Introduction

In the heart of the city lies the 100-story Fabergé Egg Museum, presenting a distinctive challenge to the world’s slickest jewel thief. The mission: procure the most valuable egg without causing a mess. Armed with only two replicas, our stealthy thief concocts a plan.

The Challenge

The task at hand: identify the highest floor from which an egg survives a fall without breaking. Each egg’s value increases with each floor, making the decision all the more crucial.

The Simple Scenario

Let’s start with a scenario involving just one replica egg. The thief would drop it from the first floor and work her way up incrementally. This process could take up to 100 attempts. However, the introduction of a second replica egg allows for a more strategic approach.

Strategic Egg Drops

Armed with two replicas, the thief optimizes her strategy. Rather than dropping the first egg from every floor, she strategically chooses larger intervals, efficiently narrowing down the range of possible critical floors.

For instance, starting with drops every 10 floors, the thief minimizes the number of attempts with the second egg. If the first egg breaks at a certain interval, the second egg helps refine the search floor by floor within that interval.

The Math Unleashed

The equation to solve the problem involves finding the smallest value of $( n )$ such that the sum of the first $( n )$ positive integers is greater than or equal to $(100)$. The sum of the first $( n )$ positive integers is given by the formula:

$[ \frac{n \cdot (n + 1)}{2} ]$

So, the equation to solve is:

$[ \frac{n \cdot (n + 1)}{2} \geq 100 ]$

To solve this inequality, you can multiply both sides by $(2)$ to get rid of the fraction:

$[ n \cdot (n + 1) \geq 200 ]$

Expand and rearrange:

$[ n^2 + n \geq 200 ]$

Now, set the inequality to zero:

$[ n^2 + n - 200 \geq 0 ]$

Solving this quadratic inequality reveals that the smallest positive integer solution for $( n )$ is $(14)$. Thus, the equation $( \frac{n \cdot (n + 1)}{2} \geq 100 )$ is solved by $( n \geq 14 )$. This means the jewel thief should start dropping the replica eggs from intervals of $(14)$ floors to minimize the number of attempts required to find the critical floor.

Why $( n(n + 1) / 2 )$?

The formula $(\frac{n \cdot (n + 1)}{2})$ comes from the sum of the first $(n)$ positive integers, and it is derived through a process known as the arithmetic series formula.

The sum of the first $(n)$ positive integers, denoted by $(S_n)$, can be expressed as:

$[S_n = 1 + 2 + 3 + \ldots + n.]$

This is an arithmetic series with a common difference of 1. Now, there is a neat trick to find the sum. You can pair the first and last terms, the second and second-to-last terms, and so on:

$[S_n = 1 + n + 2 + (n - 1) + 3 + (n - 2) + \ldots.]$

Each pair sums up to $(n + 1)$, and there are $(n)$ such pairs because there are $(n)$ terms in the series. Therefore, the sum is given by:

$[S_n = n \cdot (n + 1).]$

However, this sum includes each number twice (since we’re pairing them up), so we need to divide by 2 to get the actual sum:

$[S_n = \frac{n \cdot (n + 1)}{2}.]$

This formula is a convenient way to express the sum of the first $(n)$ positive integers and is widely used in mathematics and computer science. In the context of the two-egg problem, this formula is used to determine the minimum value of $(n)$ such that $(S_n \geq N)$, where $(N)$ is the total number of floors in the building.

Time Complexity

The time complexity of the solution to the two-egg problem is $(O(\sqrt{N}))$, where $(N)$ is the number of floors in the building. The process involves finding the sum of the first $(N)$ positive integers and solving a quadratic inequality. These operations have a time complexity of $(O(1))$, making the overall time complexity $(O(\sqrt{N}))$.

Solution for Finding the Minimum Number of Steps

import math

class Solution:
    def twoEggDrop(self, total_floors: int) -> int:
        # Calculate the initial interval using the O(sqrt(N)) solution
        initial_interval = self.calculate_initial_interval(total_floors)

        # Determine the number of drops needed
        drops = self.calculate_drops(initial_interval, total_floors)

        return drops

    def calculate_initial_interval(self, n: int) -> int:
        # Calculate the initial interval using the O(sqrt(N)) solution
        return math.ceil(math.sqrt(2 * n))

    def calculate_drops(self, initial_interval: int, n: int) -> int:
        # Use the quadratic formula to calculate the number of drops
        return math.ceil((-1 + math.sqrt(1 + 8 * n)) / 2)

# Example usage
sol = Solution()
result = sol.twoEggDrop(100)
print("The breaking point is at or below floor:", result)

Real-World Question

Question: Given two crystal balls and a list indicating the floors where the ball breaks ([false, false, true, true, true]), how do you adapt this strategy to find the breaking point with minimal drops?

Answer:

import math

def find_breaking_point(breaks):
    min_step = math.floor(math.sqrt(len(breaks)))
    i = min_step

    while i < len(breaks):
        if breaks[i]:
            # If the breaking point is found, refine the search
            # Ensure that the end index does not go beyond the array bounds
            return refine_search(breaks, i - min_step, min(i, len(breaks)))
        i += min_step

    return -1

def refine_search(breaks, start, end):
    for i in range(start, end):
        if breaks[i]:
            # Return the index where the breaking point is found
            return i

    return -1

# Example usage
breaks = [False, False, True, True, True]
result = find_breaking_point(breaks)
print("The breaking point is at or below floor:", result)

References