Leetcode #887: Super Egg Drop
In this guide, we solve Leetcode #887 Super Egg Drop in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Math, Binary Search, Dynamic Programming
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
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.
Python Solution
class Solution:
def superEggDrop(self, k: int, n: int) -> int:
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)
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.