<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:
    const double MIN_STEP = 0.000001;      // With 0.00001 not AC
    
    const int dx[4] = {0, 0,  1,-1};
    const int dy[4] = {1, -1, 0, 0};
    
    double totalDist(vector<vector<int>>& positions, double cx, double cy) {
        double dist = 0;
        for (auto p : positions) {
            dist += hypot(p[0] - cx, p[1] - cy);
        }
        return dist;
    }
    
    double getMinDistSum(vector<vector<int>>& positions) {
        int n = (int)positions.size();
        double cx = 0, cy = 0;
        for (auto p : positions) { cx += p[0], cy += p[1]; }
        cx /= n, cy /= n;
        
        pair<double, double> minDistCenter = {cx, cy};
        double minDist = totalDist(positions, cx, cy);
        //printf("cx = %.4lf, cy = %.4lf, minDist = %.4lf\n", minDistCenter.first, minDistCenter.second, minDist);
        
        double step = 50.0;  // Because max value of x, y could be 100. So half of that
        while (step > MIN_STEP) {
            pair<double, double> tempCenter = minDistCenter;
            double tempDist = minDist;
            
            for (int k = 0; k < 4; k++) {
                double xx = minDistCenter.first + dx[k] * step;
                double yy = minDistCenter.second + dy[k] * step;
                double d = totalDist(positions, xx, yy);
                //printf("d = %.4lf\n", d);
                if (d < minDist) {
                    tempCenter = {xx, yy};
                    tempDist = d;
                }
            }
            if (minDistCenter == tempCenter) step /= 2;
            minDistCenter = tempCenter;
            minDist = tempDist;
        }
        //printf("minDist = %.4lf\n", minDist);
        return minDist;
    }
};
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