LeetCode Challenge Day 16 — 1518. Water Bottles

Hey folks

This is Day 16 of my LeetCode streak 🚀.
Today’s problem is 1518. Water Bottles — we’re given some initial full bottles and a rule for exchanging empty ones. The goal is to figure out the maximum number of bottles we can drink.

It’s a classic simulation-style problem with a neat greedy loop.

📌 Problem Statement

You are given two integers:

• numBottles → the number of full water bottles you initially have.
• numExchange → the number of empty bottles required to exchange for one new full bottle.

Each time you drink a bottle, it becomes empty. You can keep exchanging empty bottles for new full bottles until no further exchanges are possible.

Return the maximum number of bottles you can drink.

Examples

• Input: numBottles = 9, numExchange = 3
  Output: 13
  Explanation: Drink 9 → exchange for 3 → drink 3 → exchange for 1 → drink 1 → total = 13.

• Input: numBottles = 15, numExchange = 4
  Output: 19
  Explanation: Drink 15 → exchange for 3 → drink 3 → exchange for 1 → drink 1 → total = 19.

Constraints

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

💡 Intuition

The problem mimics real life:

• Drink all bottles you have.
• Count the empty ones.
• Trade empties for full bottles as long as possible.

This can be directly simulated with a loop — each exchange reduces empty bottles and adds new full ones.

🔑 Approach

1. Start with totalDrunk = numBottles.
2. Keep a counter of empty bottles.
3. While you have at least numExchange empties:
   • Exchange them for new full bottles.
   • Add the new bottles to totalDrunk.
   • Update the empty count as (empties % numExchange) + newBottles.
4. Once no more exchanges are possible, return totalDrunk.

⏱️ Complexity Analysis

• Time complexity: O(log numBottles) — since bottles reduce quickly per exchange.
• Space complexity: O(1) — only a few variables are used.

🧑‍💻 Code (JavaScript)

/**
 * @param {number} numBottles
 * @param {number} numExchange
 * @return {number}
 */
var numWaterBottles = function(numBottles, numExchange) {
  let totalDrunk = numBottles;
  let empty = numBottles;

  while (empty >= numExchange) {
    let newBottles = Math.floor(empty / numExchange);
    totalDrunk += newBottles;
    empty = (empty % numExchange) + newBottles;
  }

  return totalDrunk;
};

// ✅ Quick tests
console.log(numWaterBottles(9, 3));  // 13
console.log(numWaterBottles(15, 4)); // 19

🧪 Edge Cases

• numBottles < numExchange → no exchange possible, answer = numBottles.

• Exactly divisible → the loop continues until 1 empty remains.

• Small inputs → works fine since constraints are tiny.

🎥 Reflections

This problem highlights how simple simulation loops can solve real-world style challenges.
It’s not always about math shortcuts — sometimes just carefully simulating step by step is the cleanest approach.

That’s it for Day 16 of my LeetCode journey!
On to the next challenge 🔥

Happy Coding 👨‍💻