We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsCompaniesDisclaimersPrivacyTermsContactAbout Us

Egg Drop With 2 Eggs and N Floors

Number: 2031

Difficulty: Medium

Paid? No

Companies: Amazon, DE Shaw

Problem Description

Given two eggs and a building with n floors, determine the minimum number of moves required to find with certainty the highest floor f from which an egg can be dropped without breaking. An egg will break if dropped from any floor greater than f, and will not break if dropped from any floor less than or equal to f.

Key Insights

  • With only 2 eggs, we must optimize our strategy since a broken egg limits further testing to sequential searches.
  • By dropping the first egg from progressively higher floors with decreasing increments, we minimize the worst-case number of moves.
  • The optimal strategy is based on finding the smallest number k such that 1 + 2 + ... + k >= n (i.e., k*(k+1)/2 >= n).
  • This approach ensures that if the first egg breaks, the remaining floors are small enough to check sequentially with the second egg.

Space and Time Complexity

Time Complexity: O(√n), since k is approximately O(√n) in the worst-case scenario. Space Complexity: O(1), as only a constant amount of extra space is used.

Solution

We use an incremental strategy where the first egg is dropped from floors determined by a decreasing step value. Specifically, the first drop is from floor k, then floor k+(k-1), then k+(k-1)+(k-2), and so on until the sum reaches or exceeds n. If the first egg breaks at a given drop, we use the second egg to check sequentially the floors between the previous drop and the current drop. This strategy guarantees that in the worst-case scenario, the total number of moves required is k, where k is the smallest integer such that k(k+1)/2 is at least n.

The main trick is recognizing the triangular number series and solving for the minimal k which satisfies the condition. This approach eliminates the complex dynamic programming recurrences and yields an efficient and optimal solution.

Code Solutions

# Function to determine the minimum number of moves required using 2 eggs.
def twoEggDrop(n):
    moves = 0         # Number of moves (or drops) made.
    accumulated = 0   # Accumulated number of floors that can be checked.
    
    # Increment moves until the accumulated floors cover n.
    while accumulated < n:
        moves += 1
        accumulated += moves   # Add current move count to the accumulated sum.
    
    return moves

# Example usage:
if __name__ == "__main__":
    test_floor = 100
    result = twoEggDrop(test_floor)
    print("Minimum moves for", test_floor, "floors is:", result)  # Expected output: 14
← Back to All Questions