Water Bottles - Greedy - Easy - LeetCode
💻 coding

Water Bottles - Greedy - Easy - LeetCode

1 min read 193 words
1 min read
ShareWhatsAppPost on X
  • 1You can drink a maximum number of water bottles by exchanging empty ones for full ones.
  • 2The algorithm calculates total bottles consumed using a while loop and modular arithmetic.
  • 3Time complexity is O(n) while space complexity is O(1), making it efficient.

AI-generated summary · May not capture all nuances

Key Insight
AskGif

"You can drink a maximum number of water bottles by exchanging empty ones for full ones."

Water Bottles - Greedy - Easy - LeetCode

Given numBottles full water bottles, you can exchange numExchange empty water bottles for one full water bottle.

The operation of drinking a full water bottle turns it into an empty bottle.

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. Example 3:

Input: numBottles = 5, numExchange = 5 Output: 6 Example 4:

Input: numBottles = 2, numExchange = 3 Output: 2

Constraints:

1 <= numBottles <= 100 2 <= numExchange <= 100

public class Solution {
 public int NumWaterBottles(int numBottles, int numExchange) {
 int total = 0;
 int spare = 0;
 while(numBottles>0){
 total += numBottles;
 numBottles += spare; 
 spare = numBottles%numExchange;
 numBottles = numBottles/numExchange;
 }
 
 return total;
 }
}

Time Complexity: O(n)

Space Complexity: O(1)

Enjoyed this article?

Share it with someone who'd find it useful.

ShareWhatsAppPost on X

sumitc91

Published on 5 October 2020 · 1 min read · 193 words

Part of AskGif Blog · coding

You might also like