<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://assets.leetcode.com/uploads/2020/06/25/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://assets.leetcode.com/uploads/2020/06/25/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://assets.leetcode.com/uploads/2020/06/25/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>

class Solution {
public:

    double distance(vector<vector<int>>& positions, double x, double y) {
        double d = 0;
        for (auto p : positions) {
            double td = (x - p[0])*(x - p[0]) + (y - p[1])*(y - p[1]);
            td = sqrt(td);
            d += td;
        }
        return d;
    }

    double getMinDistSum(vector<vector<int>>& positions) {
        double x = 0, y = 0;
        for (auto p : positions) {
            x += p[0];
            y += p[1];
        }
        x /= positions.size(); 
        y /= positions.size();
        double ans = distance(positions, x, y);
        double delta = 100;
        while (delta >= 1e-6) {
            vector<pair<int, int>> directions{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
            bool zoom = true;
            for (auto d : directions) {
                double dx = x + d.first*delta;
                double dy = y + d.second*delta;
                double td = distance(positions, dx, dy);
                if (td < ans) {
                    ans = td;
                    x = dx;
                    y = dy;
                    zoom = false;
                }
            }
            if (zoom) delta /= 10;
        }
        return ans;
    }
};
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