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Watering Plants

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

Solution Explanation: Watering Plants

The problem involves calculating the minimum steps required to water n plants arranged in a row, given a watering can with a specific capacity. We start at the river (position -1) and must water plants from left to right. If the water in the can is insufficient to water the next plant, we must return to the river to refill.

Approach:

The optimal approach is a simulation. We iterate through the plants, keeping track of the remaining water in the can. For each plant:

  1. Check Water Level: If the current water level is sufficient to water the plant (water >= plants[i]), we subtract the plant's water requirement from the current water level and increment the step count by 1 (for watering the plant).

  2. Refill: If the water level is insufficient, we must return to the river and refill. The number of steps to do this is calculated as follows:

    • i * 2 + 1: i steps to return to the river, i steps to get back to the current plant, and 1 step to water the plant.
    • We then refill the can (water = capacity - plants[i]) and continue.

Time and Space Complexity:

  • Time Complexity: O(n), where n is the number of plants. We iterate through the plants once.
  • Space Complexity: O(1). We use a constant number of variables to track the water level and step count.

Code Implementation (Python):

class Solution:
    def wateringPlants(self, plants: List[int], capacity: int) -> int:
        ans, water = 0, capacity  # Initialize steps and water level
        for i, p in enumerate(plants):
            if water >= p:  # Sufficient water
                water -= p
                ans += 1
            else:  # Insufficient water - refill
                water = capacity - p
                ans += i * 2 + 1
        return ans

Code Implementation (Java):

class Solution {
    public int wateringPlants(int[] plants, int capacity) {
        int ans = 0, water = capacity; 
        for (int i = 0; i < plants.length; ++i) {
            if (water >= plants[i]) { 
                water -= plants[i];
                ans++;
            } else { 
                water = capacity - plants[i];
                ans += i * 2 + 1;
            }
        }
        return ans;
    }
}

The other language implementations (C++, Go, TypeScript, Rust, C) follow a similar structure, adapting the syntax appropriately for each language. The core logic remains consistent across all implementations.