<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:
    vector<vector<int>>pos;
    double find(double xpos, double ypos){
        double ans=0;
        for(vector<int> p: pos){
            double dx = ((double)xpos - (double)(p[0]));
            double dx_squared = dx * dx;
            double dy = ((double)ypos - (double)(p[1]));
            double dy_squared = dy * dy;
            ans += (double)sqrt(dx_squared + dy_squared);
        }
        return ans;
    }
    double getMinDistSum(vector<vector<int>>& position) {
        pos = position;
        int n = pos.size();
        double xsum=0, ysum=0;
        for(int i=0;i<n;i++){
            xsum += pos[i][0];
            ysum += pos[i][1];
        }

        double xpos = xsum/(double)n, ypos = ysum/(double)n;
        double delta = 100, ans = find(xpos, ypos);
        int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
        bool zoom;
        while(delta >= 5e-7){
            zoom = 1;
            for(int i=0;i<4;i++){
                double cx = xpos + delta * dx[i];
                double cy = ypos + delta * dy[i];
                double dist = find(cx, cy);
                if(dist < ans){
                    ans = dist;
                    xpos = cx;
                    ypos = cy;
                    zoom = 0;
                    break;
                }
            }
            if(zoom){
                delta = (double) delta / (double)2.0;
            }
        }
        return ans;
    }
};
23800