<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 manDis(vector<int>& p1, vector<int>& p2){
        int x1 = p1[0];
        int y1 = p1[1];
        int x2 = p2[0];
        int y2 = p2[1];
        return abs(x1 - x2) + abs(y1 - y2);
    }
    int minCostConnectPoints(vector<vector<int>>& points) {
        // solve using prim's, already solved using Kruskal's
        int ans = 0;
        // start the min spanning tree search at node 0
        vector<int> dis (points.size(), INT_MAX);
        dis[0] = 0; // distance of 0 means the point is already in the minimum spanning tree, already searched
        for(int i = 1; i < points.size(); ++i){
            int d = manDis(points[0], points[i]);
            dis[i] = d;
        }
        int tree_sz = 1;
        while(tree_sz < points.size()){
            int min_d = INT_MAX;
            int idx = -1;
            for(int i = 0; i < dis.size(); ++i){
                if(dis[i] == 0) continue; // node already in the tree!
                if(dis[i] < min_d){
                    min_d = dis[i];
                    idx = i;
                }
            }
            // recompute minimum distance of the chosen point's adjacent cells!
            tree_sz++;
            ans += min_d;
            dis[idx] = 0;
            for(int i = 0; i < dis.size(); ++i){
                dis[i] = min(dis[i], manDis(points[i], points[idx]));
            }
        }
        return ans;
    }
};
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