leetcode/1706_min-cost-to-connect-all-points/python3/solution.py

69 lines
2.5 KiB
Python

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