<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 {
    double last_x = -100000000;
    double last_y = -100000000;
    double last_dist;

    double dist(double x1, double y1, double x2, double y2){
        double dx = x1 - x2;
        double dy = y1 - y2;
        return sqrt(dx*dx + dy*dy);
    }

    double geom_dist(double x, double y, vector<vector<int>>& positions){
        if ((last_x == x) && (last_y == y)){
            return last_dist;
        }
        double out = 0;
        for (auto& pos : positions){
            out += dist(x,y,pos[0],pos[1]);
        }
        last_dist = out;
        last_x = x;
        last_y = y;
        return out;
    }

    bool step(double& x, double& y, vector<vector<int>>& positions, double step = 1){
        bool moved = false;
        double f0 = geom_dist(x,y,positions);
        if (geom_dist(x+step,y,positions) < f0){
            x += step;
            moved = true;
        } else if (geom_dist(x-step,y,positions) < f0){
            x -= step;
            moved = true;
        }
        if (geom_dist(x,y+step,positions) < f0){
            y += step;
            moved = true;
        } else if (geom_dist(x,y-step,positions) < f0){
            y -= step;
            moved = true;
        }
        return moved;
    }
    
    void center_of_mass(double& x, double& y, vector<vector<int>>& positions){
        x = 0;
        y = 0;
        for (auto& pos : positions){
            x += pos[0];
            y += pos[1];
        }
        double mul = 1.0 / positions.size();
        x *= mul;
        y *= mul;
    }

public:
    double getMinDistSum(vector<vector<int>>& positions) {
        double x,y;
        center_of_mass(x,y,positions);

        double step_size = 10;
        for (int i = 0; i < 50; i++){
            while(step(x,y,positions,step_size))continue;
            step_size *= 0.5;
        }

        return geom_dist(x,y,positions);
    }
};
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