What is the brute force approach and complexity for Number of Operations to Make Network Connected?

Keep a group label per computer; each join renames one whole group with time complexity O(C · N) and space complexity O(N).

What is the optimal approach and complexity for Number of Operations to Make Network Connected?

Union each cable; every successful union removes one group; the answer is groups − 1 with optimal time complexity O(C · α(N)) and space complexity O(N).

How do you identify the pattern for Number of Operations to Make Network Connected?

Look for key signals in the prompt: "connections between computers" indicates Cables merge groups, so the question is how many groups are left.; "minimum number of operations" indicates Each move joins at most two groups, so count the groups, not the spare cables.; "return -1 if impossible" indicates Fewer than `n − 1` cables in total can never connect everything..

Number of Operations to Make Network Connected

Medium·LeetCode #1319 Pro OnlySlight Variation

LeetCode

Problem Statement

There are n computers, numbered 0 to n − 1. Some pairs are joined by a cable. You get the cables as connections, where connections[i] = [a, b] joins computers a and b directly. Two computers can talk if a chain of cables joins them.

You can unplug any cable and plug it into a different pair of computers. Return the minimum number of moves (cables unplugged and plugged in again) so that every computer can talk to every other one. If it cannot be done, return -1.

Input

n, the number of computers, and connections, the list of cables.

Output

The minimum number of cable moves, or -1 if it is not possible.

Example 1

Input: n = 4, connections = [[0, 1], [0, 2], [1, 2]]

Output: 1

Computers 0, 1 and 2 are linked, and computer 3 is alone. Move cable 1–2 (it closes a loop, so nothing breaks) to join 3 to the group.

Example 2

Input: n = 6, connections = [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3]]

Output: 2

Computers 0 to 3 form one group, and 4 and 5 are alone. The spare cables 1–2 and 1–3 close loops, so each can join one of the lone computers: 2 moves.

Example 3

Input: n = 6, connections = [[0, 1], [0, 2], [0, 3], [1, 2]]

Output: -1

Connecting 6 computers needs at least 5 cables, and there are only 4.

Constraints

• 1 ≤ n ≤ 10^5
• 1 ≤ connections.length ≤ min(n · (n − 1) / 2, 10^5)
• 0 ≤ aᵢ, bᵢ < n
• aᵢ ≠ bᵢ
• No two computers are connected by more than one cable.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“connections between computers”

→ Cables merge groups, so the question is how many groups are left.

“minimum number of operations”

→ Each move joins at most two groups, so count the groups, not the spare cables.

“return -1 if impossible”

→ Fewer than `n − 1` cables in total can never connect everything.

Approach Complexity Ladder

Brute Force — relabel the whole group on every joinO(C · N)O(N)

Keep a group label per computer; each join renames one whole group

Optimal — Union-Find: count the groupsO(C · α(N))O(N)

Union each cable; every successful union removes one group; the answer is groups − 1

Brute Force Intuition

Give every computer a group label, starting with its own number. For each cable whose ends have different labels, rename every computer with one label to the other label, and count one fewer group. Every join costs a full pass over all n computers.

Recognition rule for this sub-pattern:

If both ends of an edge already share a root, this edge closes a cycle.

Remember this · Union-Find (DSU)

Trigger

Edges arrive one by one, "are these connected?", merge equivalent items, count components, extra edge creating a cycle.

Think

Am I repeatedly merging groups and asking whether two things are in the same group?

Key question

What are the elements, and what does it mean to union two of them?

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