<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>

def manhattan(p, q):
    return sum(abs(pi - qi) for pi, qi in zip(p, q))


class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        # connect 2 closest points
        # find a point that is closest to a point in the connected set and connect them
        n = len(points)
        if n <= 1:
            return 0
        dist = [[0] * n for i in range(n)]
        for i in range(n):
            for j in range(i + 1, n):
                dist[i][j] = dist[j][i] = manhattan(points[i], points[j])
        p1, p2 = min(
            ((i, j) for i in range(n) for j in range(i + 1, n)),
            key=lambda idx: dist[idx[0]][idx[1]]
        )
        is_connected = [False] * n
        is_connected[p1] = is_connected[p2] = True
        connected_count = 2
        connection_cost = dist[p1][p2]
        connection_candidates = [
            p1 if dist[p1][i] < dist[p2][i] else p2 for i in range(n)
        ]
        
        while connected_count < n:
            next_conn = min(
                (i for i in range(n) if not is_connected[i]),
                key=lambda i: dist[connection_candidates[i]][i]
            )
            is_connected[next_conn] = True
            connection_cost += dist[connection_candidates[next_conn]][next_conn]
            connected_count += 1
            for i in range(n):
                if is_connected[i]:
                    continue
                cc = connection_candidates[i]
                if dist[next_conn][i] < dist[cc][i]:
                    connection_candidates[i] = next_conn
        
        return connection_cost

            

        

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