45 lines
1.3 KiB
Python
45 lines
1.3 KiB
Python
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# Time: O(E · 𝛼(n)) ; 𝛼(n) is inverse Ackermann function
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# Space: O(V)
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class Solution:
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def countComponents(self, n: int, edges: List[List[int]]) -> int:
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parent = [i for i in range(n)]
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# Or like sizes of the components
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rank = [1] * n
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def find(node):
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# Parent of node could be itself (i.e, no parent)
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current_node = node
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while current_node != parent[current_node]:
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# Path compression to speed up union-find
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parent[current_node] = parent[parent[current_node]]
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current_node = parent[current_node]
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return current_node
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def union(a, b):
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parent_a, parent_b = find(a), find(b)
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if parent_a == parent_b:
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return 0
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if rank[parent_a] > rank[parent_b]:
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parent[parent_a] = parent_b
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rank[parent_b] += rank[parent_a]
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else:
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parent[parent_b] = parent_a
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rank[parent_a] += rank[parent_b]
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return 1
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result = n
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for a, b in edges:
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result -= union(a, b)
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return result
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