How do you solve Super Egg Drop on LeetCode?

Super Egg Drop (LeetCode #887) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(kn²) time complexity and O(kn) space complexity. Key insight: Understanding the trade-off between breaking and not breaking eggs is crucial.

What is the brute force solution for Super Egg Drop?

The brute force approach for Super Egg Drop is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we drop the egg from each floor starting from the first until it breaks. This guarantees that we will find the critical floor, but it is inefficient as it may require many drops, especially with more floors. The optimal Optimal Solution approach improves this to O(kn²).

What algorithm or data structure is used to solve Super Egg Drop?

Super Egg Drop uses the following concepts: Math, Binary Search, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Binary Search.

What are the interview tips for Super Egg Drop?

Explain your thought process clearly; interviewers want to see your reasoning. Don't rush into coding; take time to outline your approach first. Practice similar problems to recognize patterns and improve your problem-solving speed.

What are common mistakes when solving Super Egg Drop?

Students often forget to initialize the base cases in DP. Some may try to optimize too early without understanding the problem structure.

#887

Super Egg Drop

Hard

MathBinary SearchDynamic ProgrammingDynamic ProgrammingBinary Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(kn²)
Space	O(1)	O(kn)

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Intuition

Time O(kn²)Space O(kn)

Using dynamic programming, we can minimize the number of moves by strategically choosing which floor to drop from. This approach balances the worst-case scenarios of breaking and not breaking eggs, leading to fewer total drops.

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Algorithm

3 steps

1Step 1: Create a DP table where dp[i][j] represents the minimum number of moves needed with i eggs and j floors.
2Step 2: Initialize the base cases: dp[i][0] = 0 (0 floors) and dp[1][j] = j (1 egg).
3Step 3: Fill the DP table using the formula: dp[i][j] = min(1 + max(dp[i-1][x-1], dp[i][j-x])) for x in 1 to j.

solution.py14 lines

1def superEggDrop(k, n):
2    dp = [[0] * (n + 1) for _ in range(k + 1)]
3    for j in range(1, n + 1):
4        dp[1][j] = j
5    for i in range(2, k + 1):
6        for j in range(1, n + 1):
7            dp[i][j] = float('inf')
8            for x in range(1, j + 1):
9                res = 1 + max(dp[i-1][x-1], dp[i][j-x])
10                dp[i][j] = min(dp[i][j], res)
11    return dp[k][n]
12
13# Example usage
14print(superEggDrop(2, 6))  # Output: 3

ℹ

Complexity note: The time complexity is O(kn²) because we fill a DP table of size k x n, and for each entry, we may check all floors, leading to a quadratic factor.

1Understanding the trade-off between breaking and not breaking eggs is crucial.
2Dynamic programming helps optimize the number of moves by reusing previously computed results.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.