2079

2079. Watering Plants (Medium)

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

Companies:
Google

Related Topics:
Array

Solution 1. Simulation

1. When we can water A[i], consume it water -= A[i] and increment i.
2. If we haven't reached the last plant, add 2 * i to the answer; otherwise, add i.
3. Refill water. Go back to step 1 if i < N.

// OJ: https://leetcode.com/problems/watering-plants/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int wateringPlants(vector<int>& A, int C) {
        int N = A.size(), ans = 0, water = C;
        for (int i = 0; i < N; water = C) { // refill water after each round
            for (; i < N && water - A[i] >= 0; ++i) water -= A[i]; // When we can water `A[i]`, consume it and increment `i`.
            ans += (i == N ? 1 : 2) * i; // If we haven't reached the last plant, add `2 * i` to the answer; otherwise, add `i`.
        }
        return ans;
    }
};

Discuss

https://leetcode.com/problems/watering-plants/discuss/1589015/C%2B%2B-O(N)-One-Pass