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 {
    public double getMinDistSum(int[][] positions) {
        int n = positions.length;
        double x = 0, y = 0;
        for (int[] p : positions) {
            x += p[0];
            y += p[1];
        }
        x /= n;
        y /= n;
        double decay = 0.999;
        double eps = 1e-6;
        double alpha = 0.5;
        while (true) {
            double gradX = 0, gradY = 0;
            double dist = 0;
            for (int[] p : positions) {
                double a = x - p[0], b = y - p[1];
                double c = Math.sqrt(a * a + b * b);
                gradX += a / (c + 1e-8);
                gradY += b / (c + 1e-8);
                dist += c;
            }
            double dx = gradX * alpha, dy = gradY * alpha;
            if (Math.abs(dx) <= eps && Math.abs(dy) <= eps) {
                return dist;
            }
            x -= dx;
            y -= dy;
            alpha *= decay;
        }
    }
}