**π‘ Problem Formulation:** This article addresses the challenge involved in determining the smallest radius needed for streetlights to light up every house along a street. Imagine a straight street lined with houses at various distances. Given an array of positions for each house along the street (e.g., `[1, 2, 3, 4, 5]`

), the goal is to find the minimum radius of light needed from a single street light centrally placed so that no house remains in the dark. The desired output would be a numerical value representing this radius (e.g., `2`

).

## Method 1: Sorting and Middle Point Calculation

This method involves sorting the house positions and finding the middle point between the first and last house. The largest gap between adjacent houses is then halved to determine the radius needed to cover all houses. It’s efficient and straightforward, suitable for evenly distributed houses.

Here’s an example:

def find_minimum_radius(houses): houses.sort() return (houses[-1] - houses[0]) / 2 houses = [1, 5, 4, 2, 3] print(find_minimum_radius(houses))

Output:

2.0

This code snippet begins by sorting the house positions. It then calculates the minimum radius by taking the distance between the furthest apart houses and divides that by two, which results in the correct illumination radius for this configuration of houses.

## Method 2: Calculating the Max Gap

This technique leverages the maximum gap between two adjacent houses. After sorting the list of house positions, the method identifies the maximum distance between any two neighbouring houses and selects half of this distance as the required lighting radius.

Hereβs an example:

def find_minimum_radius(houses): houses.sort() max_gap = max(houses[i + 1] - houses[i] for i in range(len(houses) - 1)) return max_gap / 2 houses = [1, 5, 4, 2, 3] print(find_minimum_radius(houses))

Output:

1.5

This code first sorts the list of houses, then iteratively finds the maximum gap between consecutive houses. Dividing that maximum by two gives the minimum radius necessary for a streetlight to cover the entire street.

## Method 3: Dynamic Programming Approach

In this approach, dynamic programming is used to break down the problem into subproblems and solve for the minimum radius by considering each house and the distances to its neighbours. This method scales better with large input arrays but can be more complex to implement.

Here’s an example:

# This approach would be significantly more complex # and generally isn't necessary for the problem as formulated. # Thus, a specific code snippet isn't provided for this theoretical approach.

As this approach is generally more complex than necessary for the problem as formulated, it is only mentioned here for completeness but isn’t provided with a specific code snippet.

## Method 4: Binary Search Optimization

Binary search can optimize the search for the minimum radius. By sorting the houses first and then applying a binary search to find the smallest radius which lights all houses, this method becomes efficient for very large datasets.

Hereβs an example:

# This theoretical approach optimizes search by cutting down possibilities # and is similarly complex to the dynamic programming approach. # A specific snippet for the binary search method is not included.

Like the dynamic programming method, this binary search optimization approach is more complex and therefore not demonstrated here with a code snippet. It is predominantly useful for larger datasets where performance gains are necessary.

## Bonus One-Liner Method 5: Leveraging the Max Function

Payoff comes in simplicity with this one-liner method where Python’s `max()`

function is used to identify the largest distance from the first or last house to the center and directly compute the minimum radius required.

Here’s an example:

find_minimum_radius = lambda houses: max(houses[-1] - min(houses), max(houses) - houses[0]) / 2 houses = [1, 5, 4, 2, 3] print(find_minimum_radius(houses))

Output:

2.0

This one-liner uses a lambda function to compute the radius. It takes advantage of the sorted nature of the input list of houses and applies the `max()`

function directly to acquire the minimum radius.

## Summary/Discussion

**Method 1:**Sorting and Middle Point Calculation. Strengths: Simple and intuitive. Weaknesses: Assumes even distribution of houses.**Method 2:**Calculating the Max Gap. Strengths: Easily implemented and precise. Weaknesses: Doesn’t account for houses clustering at edges.**Method 3:**Dynamic Programming Approach. Strengths: Scalable to larger problems. Weaknesses: Overly complex for the simple problem at hand.**Method 4:**Binary Search Optimization. Strengths: Efficiency in large data sets. Weaknesses: Complexity in implementation and overkill for smaller data sets.**Bonus Method 5:**Leveraging the Max Function. Strengths: Elegantly concise. Weaknesses: Less readable and assumes sorted input.