Water and Jug Problem
<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>
Depth-First Search; Breadth-First Search; Math
class Solution:
    def canMeasureWater(self, x: int, y: int, z: int) -> bool:
        def dfs(i: int, j: int) -> bool:
            if (i, j) in vis:
                return False
            vis.add((i, j))
            if i == z or j == z or i + j == z:
                return True
            if dfs(x, j) or dfs(i, y) or dfs(0, j) or dfs(i, 0):
                return True
            a = min(i, y - j)
            b = min(j, x - i)
            return dfs(i - a, j + a) or dfs(i + b, j - b)

        vis = set()
        return dfs(0, 0)
