Introduction

In this blog post, we will explore how to solve the LeetCode problem "1884" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 minimum number of moves that you need to determine with certainty what the value of f is. Example 1: Input: n = 2 Output: 2 Explanation: We can drop the first egg from floor 1 and the second egg from floor 2. If the first egg breaks, we know that f = 0. If the second egg breaks but the first egg didn't, we know that f = 1. Otherwise, if both eggs survive, we know that f = 2. Example 2: Input: n = 100 Output: 14 Explanation: One optimal strategy is: - Drop the 1st egg at floor 9. If it breaks, we know f is between 0 and 8. Drop the 2nd egg starting from floor 1 and going up one at a time to find f within 8 more drops. Total drops is 1 + 8 = 9. - If the 1st egg does not break, drop the 1st egg again at floor 22. If it breaks, we know f is between 9 and 21. Drop the 2nd egg starting from floor 10 and going up one at a time to find f within 12 more drops. Total drops is 2 + 12 = 14. - If the 1st egg does not break again, follow a similar process dropping the 1st egg from floors 34, 45, 55, 64, 72, 79, 85, 90, 94, 97, 99, and 100. Regardless of the outcome, it takes at most 14 drops to determine f. Constraints: 1 <= n <= 1000

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: The problem is solved using dynamic programming. We create a 2D DP table where dp[i][j] stores the minimum number of moves required to find f with i eggs and j floors.

• State Transition: The core idea is to consider dropping the egg from each possible floor k. If the egg breaks, we are left with i-1 eggs and k-1 floors. If the egg doesn't break, we have i eggs and n-k floors. We take the worst-case scenario (maximum of these two) and add 1 (for the current drop). We then minimize this value over all possible k.

• Base Cases: When we have only one egg, the number of moves is simply the number of floors. When we have zero floors, the number of moves is zero.

• Complexity:

  • Runtime Complexity: O(n²)
  • Storage Complexity: O(n)

Code

    def egg_drop(n: int) -> int:
    """
    Calculates the minimum number of moves to determine the critical floor f
    using two eggs and n floors.
    """

    dp = [0] * (n + 1)

    m = 0
    while dp[n] == 0:
        m += 1
        for x in range(n, 0, -1):
            dp[x] = dp[x - 1] + m

    return m