Problem Description

You have two water jugs with capacities of x liters and y liters, along with an unlimited water supply. Your goal is to determine if it's possible to have exactly target liters of water in total across both jugs.

You can perform these operations:

• Fill either jug completely from the water supply
• Empty either jug completely
• Pour water from one jug to another until either the receiving jug is full or the source jug is empty

The problem asks you to return true if you can achieve exactly target liters in total (counting water in both jugs combined), and false otherwise.

For example, with a 3-liter jug and a 5-liter jug, you can reach a target of 4 liters through a series of fill, empty, and pour operations. The key insight is that through these operations, you can measure various amounts of water that might not be directly equal to either jug's capacity.

The challenge is to determine whether any sequence of these three operations can result in having exactly target liters of water across the two jugs. This is a classic water jug puzzle that tests your ability to explore different states and combinations of water levels in both jugs.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: This problem can be modeled as a graph where each state (i, j) represents the amount of water in jug1 and jug2 respectively. The edges represent the possible operations (fill, empty, pour) that transition from one state to another.

Is it a tree?

• No: The state space is not a tree structure. States can be revisited through different sequences of operations, and there can be multiple paths to reach the same state.

Is the problem related to directed acyclic graphs?

• No: The state graph can have cycles. For example, you can fill a jug, empty it, and return to the original state, creating a cycle.

Is the problem related to shortest paths?

• No: We don't need to find the shortest sequence of operations. We only need to determine if the target is reachable at all.

Does the problem involve connectivity?

• No: We're not checking if all states are connected or finding connected components.

Does the problem have small constraints?

• Yes: The constraints are relatively small (1 ≤ x, y, target ≤ 10³), making the state space manageable for exploration.

DFS/backtracking?

• Yes: With small constraints and the need to explore different states to find if the target is reachable, DFS with backtracking is suitable.

Conclusion: The flowchart suggests using DFS/backtracking to explore all possible states (i, j) representing water amounts in the two jugs, checking if we can reach a state where the total equals the target.

Intuition

The key insight is to think of this problem as exploring different states of water in the two jugs. Each state can be represented as (i, j) where i is the amount of water in the first jug and j is the amount in the second jug.

Starting from the initial state (0, 0) where both jugs are empty, we can perform various operations to transition to new states. Each operation creates a new state:

• Filling jug1 takes us from (i, j) to (x, j)
• Filling jug2 takes us from (i, j) to (i, y)
• Emptying jug1 takes us from (i, j) to (0, j)
• Emptying jug2 takes us from (i, j) to (i, 0)
• Pouring from one jug to another creates more complex transitions

The question becomes: can we reach any state where i == target or j == target or i + j == target?

This is essentially a graph traversal problem where we need to explore all reachable states from (0, 0). DFS is perfect for this because:

1. We need to explore all possible paths until we find the target
2. We can use recursion to naturally handle the branching of different operations
3. We need to track visited states to avoid infinite loops (since we can return to previous states)

The most intricate part is handling the pour operation. When pouring from jug1 to jug2, we can only pour until either jug1 is empty or jug2 is full. The amount transferred is min(i, y - j). Similarly, when pouring from jug2 to jug1, the amount is min(j, x - i).

By recursively exploring all possible operations from each state and marking visited states, we can determine if the target is achievable. If at any point we reach a state where the total water equals the target, we return true.

Solution Approach

The solution implements a DFS approach with memoization to explore all possible states of water in the two jugs.

Core Data Structure:

• A set called vis to track visited states and prevent infinite loops
• Each state is represented as a tuple (i, j) where i is water in jug1 and j is water in jug2

DFS Function Design:

The recursive function dfs(i, j) checks if we can reach the target from state (i, j):

1. Base Cases:

   • If (i, j) has been visited, return False to avoid cycles
   • If i == z or j == z or i + j == z, we've found the target, return True

2. Mark Current State: Add (i, j) to the visited set

3. Explore All Operations:

   • Fill jug1: Call dfs(x, j) - fill jug1 to capacity x
   • Fill jug2: Call dfs(i, y) - fill jug2 to capacity y
   • Empty jug1: Call dfs(0, j) - empty jug1 completely
   • Empty jug2: Call dfs(i, 0) - empty jug2 completely

4. Pour Operations:

   • Pour jug1 → jug2: Calculate amount a = min(i, y - j)
     • This is the minimum of water in jug1 and remaining capacity in jug2
     • New state: (i - a, j + a)
   • Pour jug2 → jug1: Calculate amount b = min(j, x - i)
     • This is the minimum of water in jug2 and remaining capacity in jug1
     • New state: (i + b, j - b)

Algorithm Flow:

1. Initialize empty visited set
2. Start DFS from state (0, 0) (both jugs empty)
3. At each state, check if target is reached
4. If not, try all six possible operations
5. Return True if any path leads to the target

The algorithm efficiently explores the state space using DFS, pruning paths that lead to already-visited states. The time complexity depends on the number of unique states, which is bounded by O(x * y), and each state is visited at most once.

Example Walkthrough

Let's walk through a concrete example with x = 3 (jug1 capacity), y = 5 (jug2 capacity), and target = 4.

Initial State: (0, 0) - both jugs are empty

Step 1: From (0, 0), we explore all possible operations:

• Fill jug1: go to (3, 0)
• Fill jug2: go to (0, 5)
• Empty operations have no effect since jugs are already empty
• Pour operations have no effect since there's no water to pour

Let's follow the path through filling jug2 first.

