Problem Description

In the given problem, you have two jugs with specific capacities - one can hold jug1Capacity liters and the other can hold jug2Capacity liters. With an unlimited water supply, your task is to figure out a way to measure exactly targetCapacity liters of water by either filling, emptying, or transferring water between the two jugs. The goal is to have the combined water in both jugs equal the targetCapacity. The operations you can perform are:

1. Filling up either jug to its full capacity with water.
2. Emptying any of the jugs completely.
3. Pouring water from one jug into the other until the jug from which you are pouring is empty, or the jug being filled is full.

You are asked to determine if it is possible to achieve the targetCapacity through these operations.

Flowchart Walkthrough

To analyze the leetcode 365. Water and Jug Problem using the Flowchart, let's consider each step to determine the appropriate algorithm:

Is it a graph?

• Yes: This problem can be represented as a graph where each state (representing a specific amount of water in each jug) is a node, and transitions between these states (pouring water from one jug to another) are the edges.

Is it a tree?

• No: The state graph for the water and jug problem is not strictly hierarchical and can have cycles, e.g., pouring water back and forth.

Is the problem related to directed acyclic graphs (DAGs)?

• No: While the problem involves directed state transitions, it's not acyclic as states can repeat by returning to previous configurations.

Is the problem related to shortest paths?

• No: The problem does not involve finding the shortest path; instead, it involves finding any path that leads to a particular state (achieving the target amount of water).

Does the problem involve connectivity?

• Yes: The key challenge is determining if there is a series of transitions (connectivity between states) that leads to the target state.

Does the problem have small constraints?

• Yes: Although theoretically, the range of states is large as defined by the jug capacities, the search space is often feasible to handle with small inputs.

Conclusion: The appropriate approach suggested by the flowchart is DFS/backtracking, as it involves exploring different state transitions and backtracking when a certain path doesn't lead to the solution, which fits the nature of the Water and Jug Problem.

Intuition

The solution uses a property from number theory related to the Greatest Common Divisor (GCD). The core intuition hinges on the idea that a certain amount of water can only be measured if it is a multiple of the GCD of the two jug capacities.

This is based on the mathematical theorem which states that for any two integers, a and b, and their GCD g, any integer of the form ax + by is a multiple of g. Here, x and y can be any integer, including negative integers, zero, or positive integers.

In terms of the jugs problem:

• a and b represent the capacities of the two jugs.
• g represents the GCD of the two capacities.
• x and y represent the number of times we fill or empty a jug.

Therefore, we can measure targetCapacity if and only if it's a multiple of the GCD of the two jug capacities.

We should check that the total capacity of the two jugs is at least the targetCapacity, because if it isn't, there's no way to measure out that amount of water. Additionally, if either jug has a capacity of 0, the only achievable measurements are either 0 or the combined capacity of both jugs.

In the provided solution, we check that the sum of both jugs' capacities is greater than or equal to the targetCapacity. If not, we return False because the target cannot be measured. If one jug is of 0 capacity, we then check if it's possible to measure the target - it can only be measured if the target is 0 or equal to the sum of both capacities. Lastly, we use the % operator to check if targetCapacity is divisible by the GCD of the two jug capacities, which is calculated using the built-in gcd function.

Learn more about Depth-First Search, Breadth-First Search and Math patterns.

Solution Approach

The solution provided is quite straightforward and primarily relies on mathematical reasoning. Here's a step-by-step breakdown of the implementation strategy using the algorithms, data structures, or patterns referenced in the code:

1. Initial Checks: First, the implementation checks if the combined capacity of both jugs is less than the targetCapacity. If this is true, it immediately returns False, as it is impossible to measure more water than the total volume that both jugs can hold.

2. Handle Zero Capacity Case: The code then checks if either of the jugs has a capacity of 0. If one jug has zero capacity, then there are only two possibilities to achieve the targetCapacity: either it is exactly 0 (which is always achievable), or it equals the non-zero capacity of the other jug. This is done by checking if targetCapacity is 0 or if targetCapacity equals jug1Capacity + jug2Capacity.

3. Mathematical Theorem Application: The main part of the solution relies on the property of the greatest common divisor (GCD). According to the theorem, a linear combination of two numbers a and b (ax + by for some integers x and y) can make any multiple of the GCD of a and b. In the context of our jugs, this means that we can only measure quantities that are multiples of the GCD of the two jug capacities.

4. Using Python's gcd Function: The code uses Python's built-in gcd function from the [math](/problems/math-basics) module to find the GCD of the capacities of the jugs.

