2079. Watering Plants

Description

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

Solutions

Solution 1: Simulation

We can simulate the process of watering the plants. We use a variable \(\textit{water}\) to represent the current amount of water in the watering can, initially \(\textit{water} = \textit{capacity}\).

We traverse the plants. For each plant:

• If the current amount of water in the watering can is enough to water this plant, we move forward one step, water this plant, and update \(\textit{water} = \textit{water} - \textit{plants}[i]\).
• Otherwise, we need to return to the river to refill the watering can, walk back to the current position, and then move forward one step. The number of steps we need is \(i \times 2 + 1\). Then we water this plant and update \(\textit{water} = \textit{capacity} - \textit{plants}[i]\).

Finally, return the total number of steps.

The time complexity is \(O(n)\), where \(n\) is the number of plants. The space complexity is \(O(1)\).

Python3JavaC++GoTypeScriptRustC

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

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 += 1;
            } else {
                water = capacity - plants[i];
                ans += i * 2 + 1;
            }
        }
        return ans;
    }
}

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

func wateringPlants(plants []int, capacity int) (ans int) {
    water := capacity
    for i, p := range plants {
        if water >= p {
            water -= p
            ans++
        } else {
            water = capacity - p
            ans += i*2 + 1
        }
    }
    return
}

function wateringPlants(plants: number[], capacity: number): number {
    let [ans, water] = [0, capacity];
    for (let 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;
}

impl Solution {
    pub fn watering_plants(plants: Vec<i32>, capacity: i32) -> i32 {
        let mut ans = 0;
        let mut water = capacity;
        for (i, &p) in plants.iter().enumerate() {
            if water >= p {
                water -= p;
                ans += 1;
            } else {
                water = capacity - p;
                ans += (i as i32) * 2 + 1;
            }
        }
        ans
    }
}

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

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