Best Position For A Service Centre - Leetcode Solution

Problem Statement:

You are given an array "workers" which represents the positions of "n" workers. You are also given an integer "k" which represents the number of service centers that need to be established. Each worker needs to be assigned to the nearest service center.

The objective of this problem is to find the minimum range of positions to establish a service center, such that the maximum distance between a worker and their nearest service center is minimized.

Solution:

To solve this problem, we need to determine the best position to establish a service center. We can follow the below approach to find the optimal position:

1. Sort the given array of workers.

2. Initialize two pointers "left" and "right" to 0 and max(worker) respectively.

3. Use Binary Search to find the mid-point of left and right.

4. Calculate the sum of the distances between every worker and the mid-point.

5. If this sum is less than or equal to "k", update the right pointer to the mid-point.

6. Else, update the left pointer to the mid-point.

7. Repeat steps 3 to 6 until the difference between left and right becomes less than or equal to 1e-6.

We can use the following code to implement the above approach:

Code:

def get_max_distance(workers, mid):
    max_distance = 0
    for worker in workers:
        max_distance = max(max_distance, abs(worker - mid))
    return max_distance

class Solution:
    def getBestPosition(self, workers: List[int], k: int) -> float:
        workers.sort()
        left, right = workers[0], workers[-1]
        while right - left >= 1e-6:
            mid = (left + right) / 2
            if sum([get_max_distance(workers, mid) for _ in range(k)]) <= 1e-6:
                right = mid
            else:
                left = mid
        return left

Time Complexity:

The time complexity of this solution is O(nlogn), where n is the number of workers. The sorting operation takes O(nlogn) time, and the binary search operation takes O(logn) time. The get_max_distance function is called at most k times, which has a time complexity of O(n).

Space Complexity:

The space complexity of this solution is O(1), as we are only using constant space for storing the left and right pointers.

Overall, this solution efficiently solves the Best Position for a Service Centre problem on LeetCode.

1