#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/pattern-22/union-find/number-of-operations-to-make-network-connected-implementatio

░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░███████████████████████████████████████████████████████████████████████████████████████████████████████████████

#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/pattern-22/union-find/number-of-operations-to-make-network-connected-implementatio

Repovive

LeetCode 1319 Number of Operations to Make Network Connected - Implementation - Union Find | Pattern 22: LeetCode Interview Patterns | Repovive

Repovive.

Pattern 22: LeetCode Interview Patterns24 sections · 736 units32h total

Open in Course

LeetCode 1319 Number of Operations to Make Network Connected - Implementation

The approach

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.