How do you solve Water Bottles II on LeetCode?

Water Bottles II (LeetCode #3100) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Exchanging empty bottles increases the number of full bottles, allowing for more drinking.

What is the brute force solution for Water Bottles II?

The brute force approach for Water Bottles II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach simulates the process of drinking and exchanging bottles step by step. We keep track of the number of full and empty bottles until we can no longer exchange empty bottles for full ones. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Water Bottles II?

Water Bottles II uses the following concepts: Math, Simulation. The recommended patterns to study are: Simulation, Greedy.

What are the interview tips for Water Bottles II?

Always clarify the problem statement and constraints before diving into coding. Think about edge cases, such as when numBottles is less than numExchange. Explain your thought process clearly while coding to demonstrate your understanding.

What are common mistakes when solving Water Bottles II?

Forgetting to update numExchange after each exchange. Not correctly calculating the remaining empty bottles after an exchange.

#3100

Water Bottles II

Medium

MathSimulationSimulationGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution efficiently calculates the total number of bottles drunk without simulating each step. It uses a loop to keep track of empty bottles and adjusts the numExchange dynamically.

⚙️

Algorithm

5 steps

1Step 1: Initialize total_drunk to numBottles and empty_bottles to numBottles.
2Step 2: While empty_bottles is greater than or equal to numExchange, perform the following:
3Step 3: Calculate full_bottles from empty_bottles and update total_drunk.
4Step 4: Update empty_bottles for the next iteration.
5Step 5: Increase numExchange by 1.

solution.py9 lines

1def maxWaterBottles(numBottles, numExchange):
2    total_drunk = numBottles
3    empty_bottles = numBottles
4    while empty_bottles >= numExchange:
5        full_bottles = empty_bottles // numExchange
6        total_drunk += full_bottles
7        empty_bottles = empty_bottles % numExchange + full_bottles
8        numExchange += 1
9    return total_drunk

ℹ

Complexity note: The complexity is O(n) because we only loop through the number of empty bottles, and each operation inside the loop is constant time.

1Exchanging empty bottles increases the number of full bottles, allowing for more drinking.
2The number of exchanges needed increases with each successful exchange, making it essential to track both full and empty bottles.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.