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Weekly Contest 268 Q2
Array
Simulation

2079. Watering Plants

中文文档

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 $\text{water}$ to represent the current amount of water in the watering can, initially $\text{water} = \text{capacity}$.

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 $\text{water} = \text{water} - \text{plants}[i]$.
  • 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 $\text{water} = \text{capacity} - \text{plants}[i]$.

Finally, return the total number of steps.

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

Python3

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

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

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Rust

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
    }
}

C

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;
}