Is Water Bottles II easy or hard?

Water Bottles II is typically classified as Medium difficulty. The base logic is straightforward simulation, but recognizing the mathematical shortcut and reasoning about increasing exchange cost adds an extra layer of difficulty.

Water Bottles II Python/Java solution

In Python or Java, implement a loop that tracks empty bottles and the current exchange cost. While empties are at least the required cost, perform an exchange, increment the total bottles drunk, and increase the cost by one. The implementation uses only a few integer variables and constant extra space.

How to solve Water Bottles II in O(1)?

Use a mathematical observation about the increasing exchange cost. If k exchanges occur, the costs form an arithmetic sequence starting from numExchange. Compute the total empties spent using the sum of that sequence and ensure the remaining empties stay non‑negative. Solving the inequality gives the maximum k, and the final result is numBottles + k.

What is the best approach for Water Bottles II?

The most practical approach is iterative simulation. Track the number of empty bottles, repeatedly exchange them for a new bottle, and increase the exchange cost after each trade. This runs in O(k) time where k is the number of exchanges and uses O(1) space. Many candidates present this first in interviews because it mirrors the problem rules directly.

Is Water Bottles II asked at Google/Amazon/Meta?

Problems involving simulation and mathematical reasoning appear frequently in interviews at large tech companies. While this exact question may not always appear, similar exchange-process or resource-simulation problems are common screening exercises at companies like Amazon and Google.

What data structure is used in Water Bottles II?

No advanced data structure is required. The problem mainly relies on integer variables and arithmetic updates. It falls under simulation and math categories rather than structures like arrays, stacks, or hash maps.

What is the time complexity of Water Bottles II?

The standard simulation solution runs in O(k) time where k is the number of exchanges performed. Each loop iteration handles one exchange and updates the required cost. A mathematical optimization can reduce this to O(1) time by computing the number of valid exchanges using arithmetic progression formulas.

Description Hints 1 Solution Video 21:39

3100. Water Bottles II

Medium78.2% acceptance3 companies

You are given two integers numBottles and numExchange.

numBottles represents the number of full water bottles that you initially have. In one operation, you can perform one of the following operations:

Drink any number of full water bottles turning them into empty bottles.
Exchange numExchange empty bottles with one full water bottle. Then, increase numExchange by one.

Note that you cannot exchange multiple batches of empty bottles for the same value of numExchange. For example, if numBottles == 3 and numExchange == 1, you cannot exchange 3 empty water bottles for 3 full bottles.

Return the maximum number of water bottles you can drink.

Example 1:

Input: numBottles = 13, numExchange = 6
Output: 15
Explanation: The table above shows the number of full water bottles, empty water bottles, the value of numExchange, and the number of bottles drunk.

Example 2:

Input: numBottles = 10, numExchange = 3
Output: 13
Explanation: The table above shows the number of full water bottles, empty water bottles, the value of numExchange, and the number of bottles drunk.

Constraints:

1 <= numBottles <= 100
1 <= numExchange <= 100

Asked at 3 companies

Google Amazon Meta

Topics

Math Simulation

Hints open one at a time

Hint 1

Simulate the process step by step. At each step, drink numExchange bottles of water then exchange them for a full bottle. Keep repeating this step until you cannot exchange bottles anymore.

21:39

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