0970MediumMathDepth-First SearchBreadth-First SearchBézout's LemmaEuclidean AlgorithmGreatest Common DivisorExtended Euclidean Algorithm

Measuring Out an Exact Amount With Two Cans

Tracked in this browser only

Write code

Trains the technique from

LeetCode 365Water and Jug Problem

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Two cans hold x and y litres when full. There is an unlimited supply of water and a drain.

At any moment you may fill a can to the brim, empty a can out, or pour one can into the other until either the pouring can runs dry or the receiving can is full.

Return true when the two cans can be made to hold target litres between them.

Examples

Example 1

Input: x = 14, y = 21, target = 7
Output: true

The two sizes share the divisor 7, and 7 divides evenly by it and fits inside 35, so it can be measured out.

Example 2

Input: x = 2, y = 6, target = 5
Output: false

Both cans hold an even number of litres, so the total is always even and an odd target is out of reach.

Example 3

Input: x = 4, y = 6, target = 10
Output: true

The target is exactly both cans full, which is allowed, and it divides evenly by their shared divisor of 2.

Constraints

1 <= x <= 1000
1 <= y <= 1000
1 <= target <= 1000

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def can_measure_water(x: int, y: int, target: int) -> bool:

Java

public boolean canMeasureWater(int x, int y, int target)

The intended approach

Time O(log(min(x, y))) for the divisor step.Space O(1).

Every allowed move changes the water held by a whole number of can-fulls: filling adds one can's worth, emptying takes one away, and pouring moves water between the cans without changing the total. So whatever the total ever reaches, it is some whole multiple of `x` plus some whole multiple of `y`, where the multiples may be negative. The amounts reachable that way are exactly the multiples of the two cans' greatest common divisor. That gives the whole test: the target must divide evenly by that divisor, and it must also fit, since the two cans together can never hold more than `x` plus `y`. Both conditions together are enough, and the divisor comes from the usual repeated-remainder step.

Pythonentry point: can_measure_water

def can_measure_water(x, y, target):
    if target > x + y:
        return False
    a = x
    b = y
    while b:
        a, b = b, a % b
    return target % a == 0

The obvious version

Search the states outright. A state is how much each can holds, and from it every fill, empty and pour is tried, keeping the states already reached. Reports success on reaching a state whose two amounts add to the target. Costs the number of states rather than a divisor step.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_can_measure_water(x, y, target):
    from collections import deque
    start = (0, 0)
    seen = set([start])
    queue = deque([start])
    while queue:
        a, b = queue.popleft()
        if a + b == target:
            return True
        moves = [
            (x, b),
            (a, y),
            (0, b),
            (a, 0),
        ]
        pour = a if a < y - b else y - b
        moves.append((a - pour, b + pour))
        pour = b if b < x - a else x - a
        moves.append((a + pour, b - pour))
        for nxt in moves:
            if nxt not in seen:
                seen.add(nxt)
                queue.append(nxt)
    return False

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in