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Formatted question description: https://leetcode.ca/all/1518.html

1518. Water Bottles (Easy)

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

Related Topics:
Greedy

Solution 1.

// OJ: https://leetcode.com/problems/water-bottles/
// Time: O(log_exchange^bottle)
// Space: O(1)
class Solution {
public:
    int numWaterBottles(int bottle, int exchange) {
        int ans = 0, empty = 0;
        while (bottle) {
            ans += bottle;
            empty += bottle;
            int ex = empty / exchange;
            empty -= ex * exchange;
            bottle = ex;
        }
        return ans;
    }
};
  • class Solution {
        public int numWaterBottles(int numBottles, int numExchange) {
            int total = 0;
            int water = numBottles, empty = 0;
            while (water > 0) {
                total += water;
                empty += water;
                water = 0;
                int exchange = empty / numExchange;
                water += exchange;
                empty -= numExchange * exchange;
            }
            return total;
        }
    }
    
    ############
    
    class Solution {
        public int numWaterBottles(int numBottles, int numExchange) {
            int ans = numBottles;
            while (numBottles >= numExchange) {
                numBottles -= (numExchange - 1);
                ++ans;
            }
            return ans;
        }
    }
    
  • // OJ: https://leetcode.com/problems/water-bottles/
    // Time: O(log_exchange^bottle)
    // Space: O(1)
    class Solution {
    public:
        int numWaterBottles(int bottle, int exchange) {
            int ans = 0, empty = 0;
            while (bottle) {
                ans += bottle;
                empty += bottle;
                int ex = empty / exchange;
                empty -= ex * exchange;
                bottle = ex;
            }
            return ans;
        }
    };
    
  • class Solution:
        def numWaterBottles(self, numBottles: int, numExchange: int) -> int:
            ans = numBottles
            while numBottles >= numExchange:
                numBottles -= numExchange - 1
                ans += 1
            return ans
    
    
    
  • func numWaterBottles(numBottles int, numExchange int) int {
    	ans := numBottles
    	for numBottles >= numExchange {
    		numBottles -= (numExchange - 1)
    		ans++
    	}
    	return ans
    }
    

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