#887

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Super Egg Drop

Solve the Super Egg Drop problem using dynamic programming and binary search to minimize the number of moves required to find the critical floor.

math binary search dynamic programming

Problem Statement

You are given k identical eggs and a building with n floors, and you must determine the highest floor from which an egg can be dropped without breaking. If you drop an egg from a floor above this critical floor, it will break; if dropped from a floor below, it will not. The objective is to find the critical floor with the fewest moves possible.

In each move, you can drop an unbroken egg from any floor between 1 and n. If it breaks, you can no longer use that egg, but if it survives, you can reuse it in future drops. The challenge is to optimize the number of moves required to pinpoint the critical floor, considering the constraints of k eggs and n floors.

Examples

Example 1

Input: k = 1, n = 2

Output: 2

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 details omitted.

Example 3

Input: k = 3, n = 14

Output: 4

Example details omitted.

Constraints

1 <= k <= 100
1 <= n <= 104

Solution Approach

State Transition Dynamic Programming

This problem can be efficiently solved using dynamic programming (DP), where the state is defined by the number of eggs and floors. The key idea is to minimize the number of moves by making strategic drops and updating the DP table based on whether the egg breaks or survives.

Binary Search Optimization

By combining binary search with dynamic programming, we reduce the number of moves. The optimal drop point is not necessarily linear; binary search helps to determine the best middle point for each move, balancing between fewer floors to test and fewer eggs to waste.

Bottom-Up DP Approach

A bottom-up approach builds the DP table by starting with fewer eggs and floors, progressively solving for higher numbers. This way, solutions to smaller subproblems inform the final solution, making the process efficient and scalable.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(k * n log n), where k is the number of eggs and n is the number of floors. Space complexity depends on the approach used, with a typical O(k * n) for storing the DP table.

What Interviewers Usually Probe

The candidate demonstrates understanding of dynamic programming and how state transitions work for this problem.
The candidate suggests using binary search to optimize the dynamic programming solution, balancing egg usage and floor testing.
The candidate is able to explain how the solution scales with increasing egg count and floors, focusing on both time and space efficiency.

Common Pitfalls or Variants

Common pitfalls

Mistaking the problem for a linear search, leading to inefficient solutions that do not scale well.
Not recognizing that a drop could result in both an egg breaking or surviving, which requires careful state management in the dynamic programming approach.
Overcomplicating the problem by trying to test too many floors without leveraging binary search for optimal moves.

Follow-up variants

Increasing the number of eggs while keeping the number of floors fixed, or vice versa, to test scalability.
Modifying the problem to account for additional constraints, such as a limit on the total number of drops allowed.
Adjusting the solution to handle variations in how eggs are reused or the consequences of breaking them.

FAQ

What is the key to solving the Super Egg Drop problem efficiently?

The key is combining dynamic programming with binary search to minimize the number of drops needed to find the critical floor.

How do you optimize the solution for Super Egg Drop?

By using binary search to strategically choose which floors to drop the egg from, combined with a dynamic programming approach to track the minimum drops for each subproblem.

Can the Super Egg Drop problem be solved without dynamic programming?

While it's possible to brute force the solution, dynamic programming significantly reduces the number of moves and makes the solution scalable for larger inputs.

How does GhostInterview help with the Super Egg Drop problem?

GhostInterview assists by providing a clear dynamic programming-based solution, highlighting state transitions and optimization strategies like binary search.

What is the space complexity of solving the Super Egg Drop problem?

The space complexity is typically O(k * n), where k is the number of eggs and n is the number of floors, though optimizations can reduce it further.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def superEggDrop(self, k: int, n: int) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i < 1:
                return 0
            if j == 1:
                return i
            l, r = 1, i
            while l < r:
                mid = (l + r + 1) >> 1
                a = dfs(mid - 1, j - 1)
                b = dfs(i - mid, j)
                if a <= b:
                    l = mid
                else:
                    r = mid - 1
            return max(dfs(l - 1, j - 1), dfs(i - l, j)) + 1

        return dfs(n, k)

Solution 2

#### Python3

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class Solution:
    def superEggDrop(self, k: int, n: int) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i < 1:
                return 0
            if j == 1:
                return i
            l, r = 1, i
            while l < r:
                mid = (l + r + 1) >> 1
                a = dfs(mid - 1, j - 1)
                b = dfs(i - mid, j)
                if a <= b:
                    l = mid
                else:
                    r = mid - 1
            return max(dfs(l - 1, j - 1), dfs(i - l, j)) + 1

        return dfs(n, k)

Continue Practicing

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