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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.

• Java
• C++
• Python
• Go
• Javascript
• Php

• 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
  }
  

• /**
   * @param {number} numBottles
   * @param {number} numExchange
   * @return {number}
   */
  var numWaterBottles = function (numBottles, numExchange) {
      let sum = numBottles;
      while (numBottles >= numExchange) {
          numBottles = numBottles - numExchange + 1;
          sum++;
      }
      return sum;
  };
  

• class Solution {
      /**
       * @param Integer $numBottles
       * @param Integer $numExchange
       * @return Integer
       */
      function numWaterBottles($numBottles, $numExchange) {
          $sum = $numBottles;
          while ($numBottles >= $numExchange) {
              $numBottles = $numBottles - $numExchange + 1;
              $sum++;
          }
          return $sum;
      }
  }
  

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