How do you solve Egg Drop With 2 Eggs and N Floors on LeetCode?

Egg Drop With 2 Eggs and N Floors (LeetCode #1884) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Using dynamic programming allows us to break down the problem into smaller subproblems and build up the solution efficiently.

What is the brute force solution for Egg Drop With 2 Eggs and N Floors?

The brute force approach for Egg Drop With 2 Eggs and N Floors is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible floor to drop the egg from, which leads to a lot of unnecessary drops. We simply drop the first egg from each floor sequentially until it breaks, and then use the second egg to test each floor below the breaking point. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Egg Drop With 2 Eggs and N Floors?

Egg Drop With 2 Eggs and N Floors uses the following concepts: Math, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Egg Drop With 2 Eggs and N Floors?

Always clarify the problem and constraints before jumping into coding. Think about edge cases, such as when n = 0 or n = 1. Explain your thought process clearly as you work through the problem.

What are common mistakes when solving Egg Drop With 2 Eggs and N Floors?

Assuming that dropping from the middle floor is always optimal; it often isn't. Not considering the worst-case scenario when calculating the number of drops.

#1884

Egg Drop With 2 Eggs and N Floors

Medium

MathDynamic ProgrammingDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

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Intuition

Time O(n²)Space O(n)

The optimal solution uses a dynamic programming approach to minimize the number of drops. We can leverage the fact that we can drop the first egg from a calculated floor and use the results to narrow down the possible floors more efficiently.

⚙️

Algorithm

3 steps

1Step 1: Create a DP table where dp[i][j] represents the minimum number of drops needed with i eggs and j floors.
2Step 2: Initialize the base cases for 1 egg and j floors, and for i eggs and 0 floors.
3Step 3: Fill the DP table using the recurrence relation: dp[i][j] = min(1 + max(dp[i-1][x-1], dp[i][j-x])) for all x from 1 to j.

solution.py11 lines

1def eggDrop(n):
2    dp = [[0] * (n + 1) for _ in range(3)]
3    for j in range(1, n + 1):
4        dp[1][j] = j
5    for i in range(2, 3):
6        for j in range(1, n + 1):
7            dp[i][j] = float('inf')
8            for x in range(1, j + 1):
9                res = 1 + max(dp[i-1][x-1], dp[i][j-x])
10                dp[i][j] = min(dp[i][j], res)
11    return dp[2][n]

ℹ

Complexity note: The time complexity is O(n²) due to the nested loops for filling the DP table, while the space complexity is O(n) for storing the results in the DP table.

1Using dynamic programming allows us to break down the problem into smaller subproblems and build up the solution efficiently.
2The choice of where to drop the first egg can significantly affect the total number of drops needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.