Solution Explanation for Egg Drop With 2 Eggs and N Floors

This problem can be solved efficiently using dynamic programming. The core idea is to break down the problem into smaller subproblems and build up the solution iteratively.

Understanding the Problem

We have two identical eggs and a building with n floors. We need to find the minimum number of drops required to determine the floor f where an egg will break when dropped from a floor above f but will not break when dropped from a floor at or below f.

Dynamic Programming Approach

Let's define dp[i][j] as the minimum number of drops needed to find f given:

• i floors
• j eggs

Our goal is to find dp[n][2].

Base Cases:

• dp[0][j] = 0 (No floors, no drops needed)
• dp[i][1] = i (One egg, we must test each floor sequentially)
• dp[i][j] = infinity if i < 0 or j < 0

Recursive Relation

If we drop an egg from floor k (where 1 <= k <= i), two scenarios are possible:

1. The egg breaks: We're left with j-1 eggs and need to test floors 1 to k-1. The minimum drops for this are dp[k-1][j-1].
2. The egg doesn't break: We're left with j eggs and need to test floors k+1 to i. The minimum drops for this are dp[i-k][j].

To find the minimum number of drops for the current state, we need to consider the worst-case scenario (the maximum number of drops) from these two possibilities, and add 1 (for the drop we just made):

dp[i][j] = min(1 + max(dp[k-1][j-1], dp[i-k][j])) for k = 1 to i

Optimization:

Instead of a 2D dp array, we can optimize to a 1D array since we only use two eggs. dp[i] represents the minimum drops for i floors with 2 eggs. The recursive relation simplifies to:

dp[i] = min(1 + max(dp[k-1], dp[i-k])) for k = 1 to i

Code Implementation (Python)

def twoEggDrop(n):
    dp = [float('inf')] * (n + 1)  # Initialize dp array
    dp[0] = 0  # Base case: 0 floors, 0 drops
    dp[1] = 1  # Base case: 1 floor, 1 drop
 
    for i in range(2, n + 1):
        for k in range(1, i + 1):
            dp[i] = min(dp[i], 1 + max(dp[k - 1], dp[i - k]))
    return dp[n]
 
# Example usage
n = 100
print(twoEggDrop(n))  # Output: 14

Time and Space Complexity:

• Time Complexity: O(n^2). The nested loops iterate through all possible floor choices for each number of floors.
• Space Complexity: O(n). We use a 1D dp array of size n+1.

Mathematical Approach (More Efficient):

While dynamic programming works, a more efficient mathematical solution exists. For 2 eggs, the minimum drops needed is approximately the square root of n. The optimal strategy involves increasing the drop height incrementally. Finding a closed-form solution is beyond the scope of a brief explanation, but this observation indicates an upper bound for the minimum drops. A solution using this approach would have a lower time complexity.