<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 f = 1.0;

    double getMinDistSum(vector<vector<int>>& positions) {
        int n = positions.size();
        vector<int> x(n), y(n);
        
        for (int i = 0; i < n; ++i) {
            x[i] = positions[i][0];
            y[i] = positions[i][1];
        }

        double h = find(x, n);
        double k = find(y, n);
        double sum = getSum(x, y, h, k);

        while (true) {
            bool found = false;
            for (int i = -1; i <= 1; ++i) {
                for (int j = -1; j <= 1; ++j) {
                    double newh = h + i * f;
                    double newk = k + j * f;
                    double now = getSum(x, y, newh, newk);
                    if (now < sum) {
                        sum = now;
                        h = newh;
                        k = newk;
                        found = true;
                    }
                }
            }
            if (!found) {
                if (f == 1e-5)
                    break;
                f /= 10;
            }
        }

        return getSum(x, y, h, k);
    }

    double getSum(vector<int>& x, vector<int>& y, double h, double k) {
        double sum = 0;
        for (int i = 0; i < x.size(); ++i) {
            double q = pow(h - x[i], 2) + pow(k - y[i], 2);
            sum += sqrt(q);
        }
        return sum;
    }

    double find(vector<int>& a, int n) {
        int sum = 0;
        for (int i = 0; i < n; ++i)
            sum += a[i];
        double x = ( sum) / ( n);
        return x;
    }
};

10200