Dare2Solve
The problem requires determining the maximum number of water bottles that can be consumed, considering the ability to exchange empty bottles for full ones. Initially, you have a certain number of full water bottles (numBottles
). Each time you drink a bottle, it turns into an empty bottle, and once you accumulate enough empty bottles (numExchange
), you can trade them for a new full bottle.
The intuition is to continuously keep track of the number of full and empty bottles, ensuring to maximize the drinking process by always exchanging empty bottles for full ones when possible.
numBottles
).numExchange
).In terms of implementation, we can use a while loop to simulate this process, updating the count of full and empty bottles at each step.
To derive the formula numBottles + (numBottles - 1) // (numExchange - 1)
, let's break down the process and understand how it accounts for the maximum number of water bottles you can drink:
Initial Full Bottles: Start with numBottles
full bottles.
Drinking and Exchanging:
numExchange
empty bottles for one full bottle.Counting Bottles:
numBottles
full bottles.numBottles
empty bottles.numExchange
empty bottles, you get one full bottle.Instead of simulating the process, observe the relationship between the number of bottles and exchanges:
Total Bottles Drunk:
T
.Exchanges Needed:
numExchange
empty bottles.Formula Derivation:
numExchange - 1
(because one bottle is gained while numExchange
empty bottles are used up).Combining these insights, the total number of bottles drunk can be expressed as:
[ T = numBottles + floor((numbottles - 1) / (numExchange - 1)) ]
numBottles
: Represents the initial count of full bottles.(numBottles - 1) // (numExchange - 1)
: Represents the total number of additional bottles you can get through exchanges. Subtracting 1 from numBottles
and numExchange
aligns the exchange process with the available empty bottles.By deriving this formula, we avoid the iterative process and directly calculate the maximum number of bottles that can be drunk.
Time Complexity: (O(1))
Space Complexity: (O(1))
class Solution {
public:
int numWaterBottles(int numBottles, int numExchange) {
return numBottles + (numBottles - 1) / (numExchange - 1);
}
};
class Solution {
public int numWaterBottles(int numBottles, int numExchange) {
return numBottles + (numBottles - 1) / (numExchange - 1);
}
}
/**
* @param {number} numBottles
* @param {number} numExchange
* @return {number}
*/
var numWaterBottles = function(numBottles, numExchange) {
return numBottles + Math.floor((numBottles-1)/(numExchange-1));
};