Leetcode #1518: Water Bottles
In this guide, we solve Leetcode #1518 Water Bottles in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There are numBottles water bottles that are initially full of water. You can exchange numExchange empty water bottles from the market with one full water bottle.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Math, Simulation
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
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.
Python Solution
class Solution:
def numWaterBottles(self, numBottles: int, numExchange: int) -> int:
ans = numBottles
while numBottles >= numExchange:
numBottles -= numExchange - 1
ans += 1
return ans
Complexity
The time complexity is O(n) or O(1). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.