What is the time complexity of Egg Drop With 2 Eggs and N Floors?

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

What data structures are used to solve Egg Drop With 2 Eggs and N Floors?

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

What are the key insights for solving Egg Drop With 2 Eggs and N Floors?

Insight 1: The first egg is used to find a range of floors, and the second egg is used to linearly search within that range if the first egg breaks. Insight 2: The optimal strategy involves decreasing the number of floors the first egg skips by one each time it doesn't break. This balances the worst-case scenarios (egg breaks early vs. egg breaks late).

Which companies ask LeetCode 1884. Egg Drop With 2 Eggs and N Floors in interviews?

This problem is frequently asked at Citadel, Disney.

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Egg Drop With 2 Eggs and N Floors - LeetCode 1884 Solution

Q: How to solve Egg Drop With 2 Eggs and N Floors?

The solution iteratively calculates the minimum number of moves required. It starts with `moves = 1` and increments it until the sum of integers from 1 to `moves` is greater than or equal to `n`. The sum represents the number of floors that can be covered given `moves` number of drops. This leverages the mathematical pattern that allows us to derive a direct relation between the total floors and the number of moves.

Egg Drop With 2 Eggs and N Floors - Complete Solution Guide

Egg Drop With 2 Eggs and N Floors is LeetCode problem 1884, 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

You are given two identical eggs and you have access to a building with n floors labeled from 1 to n . You know that there exists a floor f where 0 <= f <= n such that any egg dropped at a floor higher than f will break , and any egg dropped at or below floor f will not break . In each move, you may take an unbroken egg and drop it from any floor x (where 1 <= x <= n ). If the egg breaks, you can no longer use it. However, if the egg does not break, you may reuse it in future moves. Return the m

Detailed Explanation

The problem asks us to find the minimum number of moves required to determine the critical floor `f` in a building with `n` floors, given two identical eggs. If an egg is dropped from a floor higher than `f`, it breaks. If it's dropped from floor `f` or below, it doesn't break. We can reuse eggs that don't break. The goal is to find `f` with certainty using the fewest possible drops.

Solution Approach

The solution iteratively calculates the minimum number of moves required. It starts with `moves = 1` and increments it until the sum of integers from 1 to `moves` is greater than or equal to `n`. The sum represents the number of floors that can be covered given `moves` number of drops. This leverages the mathematical pattern that allows us to derive a direct relation between the total floors and the number of moves.

Step-by-Step Algorithm

Step 1: Initialize `moves` to 0 and `floors_covered` to 0.
Step 2: While `floors_covered` is less than `n`:
Step 3: Increment `moves` by 1.
Step 4: Add the current value of `moves` to `floors_covered`. This represents the additional floors covered by using the current number of moves.
Step 5: Repeat steps 2-4 until `floors_covered` is greater than or equal to `n`.
Step 6: Return the final value of `moves`.

Key Insights

Insight 1: The first egg is used to find a range of floors, and the second egg is used to linearly search within that range if the first egg breaks.
Insight 2: The optimal strategy involves decreasing the number of floors the first egg skips by one each time it doesn't break. This balances the worst-case scenarios (egg breaks early vs. egg breaks late).
Insight 3: The problem can be reframed as finding the smallest `moves` such that `1 + 2 + 3 + ... + moves >= n`. This is equivalent to `moves * (moves + 1) / 2 >= n`.

Complexity Analysis

Time Complexity: O(sqrt(n))

Space Complexity: O(1)

Topics

This problem involves: Math, Dynamic Programming.

Companies

Asked at: Citadel, Disney.

Egg Drop With 2 Eggs and N Floors | LeetCode 1884

Medium

Math Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int twoEggDrop(int n) {
        int moves = 0;
        int floorsCovered = 0;
        while (floorsCovered < n) {
            moves += 1;
            floorsCovered += moves;
        }
        return moves;
    }
}

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Problem Info

Description

Problem 1884: Egg Drop With 2 Eggs and N Floors - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming

Companies

CitadelDisney

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1884

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 `moves` to 0 and `floors_covered` to 0.
2
Step 2: While `floors_covered` is less than `n`:
3
Step 3: Increment `moves` by 1.
4
Step 4: Add the current value of `moves` to `floors_covered`. This represents the additional floors covered by using the current number of moves.
5
Step 5: Repeat steps 2-4 until `floors_covered` is greater than or equal to `n`.
6
Step 6: Return the final value of `moves`.

💡

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(sqrt(n))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is derived from the fact that the loop iterates until the sum of the first 'moves' natural numbers reaches 'n'. Since the sum of the first 'k' natural numbers is k*(k+1)/2, we can approximate 'moves' as sqrt(2n). Hence, the number of iterations is proportional to the square root of n. The space complexity is constant as only a few integer variables are used.

📖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

Binary Search

Binary search can be applied to find the minimum moves required. Define a function `check(moves, n)` which calculates if it's possible to determine f with a given number of moves. Then, use binary search between 1 and n to find the minimum number of moves for which check(moves, n) is true.

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

Trade-offs: The binary search approach may be more complex to understand initially but offers a logarithmic time complexity. However, the constant factor in binary search might make it less efficient for smaller values of n compared to the iterative approach. The iterative approach is easier to understand and implement for this specific problem, given the relatively small constraint on n.

Mathematical Formula

We know that the minimum number of moves, `x`, must satisfy `x*(x+1)/2 >= n`. We can solve this quadratic inequality for `x`. In this case, x = ceil((-1 + sqrt(1 + 8n))/2).

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

Trade-offs: This approach is the most efficient as it directly computes the result using a formula. However, it might be less intuitive to derive without understanding the mathematical relationship underlying the problem.

✨

Key Insights

1Insight 1: The first egg is used to find a range of floors, and the second egg is used to linearly search within that range if the first egg breaks.
2Insight 2: The optimal strategy involves decreasing the number of floors the first egg skips by one each time it doesn't break. This balances the worst-case scenarios (egg breaks early vs. egg breaks late).
3Insight 3: The problem can be reframed as finding the smallest `moves` such that `1 + 2 + 3 + ... + moves >= n`. This is equivalent to `moves * (moves + 1) / 2 >= n`.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1. The loop runs once, and moves becomes 1. floors_covered becomes 1, which is >= n. The function returns 1.
Edge case 2: n = 1000. The loop continues until moves reaches 45. floors_covered becomes 1035, which is >= n. The function returns 45.
Edge case 3: n = 0. While the problem states that 1 <= n <= 1000, consider n = 0. The code will still correctly return 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Trying to apply a dynamic programming approach directly, without recognizing the simplified pattern that exists for only two eggs.

✗Common mistake 2: Incorrectly formulating the condition for the `while` loop, leading to either an infinite loop or an incorrect result.

✗Common mistake 3: Assuming the relationship between moves and floors covered is linear instead of understanding that the number of floors covered increases with each move.

✅

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: Determining the safe operating altitude for a drone, where altitude impacts battery life and risk of damage. This problem allows minimizing the number of test flights.

🏦Real-world scenario 2: Stress testing software or hardware to find the threshold at which it fails. In quality assurance, the problem helps efficiently define the boundary between the acceptable and unacceptable performance levels.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 24, 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

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Egg Drop With 2 Eggs and N Floors"?

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

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