What is the time complexity of Number of Operations to Make Network Connected?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Number of Operations to Make Network Connected?

This problem uses Depth-First Search, Breadth-First Search, Union Find, Graph. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Number of Operations to Make Network Connected?

Insight 1: The problem can be modeled as finding the number of connected components in a graph. Each computer is a node, and each connection is an edge. Insight 2: To connect `k` connected components, we need `k - 1` cables.

Which companies ask LeetCode 1319. Number of Operations to Make Network Connected in interviews?

This problem is frequently asked at Akuna Capital, IBM, Intuit, McKinsey, Nvidia.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Number of Operations to Make Network Connected - LeetCode 1319 Solution

Number of Operations to Make Network Connected - Complete Solution Guide

Number of Operations to Make Network Connected is LeetCode problem 1319, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There are n computers numbered from 0 to n - 1 connected by ethernet cables connections forming a network where connections[i] = [a i , b i ] represents a connection between computers a i and b i . Any computer can reach any other computer directly or indirectly through the network. You are given an initial computer network connections . You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connect

Detailed Explanation

The problem asks us to determine the minimum number of cable moves required to connect all `n` computers in a network. We are given `connections`, a list of existing cable connections between computers. We can move existing cables to connect disconnected computers. If there aren't enough cables to connect all computers, we should return -1.

Solution Approach

The solution uses the Union Find data structure to efficiently determine the number of connected components. First, we check if the number of connections is less than `n - 1`. If it is, we return -1 because it's impossible to connect all computers. Then, we initialize a Union Find data structure with `n` nodes, where each node is initially its own parent (representing a separate component). We iterate through the given `connections` and use the `union` operation to merge the connected components. Finally, we return the number of connected components minus 1, which represents the number of cable moves needed.

Step-by-Step Algorithm

Step 1: Check if the number of connections is less than `n - 1`. If so, return -1.
Step 2: Initialize a `parent` array of size `n`, where `parent[i] = i` for all `i`. This represents that each node is initially in its own connected component.
Step 3: Initialize `num_components` to `n` (initially, each node is a component).
Step 4: Iterate through the `connections` list. For each connection `(u, v)`:
Step 5: Find the root of the connected component containing `u` using the `find` operation.
Step 6: Find the root of the connected component containing `v` using the `find` operation.
Step 7: If the roots are different, merge the two connected components using the `union` operation. Decrement `num_components` by 1.
Step 8: Return `num_components - 1`. This is the number of moves needed to connect all the components.

Key Insights

Insight 1: The problem can be modeled as finding the number of connected components in a graph. Each computer is a node, and each connection is an edge.
Insight 2: To connect `k` connected components, we need `k - 1` cables.
Insight 3: We need at least `n - 1` cables to potentially connect all `n` computers. If we have fewer cables than that, it's impossible.
Insight 4: Union Find (Disjoint Set Union) data structure is an efficient approach to find connected components in a graph.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Depth-First Search, Breadth-First Search, Union Find, Graph.

Companies

Asked at: Akuna Capital, IBM, Intuit, McKinsey, Nvidia.

Number of Operations to Make Network Connected | LeetCode 1319

Medium

Depth-First Search Breadth-First Search Union Find Graph

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int makeConnected(int n, int[][] connections) {
        if (connections.length < n - 1) {
            return -1;
        }

        int[] parent = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }

        int num_components = n;

        function find = new function(parent);
        function union = new function(parent, num_components);


        for (int[] connection : connections) {
            union.union(connection[0], connection[1]);
            num_components = union.num_components;
        }

        return num_components - 1;
    }

    class function{
        int[] parent;
        int num_components;
        function(int[] parent){
            this.parent = parent;
        }
        function(int[] parent, int num_components){
            this.parent = parent;
            this.num_components = num_components;
        }

        public int find(int i) {
            if (parent[i] == i) {
                return i;
            }
            parent[i] = find(parent[i]);
            return parent[i];
        }

        public void union(int i, int j) {
            int root_i = find(i);
            int root_j = find(j);
            if (root_i != root_j) {
                parent[root_j] = root_i;
                num_components--;
            }
        }
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1319: Number of Operations to Make Network Connected - Detailed explanation and solution approaches.

