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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).
1#include <iostream>
2
3int numWaterBottles(int numBottles, int numExchange) {
4 int totalDrank = numBottles;
5 int emptyBottles = numBottles;
6 while (emptyBottles >= numExchange) {
7 int newBottles = emptyBottles / numExchange;
8 totalDrank += newBottles;
9 emptyBottles = emptyBottles % numExchange + newBottles;
10 }
11 return totalDrank;
12}
13
14int main() {
15 std::cout << numWaterBottles(9, 3) << std::endl;
16 return 0;
17}The solution iteratively updates the total count of drunk bottles by continuing to drink and exchange until no more exchanges can be made. It simulates bottle usage with each iteration.
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).
1function
This JavaScript solution conducts direct updates to total bottles with new exchanges calculated in a loop, efficiently tallying maximum drinks.