<p>You are given two jugs with capacities <code>x</code> liters and <code>y</code> liters. You have an infinite water supply. Return whether the total amount of water in both jugs may reach <code>target</code> using the following operations:</p>

<ul>
	<li>Fill either jug completely with water.</li>
	<li>Completely empty either jug.</li>
	<li>Pour water from one jug into another until the receiving jug is full, or the transferring jug is empty.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1: </strong></p>

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> x = 3, y = 5, target = 4 </span></p>

<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> true </span></p>

<p><strong>Explanation:</strong></p>

<p>Follow these steps to reach a total of 4 liters:</p>

<ol>
	<li>Fill the 5-liter jug (0, 5).</li>
	<li>Pour from the 5-liter jug into the 3-liter jug, leaving 2 liters (3, 2).</li>
	<li>Empty the 3-liter jug (0, 2).</li>
	<li>Transfer the 2 liters from the 5-liter jug to the 3-liter jug (2, 0).</li>
	<li>Fill the 5-liter jug again (2, 5).</li>
	<li>Pour from the 5-liter jug into the 3-liter jug until the 3-liter jug is full. This leaves 4 liters in the 5-liter jug (3, 4).</li>
	<li>Empty the 3-liter jug. Now, you have exactly 4 liters in the 5-liter jug (0, 4).</li>
</ol>

<p>Reference: The <a href="https://www.youtube.com/watch?v=BVtQNK_ZUJg&amp;ab_channel=notnek01" target="_blank">Die Hard</a> example.</p>
</div>

<p><strong class="example">Example 2: </strong></p>

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> x = 2, y = 6, target = 5 </span></p>

<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> false </span></p>
</div>

<p><strong class="example">Example 3: </strong></p>

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> x = 1, y = 2, target = 3 </span></p>

<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> true </span></p>

<p><strong>Explanation:</strong> Fill both jugs. The total amount of water in both jugs is equal to 3 now.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= x, y, target&nbsp;&lt;= 10<sup>3</sup></code></li>
</ul>

class Solution:
    def canMeasureWater(self, x: int, y: int, target: int) -> bool:
        # If the target is greater than the combined capacity of both jugs,
        # it's impossible
        if target > x + y:
            return False

        # Use BFS to explore all possible states
        queue = deque([(0, 0)])  # initially both jugs are empty
        visited = set((0, 0))  # to track visited states

        while queue:
            a, b = queue.popleft()

            # If either jug contains exactly the target amount of water
            if a == target or b == target or a + b == target:
                return True

            # Generate all possible next states
            next_states = set()

            # fill the first jug
            next_states.add((x, b)) 
            # fill the second just
            next_states.add((a, y))
            # empty the first jug
            next_states.add((0, b))
            # empty the second jug
            next_states.add((a, 0))
            # pour from jug1 to jug2
            next_states.add((a - min(a, y - b), b + min(a, y - b)))
            # pour from jug2 to jug1
            next_states.add((a + min(b, x - a), b - min(b, x - a)))

            # Explore the next states if they haven't been visited
            for state in next_states:
                if state not in visited:
                    visited.add(state)
                    queue.append(state)

        return False

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