Leetcode #1515: Best Position for a Service Centre
In this guide, we solve Leetcode #1515 Best Position for a Service Centre in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
A delivery company wants to build a new service center in a new city. The company knows the positions of all the customers in this city on a 2D-Map and wants to build the new center in a position such that the sum of the euclidean distances to all customers is minimum.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Geometry, Array, Math, Randomized
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
Input: positions = [[0,1],[1,0],[1,2],[2,1]]
Output: 4.00000
Explanation: As shown, you can see that choosing [xcentre, ycentre] = [1, 1] will make the distance to each customer = 1, the sum of all distances is 4 which is the minimum possible we can achieve.
Python Solution
class Solution:
def getMinDistSum(self, positions: List[List[int]]) -> float:
n = len(positions)
x = y = 0
for x1, y1 in positions:
x += x1
y += y1
x, y = x / n, y / n
decay = 0.999
eps = 1e-6
alpha = 0.5
while 1:
grad_x = grad_y = 0
dist = 0
for x1, y1 in positions:
a = x - x1
b = y - y1
c = sqrt(a * a + b * b)
grad_x += a / (c + 1e-8)
grad_y += b / (c + 1e-8)
dist += c
dx = grad_x * alpha
dy = grad_y * alpha
x -= dx
y -= dy
alpha *= decay
if abs(dx) <= eps and abs(dy) <= eps:
return dist
Complexity
The time complexity is O(n) or O(1). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.