69 lines
2.5 KiB
Python
69 lines
2.5 KiB
Python
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# Time: O(N^2·logN)
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# Space: O(N^2) ; we could be pushing N·N(N - 1) / 2 edges into the heap
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from collections import defaultdict
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from heapq import heappush, heappop
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class Solution:
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def minCostConnectPoints(self, points: List[List[int]]) -> int:
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'''
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BFS + Prim's algorithm
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'''
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N = len(points)
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neighbors = defaultdict(list)
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# Build adjacency lists by connecting each point to every other
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# point since we are building a new graph here
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for i in range(N):
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x1, y1 = points[i]
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for j in range(i + 1, N):
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x2, y2 = points[j]
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# We are considering the Manhattan Distance as stated in
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# the problem
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distance = abs(x1 - x2) + abs(y1 - y2)
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# We are going to add both the neighbor of given point as
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# well as the distance so that we can later use it to create
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# Prim's min heap
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neighbors[i].append((distance, j))
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neighbors[j].append((distance, i))
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# Perform Prim's algorithm
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# To avoid cycles
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visited = set()
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# We start at the first point, distance would be 0 since it's the first
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# point
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min_heap = [(0, 0)]
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# We need to find distances from one node to all other nodes (as long as
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# its not already visited), keep track of it in the min heap and once
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# all neighbors of a node is visited, we should have found the next node
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# to pop from the heap since it'd have the least distance.
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#
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# We need to keep doing this until we have visited all the nodes
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total = 0
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while len(visited) < N:
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curr_dist, min_point = heappop(min_heap)
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# It's possible we've already visited the node (let's say the very first node we
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# initialized the heap with) since it could end up being added to the heap again
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# while visiting other nodes. If this is the case, we skip this iteration of the
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# loop
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if min_point in visited:
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continue
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total += curr_dist
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visited.add(min_point)
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for neighbor in neighbors[min_point]:
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if neighbor[1] not in visited:
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heappush(min_heap, neighbor)
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return total
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