<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)
        adj = {i: [] for i in range(N)}  # i : list of [cost, node]
        for i in range(N):
            x1, y1 = points[i]
            for j in range(i + 1, N):
                x2, y2 = points[j]
                dist = abs(x1 - x2) + abs(y1 - y2)
                adj[i].append((dist, j))
                adj[j].append((dist, i))

        # Run Prims Algorithm 
        res = 0 
        visited = set() 
        minH = [[0, 0]] # cost, point 

        while len(visited) < N:
            cost, idx = heapq.heappop(minH)
            if idx in visited:
                continue 
            visited.add(idx)
            res += cost
            for dist, nei in adj[idx]:
                heapq.heappush(minH, [dist, nei])
        return res
            


        # dist = []
        # connected = set() 
        # for i in range(len(points)):
        #     for j in range(i + 1, len(points)):
        #         manhattan = abs(points[j][0] - points[i][0]) + abs(points[j][1] - points[i][1])
        #         dist.append([manhattan, points[i], points[j]]) 
        # dist.sort()
        # print(dist)

        # res = 0
        # for manhattan, point1, point2 in dist:
        #     if tuple(point1) in connected and tuple(point2) in connected:
        #         continue 
        #     else:
        #         res += manhattan 
        #         connected.add(tuple(point1))
        #         connected.add(tuple(point2))
        # return res 
211415