Step 2: State (0, 5) - jug2 is full

• Check: Is 0 == 4? No. Is 5 == 4? No. Is 0 + 5 == 4? No.
• Mark (0, 5) as visited
• Explore next operations:
  • Pour jug2 → jug1: amount = min(5, 3-0) = 3
  • New state: (0+3, 5-3) = (3, 2)

Step 3: State (3, 2) - jug1 has 3L, jug2 has 2L

• Check: Is 3 == 4? No. Is 2 == 4? No. Is 3 + 2 == 4? No.
• Mark (3, 2) as visited
• Explore operations:
  • Empty jug1: go to (0, 2)

Step 4: State (0, 2) - jug1 is empty, jug2 has 2L

• Check: Is 0 == 4? No. Is 2 == 4? No. Is 0 + 2 == 4? No.
• Mark (0, 2) as visited
• Explore operations:
  • Pour jug2 → jug1: amount = min(2, 3-0) = 2
  • New state: (0+2, 2-2) = (2, 0)

Step 5: State (2, 0) - jug1 has 2L, jug2 is empty

• Check: Is 2 == 4? No. Is 0 == 4? No. Is 2 + 0 == 4? No.
• Mark (2, 0) as visited
• Explore operations:
  • Fill jug2: go to (2, 5)

Step 6: State (2, 5) - jug1 has 2L, jug2 is full

• Check: Is 2 == 4? No. Is 5 == 4? No. Is 2 + 5 == 4? No.
• Mark (2, 5) as visited
• Explore operations:
  • Pour jug2 → jug1: amount = min(5, 3-2) = 1
  • New state: (2+1, 5-1) = (3, 4)

Step 7: State (3, 4) - jug1 has 3L, jug2 has 4L

• Check: Is 3 == 4? No. Is 4 == 4? Yes!
• Target found! Return True

The sequence of operations that achieved the target:

1. Fill jug2 (5L)
2. Pour jug2 into jug1 (3L to jug1, 2L remains in jug2)
3. Empty jug1
4. Pour jug2 into jug1 (2L to jug1, jug2 empty)
5. Fill jug2 (5L)
6. Pour jug2 into jug1 (1L to jug1, 4L remains in jug2)
7. Result: jug2 contains exactly 4L

Solution Implementation

Time and Space Complexity

The time complexity is O(x * y) and the space complexity is O(x * y).

The algorithm performs a depth-first search (DFS) on the state space where each state is represented by a tuple (i, j), indicating the current water amounts in the two jugs.

For time complexity: In the worst case, the algorithm explores all possible states. Since i can range from 0 to x and j can range from 0 to y, there are at most (x + 1) * (y + 1) unique states. Each state is visited at most once due to the visited set vis, resulting in O(x * y) time complexity.

For space complexity: The visited set vis stores all explored states to avoid revisiting them. In the worst case, it contains all possible (i, j) pairs, which is O(x * y). Additionally, the recursion stack depth can be at most O(x * y) in the worst case when exploring all states before finding a solution or exhausting all possibilities. Therefore, the total space complexity is O(x * y).

Note: The reference answer suggests O(x + y) complexity, which would apply to an optimized solution using the mathematical property that the problem is solvable if and only if z is divisible by gcd(x, y) and z ≤ x + y. However, the provided DFS solution explores the state space more exhaustively, leading to the higher O(x * y) complexity.

Common Pitfalls

1. Stack Overflow Due to Deep Recursion

The DFS approach can hit Python's recursion limit (typically ~1000) when dealing with large jug capacities. For example, with jugs of capacity 10,000 and 10,001, the recursive calls can exceed the stack limit before finding a solution.

Solution: Convert the recursive DFS to an iterative approach using a stack or queue:

def canMeasureWater(self, x: int, y: int, z: int) -> bool:
    if z > x + y:
        return False
    if z == 0:
        return True
  
    visited = set()
    stack = [(0, 0)]
  
    while stack:
        jug1, jug2 = stack.pop()
      
        if (jug1, jug2) in visited:
            continue
          
        visited.add((jug1, jug2))
      
        if jug1 == z or jug2 == z or jug1 + jug2 == z:
            return True
      
        # All possible next states
        states = [
            (x, jug2),  # Fill jug1
            (jug1, y),  # Fill jug2
            (0, jug2),  # Empty jug1
            (jug1, 0),  # Empty jug2
            (jug1 - min(jug1, y - jug2), jug2 + min(jug1, y - jug2)),  # Pour 1->2
            (jug1 + min(jug2, x - jug1), jug2 - min(jug2, x - jug1))   # Pour 2->1
        ]
      
        for state in states:
            if state not in visited:
                stack.append(state)
  
    return False

2. Missing Edge Cases

The current solution doesn't handle some edge cases efficiently:

• When z > x + y, it's impossible to measure z liters (you can't have more water than both jugs combined)
• When z == 0, the answer is always True (empty jugs)

Solution: Add early termination checks:

def canMeasureWater(self, x: int, y: int, z: int) -> bool:
    # Early termination for impossible cases
    if z > x + y:
        return False
    if z == 0:
        return True
    if x == 0 or y == 0:
        return z == x or z == y

3. Inefficient for Large Inputs

The DFS/BFS approach explores O(x * y) states, which becomes inefficient for large values. There's actually a mathematical solution using the greatest common divisor (GCD).

Mathematical Solution: The target z is achievable if and only if z is a multiple of GCD(x, y) and z ≤ x + y.

import math

def canMeasureWater(self, x: int, y: int, z: int) -> bool:
    if z > x + y:
        return False
    if z == 0:
        return True
    if x == 0 or y == 0:
        return z == x or z == y
  
    return z % math.gcd(x, y) == 0

This reduces the time complexity from O(x * y) to O(log(min(x, y))) for the GCD calculation.

4. Memory Issues with Large State Space

The visited set can consume significant memory when x and y are large, potentially causing memory errors.

Solution: Use the mathematical approach above, or implement a more memory-efficient state representation if the search approach is required.