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Leetcode 1518. Water Bottles

Problem Statement:

You have numBottles full water bottles. You drink one bottle of water and eventually recycle it, getting one empty bottle. For every numExchange empty bottles you can get one full water bottle.

The task is to find the maximum number of water bottles you can drink.

Clarifying Questions:

1. Q: Can the values of numBottles and numExchange be zero or negative? A: The problem constraints clarify that these values will be positive integers.

2. Q: Is there an upper limit on the values for numBottles and numExchange? A: Constraints typically ensure that these values are within a reasonable range to compute within time limits.

3. Q: Are there any other edge cases to consider (like numBottles < numExchange)? A: If numBottles is less than numExchange, you cannot exchange any bottles, so the number of bottles you drink will just be numBottles.

Strategy:

1. We initially have numBottles full bottles.
2. While we have enough empty bottles to exchange for more full bottles, we continue:
   • Drink all initially full bottles and convert them to empty bottles.
   • From these empty bottles, exchange them into as many full bottles as possible.
   • Track the total count of bottles drunk.
3. The loop continues until the number of empty bottles is less than numExchange.

Time Complexity:

The time complexity of the solution will be O(log(numBottles)) since in each iteration, the number of bottles reduces significantly.

Code:

public class Solution {
    public int numWaterBottles(int numBottles, int numExchange) {
        int totalDrunk = 0;
        int emptyBottles = 0;
        
        while (numBottles > 0) {
            // Drink all current full bottles
            totalDrunk += numBottles;
            emptyBottles += numBottles;
            numBottles = 0;
            
            // Exchange empty bottles for full bottles
            numBottles = emptyBottles / numExchange;
            emptyBottles = emptyBottles % numExchange;
        }
        
        return totalDrunk;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        // Test case 1
        System.out.println(solution.numWaterBottles(9, 3)); // Output: 13
        
        // Test case 2
        System.out.println(solution.numWaterBottles(15, 4)); // Output: 19
        
        // Test case 3
        System.out.println(solution.numWaterBottles(5, 5)); // Output: 6
        
        // Test case 4
        System.out.println(solution.numWaterBottles(2, 3)); // Output: 2
    }
}

Explanation:

• Initial Setup: We start with the initial number of numBottles full bottles.
• While Loop: As long as there are full bottles, we:
  • Add the number of full bottles to totalDrunk.
  • Convert the full bottles to empty bottles.
  • Calculate how many new full bottles we obtain from exchanging empty bottles.
  • Update the number of empty bottles remaining after exchanges.
• Termination: The loop terminates when there are not enough empty bottles to exchange for a new full bottle.
• Edge Cases: Cases where the initial number of bottles is already less than what is needed for an exchange are handled naturally by the loop terminating early.

By following this approach, we ensure an efficient and correct solution to the problem.

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