What is the time complexity of Champagne Tower?

The optimal solution has a time complexity of O(query_row^2).

What is the space complexity of Champagne Tower?

The space complexity is O(query_row).

What data structures are used to solve Champagne Tower?

This problem uses Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Champagne Tower?

Insight 1: The problem can be solved using dynamic programming by simulating the champagne flow row by row. Insight 2: We only need to keep track of the current row's champagne levels to calculate the next row's levels, allowing for space optimization.

Which companies ask LeetCode 799. Champagne Tower in interviews?

This problem is frequently asked at National Instruments.

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Champagne Tower - LeetCode 799 Solution

Champagne Tower - Complete Solution Guide

Champagne Tower is LeetCode problem 799, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

We stack glasses in a pyramid, where the first row has 1 glass, the second row has 2 glasses, and so on until the 100 th row. Each glass holds one cup of champagne. Then, some champagne is poured into the first glass at the top. When the topmost glass is full, any excess liquid poured will fall equally to the glass immediately to the left and right of it. When those glasses become full, any excess champagne will fall equally to the left and right of those glasses, and so on. (A glass at the bott

Detailed Explanation

The Champagne Tower problem simulates pouring champagne into a stack of glasses arranged in a pyramid. The goal is to determine how full a specific glass in the tower is after pouring a given amount of champagne into the topmost glass. The champagne flows downwards, and any excess in a glass is split equally between the two glasses directly beneath it. We are given the amount of champagne poured (`poured`), the row index (`query_row`), and the glass index (`query_glass`), and we need to return the amount of champagne in that glass, capped at 1.0 (since a glass can only be full).

Solution Approach

The solution uses dynamic programming to simulate the flow of champagne down the tower. It maintains an array `row_amounts` representing the amount of champagne in each glass of the current row. The algorithm iterates from row 0 up to `query_row`, calculating the champagne levels for each row. For each glass in a row, if it contains more than 1.0 unit of champagne, the excess amount is split equally between the two glasses in the next row directly below it. The `row_amounts` array is updated for each row, and finally, the amount of champagne in the target glass (`query_glass`) is returned, ensuring it's capped at 1.0.

Step-by-Step Algorithm

Step 1: Initialize `row_amounts` with the amount of champagne poured into the top glass (row 0, glass 0).
Step 2: Iterate from row 0 up to `query_row` - 1. In each iteration, calculate the champagne levels for the next row.
Step 3: For each glass in the current row, check if it contains more than 1.0 unit of champagne. If it does, calculate the excess amount.
Step 4: Distribute the excess champagne equally (excess / 2.0) to the two glasses in the next row directly below the current glass.
Step 5: Update the `row_amounts` array with the calculated champagne levels for the next row.
Step 6: After the loop finishes, return the amount of champagne in the glass at `query_row` and `query_glass`, ensuring the result is capped at 1.0.

Key Insights

Insight 1: The problem can be solved using dynamic programming by simulating the champagne flow row by row.
Insight 2: We only need to keep track of the current row's champagne levels to calculate the next row's levels, allowing for space optimization.
Insight 3: It's important to cap the amount of champagne in each glass at 1.0, representing a full glass.

Complexity Analysis

Time Complexity: O(query_row^2)

Space Complexity: O(query_row)

Topics

This problem involves: Dynamic Programming.

Companies

Asked at: National Instruments.

Champagne Tower | LeetCode 799

Medium

Dynamic Programming

Time: O(query_row^2)Space: O(query_row)

Solution Code

Available in:

class Solution {
    public double champagneTower(int poured, int query_row, int query_glass) {
        double[] row_amounts = new double[]{(double) poured};

        for (int i = 0; i < query_row; i++) {
            double[] next_row_amounts = new double[i + 2];
            for (int j = 0; j < i + 1; j++) {
                double amount = row_amounts[j];
                if (amount > 1.0) {
                    double excess = amount - 1.0;
                    next_row_amounts[j] += excess / 2.0;
                    next_row_amounts[j + 1] += excess / 2.0;
                }
            }
            row_amounts = next_row_amounts;
        }

        return Math.min(1.0, row_amounts[query_glass]);
    }
}

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Problem Info

Description

Problem 799: Champagne Tower - Detailed explanation and solution approaches.

Topics

Dynamic Programming

Companies

National Instruments

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:799

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

6 steps

1
Step 1: Initialize `row_amounts` with the amount of champagne poured into the top glass (row 0, glass 0).
2
Step 2: Iterate from row 0 up to `query_row` - 1. In each iteration, calculate the champagne levels for the next row.
3
Step 3: For each glass in the current row, check if it contains more than 1.0 unit of champagne. If it does, calculate the excess amount.
4
Step 4: Distribute the excess champagne equally (excess / 2.0) to the two glasses in the next row directly below the current glass.
5
Step 5: Update the `row_amounts` array with the calculated champagne levels for the next row.
6
Step 6: After the loop finishes, return the amount of champagne in the glass at `query_row` and `query_glass`, ensuring the result is capped at 1.0.

💡

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(query_row^2)

💾

Space Complexity

O(query_row)

📝

Why?

The time complexity is determined by the nested loop structure, where the outer loop processes each row up to `query_row`, and the inner loop processes each glass in the current row. The space complexity arises from storing the champagne levels for each glass in the current row, which requires space proportional to the row number.

📖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

Recursion with Memoization

A recursive approach with memoization could also solve this problem. We define a recursive function that calculates the amount of champagne in a given glass based on the amounts in the glasses above it. Memoization is used to avoid redundant calculations.

Time: O(query_row^2)Space: O(query_row^2)

Trade-offs: Recursion with memoization can be more intuitive for some, but might have slightly higher overhead due to function calls and uses more memory due to the memoization table. The iterative approach is generally preferred for its efficiency.

✨

Key Insights

1Insight 1: The problem can be solved using dynamic programming by simulating the champagne flow row by row.
2Insight 2: We only need to keep track of the current row's champagne levels to calculate the next row's levels, allowing for space optimization.
3Insight 3: It's important to cap the amount of champagne in each glass at 1.0, representing a full glass.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `poured` is 0. The solution correctly handles this by initializing the `row_amounts` with 0.0, resulting in all glasses being empty.
Edge case 2: `query_row` is 0 and `query_glass` is 0. The solution directly returns the minimum of 1.0 and the initial `poured` value.
Edge case 3: The requested glass at `query_row`, `query_glass` ends up containing more than 1 unit of champagne. The solution uses `min(1.0, row_amounts[query_glass])` to return 1.0 in such cases, representing a full glass.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to cap the amount of champagne in each glass at 1.0. This will lead to incorrect results because a glass can only be full.

✗Common mistake 2: Not handling the excess champagne correctly. The excess must be split equally between the two glasses directly below.

✗Common mistake 3: Incorrectly indexing the glasses in the next row. Ensure that the excess champagne is distributed to the correct glasses in the next row.

✅

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: Simulating fluid dynamics in systems with cascading or overflowing elements. For example, modeling the flow of water through a series of connected tanks or reservoirs.

🏦Real-world scenario 2: Distributing resources or rewards in a hierarchical structure where each level distributes excess resources to the level below. For instance, calculating bonuses distributed down a management hierarchy.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

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 19, 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

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

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Champagne Tower"?

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

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