How do you solve Number of Operations to Make Network Connected on LeetCode?

Number of Operations to Make Network Connected (LeetCode #1319) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n α(n)) time complexity and O(n) space complexity. Key insight: If the number of connections is less than n - 1, it's impossible to connect all computers.

What is the brute force solution for Number of Operations to Make Network Connected?

The brute force approach for Number of Operations to Make Network Connected is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach tries to connect all computers by checking every possible connection and counting how many operations are needed to connect all isolated clusters. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n α(n)).

What are the interview tips for Number of Operations to Make Network Connected?

Always clarify the problem constraints and edge cases before diving into coding. Explain your thought process and approach clearly to the interviewer. Practice implementing the Union-Find structure, as it's commonly used in graph-related problems.

What are common mistakes when solving Number of Operations to Make Network Connected?

Forgetting to check if the number of connections is sufficient to connect all computers. Misunderstanding how to count connected components correctly.

#1319

Number of Operations to Make Network Connected

Medium

Depth-First SearchBreadth-First SearchUnion-FindGraph TheoryUnion-FindGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n α(n))
Space	O(n)	O(n)

💡

Intuition

Time O(n α(n))Space O(n)

The optimal approach uses the Union-Find (Disjoint Set Union) data structure to efficiently manage and connect components. This allows us to quickly determine how many operations are needed to connect all computers.

⚙️

Algorithm

3 steps

1Step 1: Initialize a Union-Find structure to manage connected components.
2Step 2: Iterate through the connections and union the connected computers.
3Step 3: Count the number of unique components and calculate the number of operations needed.

solution.py28 lines

1class UnionFind:
2    def __init__(self, size):
3        self.parent = list(range(size))
4        self.rank = [1] * size
5    def find(self, x):
6        if self.parent[x] != x:
7            self.parent[x] = self.find(self.parent[x])
8        return self.parent[x]
9    def union(self, x, y):
10        rootX = self.find(x)
11        rootY = self.find(y)
12        if rootX != rootY:
13            if self.rank[rootX] > self.rank[rootY]:
14                self.parent[rootY] = rootX
15            elif self.rank[rootX] < self.rank[rootY]:
16                self.parent[rootX] = rootY
17            else:
18                self.parent[rootY] = rootX
19                self.rank[rootX] += 1
20
21def makeConnected(n, connections):
22    if len(connections) < n - 1:
23        return -1
24    uf = UnionFind(n)
25    for a, b in connections:
26        uf.union(a, b)
27    components = len(set(uf.find(i) for i in range(n)))
28    return components - 1

ℹ

Complexity note: The time complexity is nearly O(n) due to the efficient union-find operations, where α(n) is the inverse Ackermann function, which grows very slowly. The space complexity is O(n) for storing the parent and rank arrays.

1If the number of connections is less than n - 1, it's impossible to connect all computers.
2The number of operations needed to connect all computers is equal to the number of isolated components minus one.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.