#1518

Easy

LeetCode Problem Workspace

Water Bottles

Maximize the number of water bottles you can drink by simulating the exchange process between full and empty bottles.

Problem Statement

You are given an initial number of full water bottles, numBottles, and an exchange rate, numExchange, which determines how many empty bottles are needed to get one full bottle. Each time you drink a full bottle, it turns into an empty one, and you can exchange empty bottles for full ones as long as you have enough to do so.

Your task is to simulate the process of drinking the water bottles and exchanging the empty ones, and then return the maximum number of bottles you can drink in total. Make sure to simulate the exchange process until you can no longer exchange enough empty bottles for a full bottle.

Examples

Example 1

Input: numBottles = 9, numExchange = 3

Output: 13

You can exchange 3 empty bottles to get 1 full water bottle. Number of water bottles you can drink: 9 + 3 + 1 = 13.

Example 2

Input: numBottles = 15, numExchange = 4

Output: 19

You can exchange 4 empty bottles to get 1 full water bottle. Number of water bottles you can drink: 15 + 3 + 1 = 19.

Constraints

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

Solution Approach

Simulate the Drinking and Exchange Process

Start by drinking all the full bottles. Then, for each exchange, trade as many empty bottles as possible for full ones. Repeat the process until there aren't enough empty bottles left to exchange for one full bottle.

Track Full and Empty Bottles

Keep track of both full and empty bottles at each step. Each time you exchange bottles, update the count of full bottles and empty bottles accordingly.

Terminate When No More Exchanges Are Possible

The loop stops when you can't exchange enough empty bottles to get at least one full bottle. At that point, you have consumed all possible water bottles.

Complexity Analysis

Metric	Value
Time	O(\log_{M} N)
Space	O(1)

The time complexity is O(logM N), where M is the exchange rate and N is the number of bottles, due to the iterative process of exchanging bottles. The space complexity is O(1) since only a few variables are used to track the counts of full and empty bottles.

What Interviewers Usually Probe

Assesses the candidate's ability to simulate a real-world process through efficient code.
Tests for clear and concise iteration handling within a loop.
Evaluates the candidate's attention to detail when managing both full and empty bottles.

Common Pitfalls or Variants

Common pitfalls

Failing to correctly simulate the bottle exchange and drinking process.
Not accounting for the scenario when no more full bottles can be obtained from empty ones.
Overcomplicating the solution with unnecessary complexity or data structures.

Follow-up variants

Increase the number of empty bottles per exchange to test the candidate's ability to handle larger steps in the process.
Test with edge cases where the number of bottles is at its minimum (e.g., numBottles = 1).
Challenge with larger values for numBottles and numExchange to assess performance with bigger inputs.

FAQ

What is the key to solving the Water Bottles problem?

The key is to simulate the process of drinking full bottles and exchanging empty ones until no more exchanges can be made.

How does the pattern of Math plus Simulation apply to this problem?

The problem combines mathematical operations (tracking full and empty bottles) with simulation (iterating through exchanges) to achieve the solution.

What’s the time complexity of the Water Bottles problem?

The time complexity is O(logM N), where M is the exchange rate and N is the number of bottles.

Can the number of full bottles be zero at the start of the problem?

No, the problem constraints guarantee that you have at least one full bottle (numBottles >= 1).

How do I ensure that the simulation correctly handles all exchanges?

Keep track of both full and empty bottles throughout the simulation and continue exchanging until no more exchanges are possible.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def numWaterBottles(self, numBottles: int, numExchange: int) -> int:
        ans = numBottles
        while numBottles >= numExchange:
            numBottles -= numExchange - 1
            ans += 1
        return ans

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