Best Position for a Service Centre
<p>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 <strong>the sum of the euclidean distances to all customers is minimum</strong>.</p>

<p>Given an array <code>positions</code> where <code>positions[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> is the position of the <code>ith</code> customer on the map, return <em>the minimum sum of the euclidean distances</em> to all customers.</p>

<p>In other words, you need to choose the position of the service center <code>[x<sub>centre</sub>, y<sub>centre</sub>]</code> such that the following formula is minimized:</p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1515.Best%20Position%20for%20a%20Service%20Centre/images/q4_edited.jpg" />
<p>Answers within <code>10<sup>-5</sup></code> of the actual value will be accepted.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1515.Best%20Position%20for%20a%20Service%20Centre/images/q4_e1.jpg" style="width: 377px; height: 362px;" />
<pre>
<strong>Input:</strong> positions = [[0,1],[1,0],[1,2],[2,1]]
<strong>Output:</strong> 4.00000
<strong>Explanation:</strong> As shown, you can see that choosing [x<sub>centre</sub>, y<sub>centre</sub>] = [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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1515.Best%20Position%20for%20a%20Service%20Centre/images/q4_e3.jpg" style="width: 419px; height: 419px;" />
<pre>
<strong>Input:</strong> positions = [[1,1],[3,3]]
<strong>Output:</strong> 2.82843
<strong>Explanation:</strong> The minimum possible sum of distances = sqrt(2) + sqrt(2) = 2.82843
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= positions.length &lt;= 50</code></li>
	<li><code>positions[i].length == 2</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 100</code></li>
</ul>
Geometry; Array; Math; Randomized
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
