Watering Plants II

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ArraysTwo Pointers

Alice and Bob want to water n plants in their garden. 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.

Each plant needs a specific amount of water. Alice and Bob have a watering can each, initially full. They water the plants in the following way:

Alice waters the plants in order from left to right, starting from the 0^th plant. Bob waters the plants in order from right to left, starting from the (n - 1)^th plant. They begin watering the plants simultaneously.
It takes the same amount of time to water each plant regardless of how much water it needs.
Alice/Bob must water the plant if they have enough in their can to fully water it. Otherwise, they first refill their can (instantaneously) then water the plant.
In case both Alice and Bob reach the same plant, the one with more water currently in his/her watering can should water this plant. If they have the same amount of water, then Alice should water this plant.

Given a 0-indexed integer array plants of n integers, where plants[i] is the amount of water the i^th plant needs, and two integers capacityA and capacityB representing the capacities of Alice's and Bob's watering cans respectively, return the number of times they have to refill to water all the plants.

Example 1:

Input: plants = [2,2,3,3], capacityA = 5, capacityB = 5
Output: 1
Explanation:
- Initially, Alice and Bob have 5 units of water each in their watering cans.
- Alice waters plant 0, Bob waters plant 3.
- Alice and Bob now have 3 units and 2 units of water respectively.
- Alice has enough water for plant 1, so she waters it. Bob does not have enough water for plant 2, so he refills his can then waters it.
So, the total number of times they have to refill to water all the plants is 0 + 0 + 1 + 0 = 1.

Example 2:

Input: plants = [2,2,3,3], capacityA = 3, capacityB = 4
Output: 2
Explanation:
- Initially, Alice and Bob have 3 units and 4 units of water in their watering cans respectively.
- Alice waters plant 0, Bob waters plant 3.
- Alice and Bob now have 1 unit of water each, and need to water plants 1 and 2 respectively.
- Since neither of them have enough water for their current plants, they refill their cans and then water the plants.
So, the total number of times they have to refill to water all the plants is 0 + 1 + 1 + 0 = 2.

Example 3:

Input: plants = [5], capacityA = 10, capacityB = 8
Output: 0
Explanation:
- There is only one plant.
- Alice's watering can has 10 units of water, whereas Bob's can has 8 units. Since Alice has more water in her can, she waters this plant.
So, the total number of times they have to refill is 0.

Constraints:

n == plants.length
1 <= n <= 10⁵
1 <= plants[i] <= 10⁶
max(plants[i]) <= capacityA, capacityB <= 10⁹

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the `plants` array contain non-positive values, i.e., zero or negative numbers?
How large can the `plants` array be? What are the maximum values for the numbers of `plants`?
If both Alice and Bob need to refill their watering cans simultaneously on the same plant, who refills first (Alice or Bob)?
If Alice's and Bob's watering can capacities are zero, what is the expected output?
Is it possible for Alice's or Bob's capacity to be smaller than the water needed for any plant?

Brute Force Solution

Approach

Imagine two people watering plants from opposite ends of a line. This brute force method simulates every possible watering pattern. We'll explore all combinations of who waters which plants, figuring out the refills needed for each.

Here's how the algorithm would work step-by-step:

Consider every single plant and decide whether the first person waters it, or the second person waters it.
For each of these possibilities, carefully simulate the entire watering process from both ends.
Keep track of how many times each person has to refill their watering can for that particular arrangement.
After trying every possible arrangement of who waters which plant, compare the number of refills needed for each.
The arrangement with the smallest total number of refills is the solution.

Code Implementation

def watering_plants_brute_force(plants, capacity_alice, capacity_bob):
    minimum_refills = float('inf')

    # Iterate through all possible watering assignments using binary representation
    for i in range(2**len(plants)):
        refills = 0
        alice_water = capacity_alice
        bob_water = capacity_bob
        left = 0
        right = len(plants) - 1

        # Simulate the watering process based on the current assignment
        while left <= right:
            # Alice waters the plant if the bit is 0
            if (i >> left) & 1 == 0:
                if alice_water < plants[left]:
                    refills += 1
                    alice_water = capacity_alice

                alice_water -= plants[left]
                if left == right:
                    break
                left += 1

            # Bob waters the plant if the bit is 1
            else:
                if bob_water < plants[right]:
                    refills += 1
                    bob_water = capacity_bob

                bob_water -= plants[right]
                if left == right:
                    break
                right -= 1

        # Update the minimum refills if the current assignment is better
        minimum_refills = min(minimum_refills, refills)

    return minimum_refills

Big(O) Analysis

Time Complexity

O(2^n * n) – The provided solution iterates through every possible combination of watering plants, where each plant can be watered by either the first or second person. This generates 2^n possibilities. For each of these possibilities, the solution simulates the entire watering process, requiring a linear scan of the plants, taking O(n) time. Therefore, the overall time complexity is O(2^n * n), as it considers all 2^n combinations and performs an O(n) operation for each.

