LeetCode 2105: Watering Plants II ·

Table of Contents

Meta Data
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Difficulty: medium First Attempt: 2026-01-24

Submissions: Accepted (first try)

Approach
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Both start together and every plant takes the same time, so at step t Alice is at t and Bob at n - 1 - t: converging pointers.

Each step:

With enough water, water the plant; otherwise refill first (count +1), then water.
left == right (the middle plant of an odd-length row): whoever has more water waters it, Alice on a tie. If that person has enough there is no refill; otherwise one refill.

My Solution
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class Solution:
    def minimumRefill(self, plants: List[int], capacityA: int, capacityB: int) -> int:
        a, b = capacityA, capacityB
        left, right = 0, len(plants) - 1
        refills = 0
        while left < right:
            if a < plants[left]:
                a = capacityA
                refills += 1
            a -= plants[left]
            if b < plants[right]:
                b = capacityB
                refills += 1
            b -= plants[right]
            left += 1
            right -= 1
        if left == right and max(a, b) < plants[left]:
            refills += 1                       # middle plant: refill only if the fuller can is short
        return refills

Complexity
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Time O(n), space O(1).

Mistakes to Avoid
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Refilling with a = capacityA and forgetting to subtract this plant’s water.
Letting both water the middle plant, or always giving it to Alice.

Interview Explanation
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They move at the same speed, so Alice from the left and Bob from the right are converging pointers. At each step, if the water is not enough, refill first and count it. With an odd length they meet in the middle, where the one with more water waters it, and a refill is needed only if even that is not enough. Time O(n).