2079. Watering Plants LeetCode Solution

Last updated: February 26, 2025

In this guide, you will get 2079. Watering Plants LeetCode Solution with the best time and space complexity. The solution to Watering Plants problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Watering Plants solution in C++
Watering Plants solution in Java
Watering Plants solution in Python
Additional Resources

2079. Watering Plants LeetCode Solution image

Problem Statement of Watering Plants

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

Complexity Analysis

Time Complexity:
Space Complexity:

2079. Watering Plants LeetCode Solution in C++

class Solution:
  def wateringPlants(self, plants: list[int], capacity: int) -> int:
    ans = 0
    currCapacity = 0

    for i, plant in enumerate(plants):
      if currCapacity + plant <= capacity:
        currCapacity += plant
      else:
        currCapacity = plant  # Reset
        ans += i * 2

    return ans + len(plants)
/* code provided by PROGIEZ */

2079. Watering Plants LeetCode Solution in Java

N/A
// code provided by PROGIEZ

2079. Watering Plants LeetCode Solution in Python

N/A
 # code by PROGIEZ

Additional Resources

Happy Coding! Keep following PROGIEZ for more updates and solutions.

Table of Contents