<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 getXGrad(double x, double y, vector<vector<int>>& positions) {
        double res = 0;
        for (auto position : positions) {
            double d = pow(position[0] - x, 2) + pow(position[1] - y, 2);
            
            res += (x - position[0]) * pow(d, -0.5);
        }
        return res;
    }

    double getYGrad(double x, double y, vector<vector<int>>& positions) {
        double res = 0;
        for (auto position : positions) {
            double d = pow(position[0] - x, 2) + pow(position[1] - y, 2);
            res += (y - position[1]) * pow(d, -0.5);
        }
        return res;
    }

    double findOptimalY(double x, vector<vector<int>>& positions) {
        double lo = 0;
        double hi = 100;
        while (lo < hi - 1e-7) {
            double mid = lo + (hi - lo) / 2.0;
            double grad = getYGrad(x, mid, positions);
            if (grad > 0) {
                hi = mid;
            } else {
                lo = mid;
            }
        }
        return lo + (hi - lo) / 2.0;
    } 

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

    double getMinDistSum(vector<vector<int>>& positions) {
        double lo = 0;
        double hi = 100;
        while (lo < hi - 1e-7) {
            double mid = lo + (hi - lo) / 2.0;
            double y = findOptimalY(mid, positions);
            double grad = getXGrad(mid, y, positions);
            if (grad > 0) {
                hi = mid;
            } else {
                lo = mid;
            }
        }

        double x = lo + (lo - hi) / 2.0;
        double y = findOptimalY(x, positions);

        return getDist(x, y, positions);
    }
};
76500