Topics

Depth-First Search Breadth-First Search Union Find Graph

Companies

Akuna CapitalIBMIntuitMcKinseyNvidia

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:1319

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Check if the number of connections is less than `n - 1`. If so, return -1.
2
Step 2: Initialize a `parent` array of size `n`, where `parent[i] = i` for all `i`. This represents that each node is initially in its own connected component.
3
Step 3: Initialize `num_components` to `n` (initially, each node is a component).
4
Step 4: Iterate through the `connections` list. For each connection `(u, v)`:
5
Step 5: Find the root of the connected component containing `u` using the `find` operation.
6
Step 6: Find the root of the connected component containing `v` using the `find` operation.
7
Step 7: If the roots are different, merge the two connected components using the `union` operation. Decrement `num_components` by 1.
8
Step 8: Return `num_components - 1`. This is the number of moves needed to connect all the components.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the iterations through the connections list and the find/union operations. Path compression makes the amortized time complexity of find/union operations nearly constant. The space complexity comes from the parent array used in the Union Find data structure.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Depth-First Search (DFS)

Use DFS to find the number of connected components. For each unvisited node, perform a DFS to mark all nodes in the same component as visited. The number of times DFS is called is the number of connected components.

Time: O(V + E), where V is the number of vertices (computers) and E is the number of edges (connections).Space: O(V), for the visited set and recursion stack.

Trade-offs: DFS has similar time and space complexity as Union Find, but Union Find is often slightly faster in practice due to its efficient union and find operations. Union Find is also more concise to implement for this particular problem.

Breadth-First Search (BFS)

Use BFS to find the number of connected components. Similar to DFS, iterate through all the nodes. If a node is not visited, start a BFS from that node and explore all nodes in its connected component.

Time: O(V + E), where V is the number of vertices (computers) and E is the number of edges (connections).Space: O(V), for the visited set and queue.

Trade-offs: BFS and DFS will result in same complexity. Again, Union find is more efficient for this kind of problem.

✨

Key Insights

1Insight 1: The problem can be modeled as finding the number of connected components in a graph. Each computer is a node, and each connection is an edge.
2Insight 2: To connect `k` connected components, we need `k - 1` cables.
3Insight 3: We need at least `n - 1` cables to potentially connect all `n` computers. If we have fewer cables than that, it's impossible.
4Insight 4: Union Find (Disjoint Set Union) data structure is an efficient approach to find connected components in a graph.

📚

Key Terms Explained

For Students

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: `connections.length < n - 1`. The solution correctly handles this case by returning -1, as there are not enough cables to connect all the computers.
Edge case 2: The graph is already connected. In this case, the `union` operation will not merge any components, and the `num_components` will remain 1. The solution will return 0, indicating that no moves are needed.
Edge case 3: Single node and no connections. When n = 1 and connections is empty list, the code will return 0. This is because num_components will be 1, and 1-1 = 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if the number of connections is less than `n - 1`. This is a critical condition for the problem.

✗Common mistake 2: Not using path compression in the Union Find data structure. This can lead to a higher time complexity in some cases.

✗Common mistake 3: Incorrectly calculating the number of moves needed. It's `num_components - 1`, not `num_components`.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network connectivity analysis. Determining how many connections are needed to ensure all devices in a network can communicate with each other.

🏦Real-world scenario 2: Social network analysis. Identifying the number of distinct communities in a social network and the number of connections needed to merge them.

🎮Real-world scenario 3: Clustering problems. Union find can efficiently merge clusters, such as image segmentation or document clustering

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 30, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

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

For this medium-level Depth-First Search problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?