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Super Egg Drop

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

Solution Explanation for Super Egg Drop

The Super Egg Drop problem asks for the minimum number of moves needed to determine the critical floor (f) in a building with n floors using k eggs. The critical floor is the highest floor from which an egg won't break when dropped. Dropping an egg from a floor above f will break it, while dropping it at or below f will not.

Both solutions presented use dynamic programming, leveraging binary search for optimization in one of them. Let's analyze each approach:

This solution utilizes a recursive function dfs(i, j) where:

  • i represents the remaining number of floors to check.
  • j represents the number of eggs remaining.

The base cases are:

  • i < 1: No floors left, so 0 moves.
  • j == 1: Only one egg left; we must check each floor sequentially, hence i moves.

The core logic involves a binary search to find the optimal floor to drop the egg. For each possible floor mid, we consider two scenarios:

  1. The egg breaks: We recursively check the floors below mid - 1 using j - 1 eggs (dfs(mid - 1, j - 1)).
  2. The egg doesn't break: We recursively check the floors above mid using j eggs (dfs(i - mid, j)).

We choose mid that minimizes the maximum of these two scenarios (max(dfs(mid - 1, j - 1), dfs(i - mid, j))). The total moves for this choice of mid is incremented by 1 (the current drop). Memoization (@cache in Python, f array in other languages) is used to store results of subproblems to avoid redundant computations.

Time Complexity Analysis: The binary search in the inner loop takes O(log i) time. Since the recursive calls explore a state space of size approximately O(n*k), the overall time complexity is roughly O(n * k * log n). The memoization ensures that each state is visited only once.

Space Complexity Analysis: The space complexity is O(n*k) due to the memoization table. The recursive call stack can also contribute, but it is bounded by the depth of the recursion, which is at most O(n).

This solution is similar in concept but uses an iterative approach to fill a DP table f[i][j]. The table stores the minimum moves needed to solve the problem with i floors and j eggs. It iteratively builds up the solution from smaller subproblems.

The base cases are the same. The inner loop uses a binary search to find the optimal mid as in Solution 1, but instead of recursion, it directly accesses the values from the f table to calculate f[i][j].

Time Complexity Analysis: The outer loops iterate through O(n*k) states. The binary search within each iteration takes O(log n) time. Thus, the total time complexity is O(n * k * log n).

Space Complexity Analysis: The space complexity is O(n*k) for the DP table.

Comparison: Both solutions are essentially the same in terms of algorithmic approach and time complexity. The iterative version (Solution 2) might have slightly better performance in practice due to the avoidance of recursive function call overhead. However, the recursive version (Solution 1) is often considered more readable and easier to understand. The choice between them depends on priorities for readability versus potential small performance gains.