How do you solve Watering Plants II on LeetCode?

Watering Plants II (LeetCode #2105) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Alice and Bob can water plants simultaneously, which allows for a more efficient approach.

What is the brute force solution for Watering Plants II?

The brute force approach for Watering Plants II is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we simulate the watering process by iterating through the plants for both Alice and Bob. We keep track of their water levels and count how many times they need to refill their cans as they water the plants. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Watering Plants II?

Watering Plants II uses the following concepts: Array, Two Pointers, Simulation. The recommended patterns to study are: Two Pointers, Simulation.

What are the interview tips for Watering Plants II?

Always clarify the rules of the problem before diving into coding. Think about edge cases, such as when all plants need more water than their respective capacities. Practice simulating the process on paper before implementing the code.

What are common mistakes when solving Watering Plants II?

Not considering the case where Alice and Bob meet at the same plant. Failing to correctly track the water levels after each watering.

#2105

Watering Plants II

Medium

ArrayTwo PointersSimulationTwo PointersSimulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

We can optimize the simulation by determining the point where Alice and Bob would meet. Instead of checking each plant individually, we can calculate how much water they can use before they need to refill, allowing us to skip unnecessary iterations.

⚙️

Algorithm

5 steps

1Step 1: Initialize Alice's and Bob's water levels to their respective capacities and a refill counter to 0.
2Step 2: Use a while loop to simulate watering until both pointers meet.
3Step 3: Calculate how much water Alice can use before needing a refill and how much Bob can use.
4Step 4: If they meet, check who has more water to water the plant. If they have the same, Alice waters it.
5Step 5: Count refills as needed and return the total refills.

solution.py29 lines

1def wateringPlants(plants, capacityA, capacityB):
2    alice_water = capacityA
3    bob_water = capacityB
4    refills = 0
5    n = len(plants)
6    left = 0
7    right = n - 1
8
9    while left <= right:
10        if alice_water >= plants[left]:
11            alice_water -= plants[left]
12            left += 1
13        else:
14            refills += 1
15            alice_water = capacityA - plants[left]
16            left += 1
17
18        if left > right:
19            break
20
21        if bob_water >= plants[right]:
22            bob_water -= plants[right]
23            right -= 1
24        else:
25            refills += 1
26            bob_water = capacityB - plants[right]
27            right -= 1
28
29    return refills

ℹ

Complexity note: The time complexity is O(n) because we only traverse the plants once, and the space complexity is O(1) as we are using a constant amount of space.

1Alice and Bob can water plants simultaneously, which allows for a more efficient approach.
2Understanding the refill mechanism is crucial to minimizing the number of refills.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.