<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:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        if len(points) == 1:
            return 0

        graph = defaultdict(list)
        min_edge = (float('inf'), 0, 0)

        for i, j in itertools.combinations(range(len(points)), 2):
            a, b = points[i], points[j]
            mh_distance = abs(a[0]-b[0]) + abs(a[1]-b[1])
            graph[i].append((mh_distance, j))
            graph[j].append((mh_distance, i))

            if min_edge[0] > mh_distance:
                min_edge = (mh_distance, i, j)

        total_dist, a, b = min_edge
        pq = [*graph[a], *graph[b]]
        heapq.heapify(pq)
        visited = set()
        visited.add(a)
        visited.add(b)
        k = len(points)-2

        while pq:
            dist, curr_node = heapq.heappop(pq)
            if curr_node in visited:
                continue
            
            visited.add(curr_node)
            total_dist += dist
            k -= 1

            for d, next_node in graph[curr_node]:
                if next_node not in visited:
                    heapq.heappush(pq, (d, next_node))

        return total_dist



        # min_dist, a, b = min_edge
        # total_dist = min_dist
        # pq = [*graph[a], *graph[b]]
        # heapq.heapify(pq)
        # visited = set([a, b])
        # k = len(points)-2

        # while pq and k:
        #     dist, idx = heapq.heappop(pq)
        #     if idx in visited:
        #         continue
            
        #     total_dist += dist
        #     visited.add(idx)
        #     k -= 1

        #     for d, i in graph[idx]:
        #         if i not in visited:
        #             heapq.heappush(pq, (d, i))

        # return total_dist
138516