Leetcode #2105: Watering Plants II
In this guide, we solve Leetcode #2105 Watering Plants II in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Alice and Bob want to water n plants in their garden. The plants are arranged in a row and are labeled from 0 to n - 1 from left to right where the ith plant is located at x = i.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Two Pointers, Simulation
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: plants = [2,2,3,3], capacityA = 5, capacityB = 5
Output: 1
Explanation:
- Initially, Alice and Bob have 5 units of water each in their watering cans.
- Alice waters plant 0, Bob waters plant 3.
- Alice and Bob now have 3 units and 2 units of water respectively.
- Alice has enough water for plant 1, so she waters it. Bob does not have enough water for plant 2, so he refills his can then waters it.
So, the total number of times they have to refill to water all the plants is 0 + 0 + 1 + 0 = 1.
Python Solution
class Solution:
def minimumRefill(self, plants: List[int], capacityA: int, capacityB: int) -> int:
a, b = capacityA, capacityB
ans = 0
i, j = 0, len(plants) - 1
while i < j:
if a < plants[i]:
ans += 1
a = capacityA
a -= plants[i]
if b < plants[j]:
ans += 1
b = capacityB
b -= plants[j]
i, j = i + 1, j - 1
ans += i == j and max(a, b) < plants[i]
return ans
Complexity
The time complexity is , where is the length of the plant array. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.