Introduction

In this blog post, we will explore how to solve the LeetCode problem "2079" 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

You want to water n plants in your garden with a watering can. 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. There is a river at x = -1 that you can refill your watering can at. Each plant needs a specific amount of water. You will water the plants in the following way: Water the plants in order from left to right. After watering the current plant, if you do not have enough water to completely water the next plant, return to the river to fully refill the watering can. You cannot refill the watering can early. You are initially at the river (i.e., x = -1). It takes one step to move one unit on the x-axis. Given a 0-indexed integer array plants of n integers, where plants[i] is the amount of water the ith plant needs, and an integer capacity representing the watering can capacity, return the number of steps needed to water all the plants. Example 1: Input: plants = [2,2,3,3], capacity = 5 Output: 14 Explanation: Start at the river with a full watering can: - Walk to plant 0 (1 step) and water it. Watering can has 3 units of water. - Walk to plant 1 (1 step) and water it. Watering can has 1 unit of water. - Since you cannot completely water plant 2, walk back to the river to refill (2 steps). - Walk to plant 2 (3 steps) and water it. Watering can has 2 units of water. - Since you cannot completely water plant 3, walk back to the river to refill (3 steps). - Walk to plant 3 (4 steps) and water it. Steps needed = 1 + 1 + 2 + 3 + 3 + 4 = 14. Example 2: Input: plants = [1,1,1,4,2,3], capacity = 4 Output: 30 Explanation: Start at the river with a full watering can: - Water plants 0, 1, and 2 (3 steps). Return to river (3 steps). - Water plant 3 (4 steps). Return to river (4 steps). - Water plant 4 (5 steps). Return to river (5 steps). - Water plant 5 (6 steps). Steps needed = 3 + 3 + 4 + 4 + 5 + 5 + 6 = 30. Example 3: Input: plants = [7,7,7,7,7,7,7], capacity = 8 Output: 49 Explanation: You have to refill before watering each plant. Steps needed = 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + 6 + 6 + 7 = 49. Constraints: n == plants.length 1 <= n <= 1000 1 <= plants[i] <= 106 max(plants[i]) <= capacity <= 109

Explanation

Here's the solution to the watering plants problem:

• High-Level Approach:

  • Iterate through the plants from left to right.
  • Keep track of the current water level in the watering can.
  • If the water level is insufficient to water the next plant, return to the river (updating the steps accordingly) and refill.

• Complexity:

  • Runtime: O(n) - We iterate through the plants once.
  • Storage: O(1) - We use a constant amount of extra space.

Code

    def watering_plants(plants, capacity):
    """
    Calculates the number of steps needed to water all plants.

    Args:
        plants: A list of integers representing the water needed for each plant.
        capacity: An integer representing the capacity of the watering can.

    Returns:
        An integer representing the total number of steps needed.
    """

    n = len(plants)
    steps = 0
    current_water = capacity

    for i in range(n):
        if current_water < plants[i]:
            steps += (i + 1)  # Return to river
            steps += i + 1  # Walk to plant i
            current_water = capacity
        else:
            steps += 1  # Walk to plant i
        current_water -= plants[i]

    return steps