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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

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

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

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


  • func wateringPlants(plants []int, capacity int) int {
	ans, cap := 0, capacity
	for i, x := range plants {
		if cap >= x {
			cap -= x
			ans++
		} else {
			cap = capacity - x
			ans += i*2 + 1
		}
	}
	return ans
}

  • function wateringPlants(plants: number[], capacity: number): number {
    const n = plants.length;
    let ans = 0;
    let water = capacity;
    for (let i = 0; i < n; i++) {
        if (water < plants[i]) {
            ans += i * 2 + 1;
            water = capacity - plants[i];
        } else {
            ans++;
            water -= plants[i];
        }
    }
    return ans;
}


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

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