<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:
        '''
        n = len(points)

        # distance
        dist = [math.inf] * n
        dist[0] = 0

        # priority queue
        pq, seen = [], set()
        pq.append([0, 0]) # dist, node

        while pq:
            cur_dist, cur_node = heappop(pq)
            seen.add(cur_node)

            for next_node in range(n):
                if next_node != cur_node and next_node not in seen:
                    next_dist = abs(points[next_node][0]-points[cur_node][0]) + abs(points[next_node][1]-points[cur_node][1])
                    if next_dist < dist[next_node]:
                        heappush(pq, [next_dist, next_node])
                        dist[next_node] = next_dist
        
        return sum(dist)
        '''


        # build adj: O(N^2)
        adj = collections.defaultdict(list)
        for i in range(len(points)-1):
            for j in range(i+1, len(points)):
                w = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1])
                adj[i].append((j, w))
                adj[j].append((i, w))

        # bfs
        q = [(0, 0)] # (dist, node)
        dist_arr = [float("inf")] * len(points)
        dist_arr[0] = 0
        explored = set()

        while q:
            val, node = heapq.heappop(q)
            # explore neighbor node
            for next_node, next_val in adj[node]:
                if next_node not in explored and next_val < dist_arr[next_node]:
                    # next_val = #abs(points[next_node][0] - points[node][0]) + abs(points[next_node][1] - points[node][1])
                    #if next_val < dist_arr[next_node]:
                    # update the dist_arr and priority
                    dist_arr[next_node] = next_val
                    heapq.heappush(q, (next_val, next_node))
            explored.add(node)
            
        return sum(dist_arr)
120920