y: int, z: int) -> bool:
        

Solutions:
class Solution:

	def canMeasureWater(self, x, y, z):
		if x > y:
			(x, y) = (y, x)
		if z < 0 or z > x + y:
			return False
		if x == 0:
			return z == y or z == 0
		if z % x == 0:
			return True
		if y % x == 0:
			return False
		a = x
		b = y % x
		while a > 1 and b > 1:
			a = a % b
			(a, b) = (b, a)
		if b == 0:
			m = a
		else:
			m = b
		if z % m == 0:
			return True
		return False


class Solution:

	def canMeasureWater(self, x, y, z):

		def gcd(x, y):
			if y == 0:
				return x
			return gcd(y, x % y)
		if x == 0 and y == 0:
			return z == 0
		return z % gcd(x, y) == 0 and x + y >= z


class Solution:

	def canMeasureWater(self, x, y, z):
		if x + y < z:
			return False
		if x == z or y == z or x + y == z:
			return True
		while y:
			(x, y) = (y, x % y)
		return not z % x


class Solution:

	def canMeasureWater(self, x, y, z):
		if x + y < z:
			return False
		gcd = lambda a, b: gcd(b, a % b) if a % b else b
		return z == 0 or z % gcd(x, y) == 0


class Solution:

	def canMeasureWater(self, x, y, z):
		(a, b) = (x, y)
		while y:
			r = x % y
			x = y
			y = r
		return bool(not z or (x and z <= a + b and (not z % x)))


class Solution:

	def canMeasureWater(self, x, y, z):
		if z in [0, x, y, x + y]:
			return True
		if z > x + y or z < 0:
			return False
		if x > y:
			(x, y) = (y, x)
		if x != 0:
			temp = z % x
			for i in range(x):
				if temp == y * i % x:
					return True
		if y != 0:
			temp = z % y
			for i in range(y):
				if temp == x * i % y:
					return True
		return False