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Super Egg Drop

Difficulty: Hard

Problem Description

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.

Key Insights

  • The problem can be approached using dynamic programming to minimize the number of moves required.
  • The relation between the number of eggs, the number of floors, and the moves can be computed using a table.
  • The goal is to find the minimum number of moves required in the worst-case scenario.

Space and Time Complexity

Time Complexity: O(k * n^2)
Space Complexity: O(k * n)

Solution

The solution uses a dynamic programming approach where we maintain a table dp such that dp[i][j] represents the minimum number of moves required to find the critical floor with i eggs and j floors.

  1. If we drop an egg from a certain floor:
    • If it breaks, we need to check the floors below (f-1).
    • If it does not break, we need to check the floors above (n-f).
  2. Thus, we can compute the worst-case scenario using the relation: dp[i][j] = 1 + min(max(dp[i-1][x-1], dp[i][j-x])) for all x in 1 to j
  3. We iterate through the number of eggs and floors to fill in the dp table and return dp[k][n] as the final answer.

Code Solutions

def superEggDrop(k: int, n: int) -> int:
    # Initialize the DP table
    dp = [[0] * (n + 1) for _ in range(k + 1)]
    
    # Fill the table
    for i in range(1, k + 1):
        for j in range(1, n + 1):
            # We need to find the minimum number of moves
            dp[i][j] = float('inf')
            for x in range(1, j + 1):
                # Calculate the worst-case scenario
                worst_case = 1 + max(dp[i - 1][x - 1], dp[i][j - x])
                dp[i][j] = min(dp[i][j], worst_case)
    
    return dp[k][n]
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