<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>

# 
# Kruskals with UnionFind datastructure

#class UnionFind:
#     def __init__(self, n):
#         self.n = n
#         self.roots = [i for i in range(n)]
#         self.rank = [0 for _ in range(n)]

#     def find(self, x):
#         if self.roots[x] != x:
#             self.roots[x] = self.find(self.roots[x])
#         return self.roots[x]
    
#     def union(self, x, y):
#         rootX = self.find(x)
#         rootY = self.find(y)
#         if rootX != rootY:
#             if self.rank[rootX] > self.rank[rootY]:
#                 self.roots[rootY] = rootX
#             elif self.rank[rootX] < self.rank[rootY]:
#                 self.roots[rootX] = rootY
#             else:
#                 self.roots[rootY] = rootX
#                 self.rank[rootX] += 1
#             return True
#         return False

# class Solution:
#     def minCostConnectPoints(self, points: List[List[int]]) -> int:
#         n = len(points)
#         if n <= 1:
#             return 0
        
#         graph = UnionFind(n)

#         edgeList = []

#         for i in range(n-1):
#             for j in range(i+1, n):
#                 point1_x, point1_y = points[i]
#                 point2_x, point2_y = points[j]
#                 dist = abs(point1_x-point2_x) + abs(point1_y-point2_y)

#                 edgeList.append((dist, i, j))
#         edgeList.sort()
        
#         minCost = 0
#         edgeCount = 0
#         for edge in edgeList:
#             cost, start, end = edge
#             if graph.union(start, end):
#                 edgeCount += 1
#                 minCost += cost
#                 if edgeCount == graph.n-1:
#                     break
#         return minCost

#Primms Algo with Minheap and Sets
class Solution:
    def manhattanDistance(self, point1, point2):
        return abs(point1[0]-point2[0])+abs(point1[1]-point2[1])
    
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        n = len(points)
        if n <= 1:
            return 0
        
        adjList = defaultdict(list)

        for i in range(n-1):
            for j in range(i, n):
                dist = self.manhattanDistance(points[i], points[j])
                adjList[i] += (dist, j),
                adjList[j] += (dist, i),

        unseen = set([i for i in range(1,n)])
        minHeap = adjList[0]
        heapq.heapify(minHeap)
        
        minCost = 0
        while len(unseen) != 0:
            next_cost, next_node = heapq.heappop(minHeap)
            if next_node in unseen:
                minCost += next_cost
                unseen.remove(next_node)
            
                for neighbor in adjList[next_node]:
                    if neighbor[1] in unseen:
                        heapq.heappush(minHeap, neighbor)
        
        return minCost
            
            
                
    
        
        
        
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