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