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This 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.
Time Complexity: O(numBottles).
Space Complexity: O(1).
1def numWaterBottles(numBottles, numExchange):
2 total_drunk = numBottles
3 empty_bottles = numBottles
4 while empty_bottles >= numExchange:
5 new_bottles = empty_bottles // numExchange
6 total_drunk += new_bottles
7 empty_bottles = empty_bottles % numExchange + new_bottles
8 return total_drunk
9
10print(numWaterBottles(9, 3))The Python solution simulates the process of drinking and exchanging bottles while updating the count of total bottles drunk using a while loop.
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.
Time Complexity: O(log(numBottles)).
Space Complexity: O(1).
1def
This Python function computes the total bottles drunk using a concise loop to update the total directly with new bottles obtained from exchanges.