5. Final Check with Modulus Operator: With the GCD calculated, the final step is to check if targetCapacity is a multiple of the GCD. This is achieved by checking if the remainder of the division of targetCapacity by the GCD (targetCapacity % gcd(jug1Capacity, jug2Capacity)) is 0. If the remainder is 0, then targetCapacity is a multiple of the GCD, and the function returns True. Otherwise, if there is a non-zero remainder, it implies targetCapacity cannot be formed by any combination of the two jug capacities according to the theorem mentioned, and the function returns False.

No advanced data structures or patterns are necessary for this approach, as the problem can be solved solely through arithmetic and logical operations. The elegance of the solution lies in its use of a well-known mathematical principle to reduce what might seem like a complex, operation-based problem to a simple modulo operation.

Here is the essential part of the code encapsulating the solution logic:

class Solution:
    def canMeasureWater(
        self, jug1Capacity: int, jug2Capacity: int, targetCapacity: int
    ) -> bool:
        if jug1Capacity + jug2Capacity < targetCapacity:
            return False
        if jug1Capacity == 0 or jug2Capacity == 0:
            return targetCapacity == 0 or jug1Capacity + jug2Capacity == targetCapacity
        return targetCapacity % gcd(jug1Capacity, jug2Capacity) == 0

Each check corresponds to a logical step in our step-by-step approach, ensuring that the problem is tackled efficiently and correctly.

Example Walkthrough

Let's go through an example to illustrate the solution approach with actual numbers. Suppose we have jug1Capacity = 3 liters, jug2Capacity = 5 liters, and our targetCapacity = 4 liters. We want to find out if we can measure exactly 4 liters using these two jugs.

Here are the steps we would take according to our solution approach:

1. Initial Checks: We first check if jug1Capacity + jug2Capacity is less than targetCapacity. In our case, 3 + 5 is not less than 4, so we can proceed.

2. Handle Zero Capacity Case: Next, we check for zero capacities. Neither jug has zero capacity (jug1Capacity = 3, jug2Capacity = 5), so this condition does not apply to our example.

3. Mathematical Theorem Application: We know that we can only measure targetCapacity if it is a multiple of the GCD of jug1Capacity and jug2Capacity. So we need to find the GCD of 3 and 5.

4. Using Python's gcd Function: We use the Python gcd function to find the GCD of 3 and 5, which is 1 because 3 and 5 are co-prime numbers.

5. Final Check with Modulus Operator: We perform the final check: targetCapacity % gcd(jug1Capacity, jug2Capacity). In our example, we check 4 % 1. Since 4 is a multiple of 1, 4 % 1 equals 0. This means the remainder of dividing 4 by the GCD, which is 1, equals 0.

Since the final modulus check passes, the function would return True, indicating that it is feasible to measure exactly 4 liters using a 3-liter jug and a 5-liter jug. This solution aligns with the well-known "water jug" problem-solving approach, where the operation of transferring water between jugs leads to measuring different volumes which are multiples of the GCD of the jugs' capacities.

Solution Implementation

Time and Space Complexity

The given Python function canMeasureWater determines whether it is possible to measure exactly targetCapacity liters by using two jugs of capacities jug1Capacity and jug2Capacity. It does so using a theorem related to the Diophantine equation which states that a target capacity x can be measured using two jugs with capacities m and n if and only if x is a multiple of the greatest common divisor (GCD) of m and n.

Time Complexity:

The time complexity of the function is predominantly determined by the computation of the GCD of jug1Capacity and jug2Capacity. Here’s how the complexity breaks down:

1. The function checks if the sum of the capacities of the two jugs is less than the targetCapacity. This comparison is constant time, O(1).
2. Then, it checks if either jug has a 0 capacity, and in such cases, it also performs constant-time comparisons: O(1).
3. Finally, it calculates the GCD of the two jug capacities. The GCD is calculated using Euclid's algorithm, which has a worst-case time complexity of O(log(min(a, b))), where a and b are jug1Capacity and jug2Capacity. Since the GCD function is bounded by the smaller of the two numbers, the time complexity for this step is O(log(min(jug1Capacity, jug2Capacity))).

Therefore, the overall time complexity of the function is O(log(min(jug1Capacity, jug2Capacity))).

Space Complexity:

The space complexity of the function is determined by the space used to hold any variables and the stack space used by the recursion (if the implementation of GCD is recursive):

1. Only a fixed number of integer variables are used, and there’s no use of any data structures that scale with the input size. This contributes a constant space complexity: O(1).
2. Assuming gcd function from the math library is used, which is typically implemented iteratively, the space complexity remains constant as there are no recursive calls stacking up.

Therefore, the overall space complexity of the function is O(1) constant space.

Learn more about how to find time and space complexity quickly using problem constraints.