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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 i^{th}
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 xaxis.
Given a 0indexed integer array plants
of n
integers, where plants[i]
is the amount of water the i^{th}
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] <= 10^{6}
max(plants[i]) <= capacity <= 10^{9}
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 } }