LeetCode #3100 — MEDIUM

Water Bottles II

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

You are given two integers numBottles and numExchange.

numBottles represents the number of full water bottles that you initially have. In one operation, you can perform one of the following operations:

Drink any number of full water bottles turning them into empty bottles.
Exchange numExchange empty bottles with one full water bottle. Then, increase numExchange by one.

Note that you cannot exchange multiple batches of empty bottles for the same value of numExchange. For example, if numBottles == 3 and numExchange == 1, you cannot exchange 3 empty water bottles for 3 full bottles.

Return the maximum number of water bottles you can drink.

Example 1:

Input: numBottles = 13, numExchange = 6
Output: 15
Explanation: The table above shows the number of full water bottles, empty water bottles, the value of numExchange, and the number of bottles drunk.

Example 2:

Input: numBottles = 10, numExchange = 3
Output: 13
Explanation: The table above shows the number of full water bottles, empty water bottles, the value of numExchange, and the number of bottles drunk.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: You are given two integers numBottles and numExchange. numBottles represents the number of full water bottles that you initially have. In one operation, you can perform one of the following operations: Drink any number of full water bottles turning them into empty bottles. Exchange numExchange empty bottles with one full water bottle. Then, increase numExchange by one. Note that you cannot exchange multiple batches of empty bottles for the same value of numExchange. For example, if numBottles == 3 and numExchange == 1, you cannot exchange 3 empty water bottles for 3 full bottles. 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

13
6

Example 2

10
3

Core Insight

What unlocks the optimal approach

Simulate the process step by step. At each step, drink <code>numExchange</code> bottles of water then exchange them for a full bottle. Keep repeating this step until you cannot exchange bottles anymore.

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 #3100: Water Bottles II
class Solution {
    public int maxBottlesDrunk(int numBottles, int numExchange) {
        int ans = numBottles;
        while (numBottles >= numExchange) {
            numBottles -= numExchange;
            ++numExchange;
            ++ans;
            ++numBottles;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3100: Water Bottles II
func maxBottlesDrunk(numBottles int, numExchange int) int {
	ans := numBottles
	for numBottles >= numExchange {
		numBottles -= numExchange
		numExchange++
		ans++
		numBottles++
	}
	return ans
}

# Accepted solution for LeetCode #3100: Water Bottles II
class Solution:
    def maxBottlesDrunk(self, numBottles: int, numExchange: int) -> int:
        ans = numBottles
        while numBottles >= numExchange:
            numBottles -= numExchange
            numExchange += 1
            ans += 1
            numBottles += 1
        return ans

// Accepted solution for LeetCode #3100: Water Bottles II
impl Solution {
    pub fn max_bottles_drunk(mut num_bottles: i32, mut num_exchange: i32) -> i32 {
        let mut ans = num_bottles;

        while num_bottles >= num_exchange {
            num_bottles -= num_exchange;
            num_exchange += 1;
            ans += 1;
            num_bottles += 1;
        }

        ans
    }
}

// Accepted solution for LeetCode #3100: Water Bottles II
function maxBottlesDrunk(numBottles: number, numExchange: number): number {
    let ans = numBottles;
    while (numBottles >= numExchange) {
        numBottles -= numExchange;
        ++numExchange;
        ++ans;
        ++numBottles;
    }
    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.