What is the time complexity of Water Bottles?

The optimal solution has a time complexity of O(log(n)).

What data structures are used to solve Water Bottles?

This problem uses Math, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Water Bottles?

Insight 1: The problem can be solved iteratively by repeatedly exchanging empty bottles for full ones until no more exchanges are possible. Insight 2: Using integer division (`//` or `/`) and modulo operator (`%` or `%`) efficiently handles the exchange process and tracking of remaining empty bottles.

Which companies ask LeetCode 1518. Water Bottles in interviews?

This problem is frequently asked at Amadeus, HiLabs.

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Water Bottles - LeetCode 1518 Solution

Water Bottles - Complete Solution Guide

Water Bottles is LeetCode problem 1518, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks you to calculate the maximum number of water bottles you can drink given a starting number of full water bottles and the number of empty bottles required to exchange for a new full bottle. You drink a bottle to empty it, then exchange empty bottles for full ones until you have fewer empty bottles than needed for an exchange. The input consists of two integers: `numBottles` (the initial number of full bottles) and `numExchange` (the number of empty bottles needed for an exchange). The output is a single integer representing the total number of bottles drunk.

Solution Approach

The provided code uses an iterative approach. It initializes `total_drunk` (or `totalBottles`) with the initial number of bottles. It then enters a `while` loop that continues as long as the number of empty bottles (`empty_bottles` or `emptyBottles`) is greater than or equal to the number of bottles needed for an exchange (`numExchange`). Inside the loop, it calculates the number of new bottles obtained from the exchange, adds them to the `total_drunk`, and updates the number of empty bottles. The loop terminates when no more exchanges are possible, and the total number of bottles drunk is returned.

Step-by-Step Algorithm

Step 1: Initialize `total_drunk` (the total number of bottles drunk) to `numBottles`. Initialize `empty_bottles` (the number of empty bottles) to `numBottles`.
Step 2: Enter a `while` loop that continues as long as `empty_bottles` is greater than or equal to `numExchange`.
Step 3: Calculate the number of new bottles obtained from the exchange: `new_bottles = empty_bottles // numExchange`.
Step 4: Add the `new_bottles` to `total_drunk`.
Step 5: Update `empty_bottles`: `empty_bottles = empty_bottles % numExchange + new_bottles` (the remaining empty bottles plus the newly obtained bottles).
Step 6: Repeat steps 3-5 until the `while` loop condition is false.
Step 7: Return `total_drunk`.

Key Insights

Insight 1: The problem can be solved iteratively by repeatedly exchanging empty bottles for full ones until no more exchanges are possible.
Insight 2: Using integer division (`//` or `/`) and modulo operator (`%` or `%`) efficiently handles the exchange process and tracking of remaining empty bottles.
Insight 3: The loop condition `emptyBottles >= numExchange` ensures that the exchange process stops when there aren't enough empty bottles left for another exchange.

Complexity Analysis

Time Complexity: O(log(n))

Space Complexity: O(1)

Topics

This problem involves: Math, Simulation.

Companies

Asked at: Amadeus, HiLabs.

Water Bottles | LeetCode 1518

Easy

Math Simulation

Time: O(log(n))Space: O(1)

Solution Code

Available in:

class Solution {
    public int numWaterBottles(int numBottles, int numExchange) {
        int totalBottles = numBottles;
        int emptyBottles = numBottles;
        while (emptyBottles >= numExchange) {
            int newBottles = emptyBottles / numExchange;
            totalBottles += newBottles;
            emptyBottles = emptyBottles % numExchange + newBottles;
        }
        return totalBottles;
    }
}

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Problem Info

Description

Problem 1518: Water Bottles - Detailed explanation and solution approaches.

Topics

Math Simulation

Companies

AmadeusHiLabs

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:1518

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize `total_drunk` (the total number of bottles drunk) to `numBottles`. Initialize `empty_bottles` (the number of empty bottles) to `numBottles`.
2
Step 2: Enter a `while` loop that continues as long as `empty_bottles` is greater than or equal to `numExchange`.
3
Step 3: Calculate the number of new bottles obtained from the exchange: `new_bottles = empty_bottles // numExchange`.
4
Step 4: Add the `new_bottles` to `total_drunk`.
5
Step 5: Update `empty_bottles`: `empty_bottles = empty_bottles % numExchange + new_bottles` (the remaining empty bottles plus the newly obtained bottles).
6
Step 6: Repeat steps 3-5 until the `while` loop condition is false.
7
Step 7: Return `total_drunk`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log(n))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is logarithmic because the number of empty bottles roughly halves with each iteration. The space complexity is constant because only a few integer variables are used, regardless of the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Mathematical Approach

Derive a mathematical formula to directly calculate the total number of bottles drunk. This avoids the iterative approach.

Time: O(1)Space: O(1)

Trade-offs: While slightly more complex to derive, this approach offers the best time complexity. It's preferable when performance is critical and the input numbers are within reasonable bounds.

✨

Key Insights

1Insight 1: The problem can be solved iteratively by repeatedly exchanging empty bottles for full ones until no more exchanges are possible.
2Insight 2: Using integer division (`//` or `/`) and modulo operator (`%` or `%`) efficiently handles the exchange process and tracking of remaining empty bottles.
3Insight 3: The loop condition `emptyBottles >= numExchange` ensures that the exchange process stops when there aren't enough empty bottles left for another exchange.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: `numBottles < numExchange`. The `while` loop will not execute, and the function will correctly return `numBottles`.
Edge case 2: `numBottles` is a multiple of `numExchange`. The loop will continue until the remaining empty bottles are less than `numExchange`, resulting in the correct calculation of the maximum number of bottles that can be drunk.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the modulo operation. Forgetting to add the newly obtained bottles to the remaining empty bottles in the update step (`empty_bottles = empty_bottles % numExchange + new_bottles`) will lead to an incorrect result.

✗Common mistake 2: Off-by-one errors in the loop condition or in updating the number of empty bottles. Carefully checking the loop termination condition and the update formula is crucial.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Modeling resource management problems, such as managing inventory where you can exchange smaller items for larger ones.

🏦Real-world scenario 2: Simulating reward systems where points or tokens can be exchanged for higher-value rewards.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 18, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Water Bottles"?

For this easy-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether simulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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