Introduction

In this blog post, we will explore how to solve the LeetCode problem "887" 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 k 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. 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: k = 1, n = 2 Output: 2 Explanation: Drop the egg from floor 1. If it breaks, we know that f = 0. Otherwise, drop the egg from floor 2. If it breaks, we know that f = 1. If it does not break, then we know f = 2. Hence, we need at minimum 2 moves to determine with certainty what the value of f is. Example 2: Input: k = 2, n = 6 Output: 3 Example 3: Input: k = 3, n = 14 Output: 4 Constraints: 1 <= k <= 100 1 <= n <= 104

Explanation

Here's the solution to the egg drop problem:

• Dynamic Programming: We use dynamic programming to store the minimum number of moves required for a given number of eggs and floors. The state dp[i][j] represents the minimum number of moves needed to find the critical floor with i eggs and j floors.

• State Transition: When dropping an egg from floor x, there are two possibilities: the egg breaks or it doesn't. If it breaks, we have i-1 eggs and x-1 floors left to check. If it doesn't break, we have i eggs and n-x floors left. We take the worst-case scenario (maximum of these two possibilities) and add 1 (for the current move) to minimize the total moves.

• Base Cases: If we have only one egg (i=1), we need to check each floor one by one, so the number of moves is j. If we have no floors (j=0), we need zero moves. If we have one floor (j=1), we need one move.

• Runtime Complexity: O(k*n*n), where k is the number of eggs, and n is the number of floors. Space Complexity: O(k*n)

Code

    def egg_drop(k: int, n: int) -> int:
    """
    You are given k 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.
    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.
    """

    dp = [[0] * (n + 1) for _ in range(k + 1)]

    # Base cases
    for j in range(1, n + 1):
        dp[1][j] = j  # If only 1 egg, we need to check each floor
    for i in range(1, k + 1):
        dp[i][0] = 0  # If no floors, no moves needed
        dp[i][1] = 1  # If only 1 floor, only 1 move needed

    # Fill the DP table
    for i in range(2, k + 1):
        for j in range(2, n + 1):
            dp[i][j] = float('inf')
            for x in range(1, j + 1):
                # Egg breaks or doesn't break, take the worst-case scenario
                res = 1 + max(dp[i - 1][x - 1], dp[i][j - x])
                dp[i][j] = min(dp[i][j], res)

    return dp[k][n]