<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:
        # choose a starting point
        # starting point must not be a part of the visited set
        # add all neighbours to minHeap
        # pick out smallest neigbour and add to minHeap 
        # mark as visited
        # add dist to ans
        adjList = {}
        for [n1, n2] in points:
            key = (n1, n2)
            adjList[key] = []
        
        def dist(n1, n2):
            return abs(n1[0] - n2[0]) + abs(n1[1] - n2[1])

        P = len(points)
        for i in range(P):
            n1 = points[i]
            for j in range(i + 1, P):
                n2 = points[j]
                d = dist(n1, n2)
                adjList[tuple(n1)].append([d, n2])
                adjList[tuple(n2)].append([d, n1])
        
        start = points[i]
        minHeap = []
        heapq.heapify(minHeap)
        heapq.heappush(minHeap, [0, start])

        count = 0
        visited = set()
        ans = 0
        while minHeap and len(visited) < P:
            [w1, n1] = heapq.heappop(minHeap)
            t1 = tuple(n1)
            if t1 in visited:
                continue
            visited.add(t1)
            count += 1
            # print("ans is", ans, "w1 is", w1)
            ans += w1
            
            for [w2, n2] in adjList[t1]:
                t2 = tuple(n2)
                # print("neighbour of", n1, n2, "has weight", w2)
                if t2 not in visited:
                    heapq.heappush(minHeap, [w2, n2])
        

        if len(visited) == P:
            return ans
        return -1




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