<p>You are given an array <code>points</code> representing integer coordinates of some points on a 2D-plane, where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>.</p>

<p>The cost of connecting two points <code>[x<sub>i</sub>, y<sub>i</sub>]</code> and <code>[x<sub>j</sub>, y<sub>j</sub>]</code> is the <strong>manhattan distance</strong> between them: <code>|x<sub>i</sub> - x<sub>j</sub>| + |y<sub>i</sub> - y<sub>j</sub>|</code>, where <code>|val|</code> denotes the absolute value of <code>val</code>.</p>

<p>Return <em>the minimum cost to make all points connected.</em> All points are connected if there is <strong>exactly one</strong> simple path between any two points.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/26/d.png" style="width: 214px; height: 268px;" />
<pre>
<strong>Input:</strong> points = [[0,0],[2,2],[3,10],[5,2],[7,0]]
<strong>Output:</strong> 20
<strong>Explanation:</strong> 
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/26/c.png" style="width: 214px; height: 268px;" />
We can connect the points as shown above to get the minimum cost of 20.
Notice that there is a unique path between every pair of points.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> points = [[3,12],[-2,5],[-4,1]]
<strong>Output:</strong> 18
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= points.length &lt;= 1000</code></li>
	<li><code>-10<sup>6</sup> &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>6</sup></code></li>
	<li>All pairs <code>(x<sub>i</sub>, y<sub>i</sub>)</code> are distinct.</li>
</ul>

class Solution {
public:
    int distance(vector<int>p1, vector<int>p2)
    {
        return abs(p1[0] - p2[0]) + abs(p1[1] - p2[1]);
    }

    int minkey(vector<vector<int>>& points, vector<int> &val, vector<bool> &v)
    {
        int n = points.size();
        int mini = INT_MAX;

        int ind = -1;
        for(int i=0; i<n; i++)
        {
            if(!v[i] && val[i] < mini)
            {
                ind = i;
                mini = val[i];
            }
        }

        return ind;
    }



    void func1(vector<vector<int>>& points, int &sum)
    {
        int n = points.size();

        vector<int> val(n, INT_MAX);
        vector<int> par(n, -1);

        vector<bool> v(n, false);


        val[0] = 0;
        par[0] = 0;

        for(int i=0; i<n; i++)
        {
            int u = minkey(points, val, v);
            v[u] = true;
            int mini = INT_MAX;
            for(int j = 0; j<n;j++)
            {
                if(!v[j] && distance(points[u], points[j]) < val[j])
                {
                    val[j] = distance(points[u], points[j]);
                    mini = val[j];
                    par[j] = u;
                }
            }

            //if(mini != INT_MAX)
                //sum+= mini;


        }

        for(int i=0; i<n;i++)
        {
            sum += distance(points[par[i]], points[i]);
        }

        // int mini = INT_MAX;
        // int ind = -1;

        // for(int i=0; i<n; i++)
        // {
        //     if(v[i])
        //     {
        //         for(int j=0; j<n; j++)
        //         {
        //             if(j!=i && !v[j] && (distance(points[i], points[j]) < mini))
        //             {
        //                 ind = j;
        //                 mini = distance(points[i], points[j]);
        //                 co++;
        //             }
        //         }
        //     }
        // }

        // if(ind != -1)
        //     v[ind] = true;

        // if(mini != INT_MAX)
        //     sum += mini;

    }

    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size();
       // vector<bool> v(n, false);

        int sum = 0;
       // int co = 0;
       func1(points, sum);

        //v[0] = true; 

        

        return sum;

        //return 0;

    }
};
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