365. Water and Jug Problem
Description
You are given two jugs with capacities x
liters and y
liters. You have an infinite water supply. Return whether the total amount of water in both jugs may reach target
using the following operations:
- Fill either jug completely with water.
- Completely empty either jug.
- Pour water from one jug into another until the receiving jug is full, or the transferring jug is empty.
Example 1:
Input: x = 3, y = 5, target = 4
Output: true
Explanation:
Follow these steps to reach a total of 4 liters:
- Fill the 5-liter jug (0, 5).
- Pour from the 5-liter jug into the 3-liter jug, leaving 2 liters (3, 2).
- Empty the 3-liter jug (0, 2).
- Transfer the 2 liters from the 5-liter jug to the 3-liter jug (2, 0).
- Fill the 5-liter jug again (2, 5).
- 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).
- Empty the 3-liter jug. Now, you have exactly 4 liters in the 5-liter jug (0, 4).
Reference: The Die Hard example.
Example 2:
Input: x = 2, y = 6, target = 5
Output: false
Example 3:
Input: x = 1, y = 2, target = 3
Output: true
Explanation: Fill both jugs. The total amount of water in both jugs is equal to 3 now.
Constraints:
1 <= x, y, target <= 103
Solutions
Solution 1: DFS
Let's denote $jug1Capacity$ as $x$, $jug2Capacity$ as $y$, and $targetCapacity$ as $z$.
Next, we design a function $dfs(i, j)$, which represents whether we can get $z$ liters of water when there are $i$ liters of water in $jug1$ and $j$ liters of water in $jug2$.
The execution process of the function $dfs(i, j)$ is as follows:
- If $(i, j)$ has been visited, return $false$.
- If $i = z$ or $j = z$ or $i + j = z$, return $true$.
- If we can get $z$ liters of water by filling $jug1$ or $jug2$, or emptying $jug1$ or $jug2$, return $true$.
- If we can get $z$ liters of water by pouring water from $jug1$ into $jug2$, or pouring water from $jug2$ into $jug1$, return $true$.
The answer is $dfs(0, 0)$.
The time complexity is $O(x + y)$, and the space complexity is $O(x + y)$. Here, $x$ and $y$ are the sizes of $jug1Capacity$ and $jug2Capacity$ respectively.
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