<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 DJS {
private: 
    vector<int> parent;
public:
    DJS(int size) { // huh - doesnt even need to know coordinates even though it only has indices, just tracks set
        parent.resize(size);
        for (int i = 0; i < size; i++) {
            parent[i] = i;
        }
    }

    int find(int x) {
        if (parent[x] != x) {
            parent[x] = find(parent[x]);
        }
        return parent[x];
    }

    void makeUnion(int a, int b) {
        int rootA = find(a);
        int rootB = find(b);
        if (rootA != rootB) {
            parent[rootA] = rootB;
        }
    }

    bool sameSet(int a, int b) {
        return find(a) == find(b);
    }
};

class Solution {
private:
    int getManhattanDistance(vector<int> edge) {
        return abs(edge[0] - edge[2]) + abs(edge[1] - edge[3]);
    }

    static bool customComparator(const vector<int>& a, const vector<int>& b) {
        return a[2] < b[2];
    }


public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        int cost = 0; 
        int n = points.size();
        DJS djs(n);
        vector<vector<int>> edges;

        for (int i = 0; i < n; i++) {
            for (int j = i+1; j < n; j++) {
                edges.push_back({i, j, getManhattanDistance({points[i][0], points[i][1], points[j][0], points[j][1]})});
            }
        }

        sort(edges.begin(), edges.end(), customComparator);

        for (auto edge : edges) {
            if (!djs.sameSet(edge[0], edge[1])) {
                cost+=edge[2];
                djs.makeUnion(edge[0], edge[1]);
            }
        }

        return cost;
    }
};
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