There are numBottles water bottles that are initially full of water. You can exchange numExchange empty water bottles from the market with one full water bottle.
The operation of drinking a full water bottle turns it into an empty bottle.
Given the two integers numBottles and numExchange, return the maximum number of water bottles you can drink.
Example 1:
Input: numBottles = 9, numExchange = 3 Output: 13 Explanation: 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 Explanation: 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 <= 1002 <= numExchange <= 100This approach uses a simple iterative simulation to solve the problem. We keep track of the current number of full bottles and empty bottles. We repeatedly drink a full bottle and calculate how many empty bottles we have. If the number of empty bottles is enough to exchange for a new full bottle, we perform the exchange and continue the loop. This process is repeated until no more exchanges can be made.
The function numWaterBottles calculates the maximum number of water bottles one can drink. It uses a loop to repeatedly simulate drinking and exchanging bottles while keeping track of total bottles drunk and current empty bottles.
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Time Complexity: O(numBottles).
Space Complexity: O(1).
This approach leverages a mathematical understanding of the problem. Instead of simulating every exchange, you can calculate how many overall bottles you'll get by using a formula. This involves understanding the number of bottles resulting from continuous exchanges until no more can be made.
In this C solution, a while loop is used to calculate the total number of bottles drunk by updating the current bottles with the new one obtained from the exchange.
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Java
Python
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Time Complexity: O(log(numBottles)).
Space Complexity: O(1).
| Approach | Complexity |
|---|---|
| Iterative Simulation | Time Complexity: O(numBottles). |
| Mathematical Approach | Time Complexity: O(log(numBottles)). |
Water Bottles - Leetcode 1518 - Python • NeetCodeIO • 9,468 views views
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