LeetCode #1518 — EASY

Water Bottles

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

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.

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

Given the two integers numBottles and numExchange, 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.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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. The operation of drinking a full water bottle turns it into an empty bottle. Given the two integers numBottles and numExchange, return the maximum number of water bottles you can drink.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

9
3

Example 2

15
4

Core Insight

What unlocks the optimal approach

Simulate the process until there are not enough empty bottles for even one full bottle of water.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1518: Water Bottles
class Solution {
    public int numWaterBottles(int numBottles, int numExchange) {
        int ans = numBottles;
        for (; numBottles >= numExchange; ++ans) {
            numBottles -= (numExchange - 1);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1518: Water Bottles
func numWaterBottles(numBottles int, numExchange int) int {
	ans := numBottles
	for ; numBottles >= numExchange; ans++ {
		numBottles -= (numExchange - 1)
	}
	return ans
}

# Accepted solution for LeetCode #1518: Water Bottles
class Solution:
    def numWaterBottles(self, numBottles: int, numExchange: int) -> int:
        ans = numBottles
        while numBottles >= numExchange:
            numBottles -= numExchange - 1
            ans += 1
        return ans

// Accepted solution for LeetCode #1518: Water Bottles
struct Solution;

impl Solution {
    fn num_water_bottles(num_bottles: i32, num_exchange: i32) -> i32 {
        let mut full = num_bottles;
        let mut empty = 0;
        let mut res = 0;
        while full > 0 {
            res += full;
            empty += full;
            full = empty / num_exchange;
            empty %= num_exchange;
        }
        res
    }
}

#[test]
fn test() {
    let num_bottles = 9;
    let num_exchange = 3;
    let res = 13;
    assert_eq!(Solution::num_water_bottles(num_bottles, num_exchange), res);
    let num_bottles = 15;
    let num_exchange = 4;
    let res = 19;
    assert_eq!(Solution::num_water_bottles(num_bottles, num_exchange), res);
    let num_bottles = 15;
    let num_exchange = 4;
    let res = 19;
    assert_eq!(Solution::num_water_bottles(num_bottles, num_exchange), res);
    let num_bottles = 5;
    let num_exchange = 5;
    let res = 6;
    assert_eq!(Solution::num_water_bottles(num_bottles, num_exchange), res);
    let num_bottles = 2;
    let num_exchange = 3;
    let res = 2;
    assert_eq!(Solution::num_water_bottles(num_bottles, num_exchange), res);
}

// Accepted solution for LeetCode #1518: Water Bottles
function numWaterBottles(numBottles: number, numExchange: number): number {
    let ans = numBottles;
    for (; numBottles >= numExchange; ++ans) {
        numBottles -= numExchange - 1;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.