<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>

#define LL long long

class Solution {
public:
    int disjoint_find(int x, vector<int> &p) {
        if (p[x] == x) {
            return x;
        }
        return p[x] = disjoint_find(p[x], p);
    }

    void disjoint_union(int x, int y, vector<int> &p) {
        x = disjoint_find(x, p);
        y = disjoint_find(y, p);
        if (x != y) {
            p[x] = y;
        }
    }

    int minCostConnectPoints(vector<vector<int>>& points) {
        vector<vector<int>> edges;
        for (int i = 0; i < points.size(); i++) {
            for (int j = 0;j < points.size(); j++) {
                if (i == j) {
                    continue;
                }
                edges.push_back({abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1]), i, j});
            }
        }

        sort(edges.begin(), edges.end());
        vector<int> p = vector<int> (edges.size());
        int ret = 0;

        for (int i = 0; i < p.size(); i++) {
            p[i] = i;
        }

        for (int i = 0; i < edges.size(); i++) {
            int x = disjoint_find(edges[i][1], p);
            int y = disjoint_find(edges[i][2], p);

            if (x == y) {
                continue;
            }

            ret += edges[i][0];
            disjoint_union(x, y, p);
        }

        return ret;
    }
};
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