LeetCode 1319 Number of Operations to Make Network Connected - Implementation

Check if enough cables. Count components. Answer = components - 1.

def makeConnected(n, connections): if len(connections) < n - 1: return -1

parent = list(range(n))

def find(x):
    if parent[x] != x:
        parent[x] = find(parent[x])
    return parent[x]

def union(x, y):
    px, py = find(x), find(y)
    if px != py:
        parent[px] = py

for a, b in connections:
    union(a, b)

components = sum(1 for i in range(n) if find(i) == i)
return components - 1

$O(m \cdot \alpha(n))$ time, $O(n)$ space.