# clean python union find with path compression

• All credit goes to this guy for awesome explanation.

``````class Solution(object):
def countComponents(self, n, edges):
ds = DisjointSet()

for i in range(n):
ds.make_set(i)

for i, j in edges:
ds.union(i, j)

return ds.num_sets

class Node(object):
def __init__(self, data, parent=None, rank=0):
self.data = data
self.parent = parent
self.rank = rank

class DisjointSet(object):
def __init__(self):
self.map = {}
self.num_sets = 0

def make_set(self, data):
node = Node(data)
node.parent = node
self.map[data] = node
self.num_sets += 1

def union(self, data1, data2):
node1 = self.map[data1]
node2 = self.map[data2]

parent1 = self.find_set_util(node1)
parent2 = self.find_set_util(node2)

if parent1.data == parent2.data:
return

if parent1.rank >= parent2.rank:
if parent1.rank == parent2.rank:
parent1.rank += 1

parent2.parent = parent1
else:
parent1.parent = parent2

self.num_sets -= 1

def find_set(self, data):
return self.find_set_util(self.map[data])

def find_set_util(self, node):
parent = node.parent

if parent == node:
return parent

node.parent = self.find_set_util(node.parent)
return node.parent
``````

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