Egg Drop With 2 Eggs and N Floors - Problem

You are given two 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.

In 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.

Input & Output

Example 1 — Small Building

$ Input: n = 2

› Output: 2

💡 Note: Drop first egg from floor 1. If it breaks, critical floor is 0 (2 moves). If it doesn't break, drop from floor 2. If it breaks, critical floor is 1, otherwise critical floor is 2 (2 moves total).

Example 2 — Medium Building

$ Input: n = 10

› Output: 4

💡 Note: Using optimal strategy: drop from floors 4, 7, 9, 10. This guarantees finding the critical floor in at most 4 moves regardless of where it is.

Example 3 — Single Floor

$ Input: n = 1

› Output: 1

💡 Note: Only one floor to test. Drop the egg from floor 1. If it breaks, critical floor is 0, otherwise it's 1. Only 1 move needed.

Constraints

1 ≤ n ≤ 1000

Visualization

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Asked in

G Google 35 M Microsoft 28 A Amazon 22 f Facebook 18

The key insight is to balance worst-case scenarios for both egg outcomes. The optimal strategy uses triangular numbers to minimize maximum drops. Best approach is the mathematical formula m = ceil((-1 + sqrt(1 + 8n)) / 2). Time: O(√n), Space: O(1)

Common Approaches

✓ Stack

⏱️ Time: N/A Space: N/A

Dynamic Programming Approach

⏱️ Time: O(n²) Space: O(n)

Use dynamic programming where dp[i] represents the minimum number of moves needed for i floors with 2 eggs. For each possible first drop at floor k, consider both outcomes (break/no break) and take the maximum, then minimize over all k.

Linear Search Strategy

⏱️ Time: O(n) Space: O(1)

Start from floor 1 and keep dropping the first egg until it breaks at some floor k. Then use the second egg to test floors k-1, k-2, ... down to 1 to find the exact critical floor.

Optimal Mathematical Strategy

⏱️ Time: O(√n) Space: O(1)

The key insight is to make the total number of drops the same regardless of where the first egg breaks. If we drop from floor x first, we need at most x-1 more drops. To minimize worst case, we solve for the smallest m where m(m+1)/2 >= n.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

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Smart Actions

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