Space Complexity

O(1) – The proposed solution explores all possible watering patterns by simulating each arrangement. It needs to keep track of the number of refills for each person in the current arrangement and the minimum number of refills found so far. It doesn't appear to create auxiliary arrays or data structures whose size depends on the number of plants, N. Therefore, the space used remains constant regardless of the input size N, and the auxiliary space complexity is O(1).

Optimal Solution

Approach

This problem involves two people watering plants from opposite ends. The key is to simulate the watering process, keeping track of how much water each person has and when they need to refill. By carefully simulating the process, we can determine the minimum number of refills needed.

Here's how the algorithm would work step-by-step:

Imagine two people starting at opposite ends of the row of plants.
Keep track of how much water each person has left in their watering can.
Go through the plants one by one, from both ends towards the middle.
If a person has enough water to water the current plant, they do so, and the amount of water in their can decreases.
If a person does NOT have enough water, they must refill their watering can completely before watering the plant. Count this as one refill.
Continue this process until both people meet at a plant in the middle, or pass each other.
If they meet at a single plant, determine who has enough water to water that plant (or if both don't, let either refill and water it).
The total number of refills is the final answer.

Code Implementation

def watering_plants_ii(plants, capacity_a, capacity_b):
    number_of_plants = len(plants)
    water_left_a = capacity_a
    water_left_b = capacity_b
    refills = 0
    left_index = 0
    right_index = number_of_plants - 1

    while left_index < right_index:
        # Water plant from left if possible, else refill.
        if water_left_a >= plants[left_index]:
            water_left_a -= plants[left_index]
        else:
            refills += 1
            water_left_a = capacity_a - plants[left_index]

        # Water plant from right if possible, else refill.
        if water_left_b >= plants[right_index]:
            water_left_b -= plants[right_index]
        else:
            refills += 1
            water_left_b = capacity_b - plants[right_index]

        left_index += 1
        right_index -= 1

    # Handle the middle plant, if there is one.
    if left_index == right_index:
        # Determine who refills if neither has enough.
        if water_left_a < plants[left_index] and water_left_b < plants[left_index]:
            refills += 1
        elif water_left_a < plants[left_index] and water_left_b >= plants[left_index]:
            pass
        elif water_left_a >= plants[left_index] and water_left_b < plants[left_index]:
            pass
        
    return refills

Big(O) Analysis

Time Complexity

O(n) – The primary operation involves traversing the plants array from both ends towards the middle using two pointers. Each pointer moves towards the center in a single pass. Therefore, each plant is visited at most once. The number of iterations is directly proportional to the number of plants (n), resulting in a linear time complexity of O(n).

Space Complexity

O(1) – The algorithm uses a fixed number of variables to store the water levels of Alice and Bob, their current positions (left and right indices), and the number of refills. The number of variables used does not depend on the number of plants, N. Therefore, the auxiliary space required is constant, resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
Empty plants array	Return 0 immediately as there are no plants to water.
Plants array with a single plant	Return 1 if capacity of either Alice or Bob is enough, else return 2.
Alice's and Bob's capacity are both zero	Return the length of the plants array, as both players need to refill for each plant.
Plants array with large number of elements and small capacities	Ensure the solution's time complexity is linear (O(n)) to avoid timeouts.
All plants require the same amount of water	The solution should still correctly alternate and refill as needed.
One person has much larger capacity than the other	The person with larger capacity may water a large number of plants without refilling, ensure that logic handles this case correctly.
Plant needs more water than either Alice or Bob's capacity	Return the number of refills as the size of the plants array.
Large plant array with small capacity requiring frequent refilling	Optimize the solution for performance due to numerous refills to avoid timeout issues.

Empty plants array

How to Handle:

Return 0 immediately as there are no plants to water.

Plants array with a single plant

How to Handle:

Return 1 if capacity of either Alice or Bob is enough, else return 2.

Alice's and Bob's capacity are both zero

How to Handle:

Return the length of the plants array, as both players need to refill for each plant.

Plants array with large number of elements and small capacities

How to Handle:

Ensure the solution's time complexity is linear (O(n)) to avoid timeouts.

All plants require the same amount of water

How to Handle:

The solution should still correctly alternate and refill as needed.

One person has much larger capacity than the other

How to Handle:

The person with larger capacity may water a large number of plants without refilling, ensure that logic handles this case correctly.

Plant needs more water than either Alice or Bob's capacity

How to Handle:

Return the number of refills as the size of the plants array.

Large plant array with small capacity requiring frequent refilling

How to Handle:

Optimize the solution for performance due to numerous refills to avoid timeout issues.