<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:
    struct Edge {
        int dist;
        int n1;
        int n2;
    };

    int find(int node, vector<int>& component) {
        if (component[node] == 0) return node;
        component[node] = find(component[node], component);
        return component[node];
    }

    int minCostConnectPoints(vector<vector<int>>& points) {
        auto compare = [](Edge* a, Edge* b) { return a->dist > b->dist; };
        priority_queue<Edge*, vector<Edge*>, decltype(compare)> pq;

        int n = points.size();
        vector<int> component(n, 0);
        vector<vector<int>> dist(n, vector<int>(n, 0));
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                dist[i][j] = dist[j][i] = abs(points[j][0] - points[i][0]) + abs(points[j][1] - points[i][1]);
                pq.push(new Edge{dist[i][j], i, j});
            }
        }

        int cost = 0;
        while (!pq.empty()) {
            Edge* current_edge = pq.top();
            int dist = current_edge->dist;
            int n1 = current_edge->n1;
            int n2 = current_edge->n2;
            pq.pop();
            int component1 = find(n1, component);
            int component2 = find(n2, component);
            if (component1 == component2) continue;
            else {
                cost += dist;
                component[component1] = component2;
            }
            delete current_edge;
        }
        return cost;
    }
};
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