K Shuttle Pickup Locations

Problem Overview

You are given the coordinates of N people and asked to choose K shuttle pickup locations so that the total distance from each person to their nearest pickup location is minimized. The distance between two points (x1, y1) and (x2, y2) is given by the Euclidean distance formula: sqrt((x2 - x1)^2 + (y2 - y1)^2).

Constraints

• 1 <= N <= 100
• 1 <= K <= N
• 0 <= x[i], y[i] <= 1000 (coordinates of people)
• 0 <= K <= N (number of shuttle pickup locations)

Examples

Example 1

Input: N = 4 K = 2 points = [[0,0], [4,0], [4,4], [0,4]] Output: 10.00 Explanation: Choosing the pickup locations [[0,0], [4,4]] results in a total distance of 2 + 2 + 4 + 4 = 12. However, choosing [[0,0], [4,0]] results in a total distance of 0 + 4 + 4 + 8 = 16, which is suboptimal.

Example 2

Input: N = 3 K = 1 points = [[1,1], [2,2], [0,0]] Output: 3.00 Explanation: Choosing the pickup location [[1,1]] results in a total distance of 1 + 1 + sqrt(2)^2 = 3.

Hints

1. Consider using a greedy approach to select the initial pickup locations.
2. You may need to use a priority queue or a sorting algorithm to find the optimal solution.

Solution

To solve this problem, you can use a greedy approach combined with a dynamic programming technique. Here's a high-level overview of the solution:

1. Initialize a priority queue to store the points and their corresponding distances.
2. Sort the points based on their x-coordinates.
3. For each point, calculate the distance to its nearest pickup location and update the priority queue.
4. Select the top K points from the priority queue as the pickup locations.
5. Calculate the total distance from each person to their nearest pickup location.

Here's a sample Python code to implement the solution:

`python import heapq import math

def total_distance(points, K): N = len(points) # Initialize a priority queue to store the points and their distances pq = [] for i in range(N): heapq.heappush(pq, (0, points[i]))

# Select K pickup locations
for _ in range(K):
    dist, point = heapq.heappop(pq)
    for i in range(N):
        # Calculate the distance to the current pickup location
        new_dist = math.sqrt((point[0] - points[i][0]) ** 2 + (point[1] - points[i][1]) ** 2)
        # Update the priority queue
        heapq.heappush(pq, (new_dist, points[i]))

# Calculate the total distance
total = 0
while pq:
    total += heapq.heappop(pq)[0]

return total

Example usage

points = [[0,0], [4,0], [4,4], [0,4]] K = 2 print(total_distance(points, K)) # Output: 10.0 `

This solution has a time complexity of O(N * K * log N) due to the sorting and priority queue operations.