Introduction

In this blog post, we will explore how to solve the LeetCode problem "2105" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. Each plant needs a specific amount of water. Alice and Bob have a watering can each, initially full. They water the plants in the following way: Alice waters the plants in order from left to right, starting from the 0th plant. Bob waters the plants in order from right to left, starting from the (n - 1)th plant. They begin watering the plants simultaneously. It takes the same amount of time to water each plant regardless of how much water it needs. Alice/Bob must water the plant if they have enough in their can to fully water it. Otherwise, they first refill their can (instantaneously) then water the plant. In case both Alice and Bob reach the same plant, the one with more water currently in his/her watering can should water this plant. If they have the same amount of water, then Alice should water this plant. Given a 0-indexed integer array plants of n integers, where plants[i] is the amount of water the ith plant needs, and two integers capacityA and capacityB representing the capacities of Alice's and Bob's watering cans respectively, return the number of times they have to refill to water all the plants. Example 1: 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. Example 2: Input: plants = [2,2,3,3], capacityA = 3, capacityB = 4 Output: 2 Explanation: - Initially, Alice and Bob have 3 units and 4 units of water in their watering cans respectively. - Alice waters plant 0, Bob waters plant 3. - Alice and Bob now have 1 unit of water each, and need to water plants 1 and 2 respectively. - Since neither of them have enough water for their current plants, they refill their cans and then water the plants. So, the total number of times they have to refill to water all the plants is 0 + 1 + 1 + 0 = 2. Example 3: Input: plants = [5], capacityA = 10, capacityB = 8 Output: 0 Explanation: - There is only one plant. - Alice's watering can has 10 units of water, whereas Bob's can has 8 units. Since Alice has more water in her can, she waters this plant. So, the total number of times they have to refill is 0. Constraints: n == plants.length 1 <= n <= 105 1 <= plants[i] <= 106 max(plants[i]) <= capacityA, capacityB <= 109

Explanation

Here's the breakdown of the solution:

• Simultaneous Watering: Simulate Alice and Bob watering plants from opposite ends, updating their water levels at each step.
• Refill Logic: If a person doesn't have enough water for the current plant, increment the refill count and refill their can to full capacity before watering.

• Middle Plant Handling: When Alice and Bob meet at the middle plant (if n is odd), determine who waters the plant based on their current water levels.

• Runtime Complexity: O(n), where n is the number of plants.

• Storage Complexity: O(1) - constant extra space.

Code

    def watering_plants(plants, capacityA, capacityB):
    n = len(plants)
    alice_water = capacityA
    bob_water = capacityB
    refills = 0
    left = 0
    right = n - 1

    while left < right:
        if alice_water < plants[left]:
            refills += 1
            alice_water = capacityA
        alice_water -= plants[left]
        left += 1

        if bob_water < plants[right]:
            refills += 1
            bob_water = capacityB
        bob_water -= plants[right]
        right -= 1

    if left == right:
        if alice_water == bob_water:
            if alice_water < plants[left]:
                refills += 1
        elif alice_water > bob_water:
            if alice_water < plants[left]:
                refills += 1
        else:
            if bob_water < plants[left]:
                refills += 1
    return refills