What is the time complexity of Count Unreachable Pairs of Nodes in an Undirected Graph?

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

What data structures are used to solve Count Unreachable Pairs of Nodes in an Undirected Graph?

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 Count Unreachable Pairs of Nodes in an Undirected Graph?

Insight 1: We can use the Union-Find algorithm to efficiently determine the connected components in the graph. Insight 2: Once we know the size of each connected component, we can calculate the number of unreachable pairs by multiplying the size of each component with the sum of the sizes of all the previous components.

Which companies ask LeetCode 2316. Count Unreachable Pairs of Nodes in an Undirected Graph in interviews?

This problem is frequently asked at Commvault.

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Count Unreachable Pairs of Nodes in an Undirected Graph - LeetCode 2316 Solution

Count Unreachable Pairs of Nodes in an Undirected Graph - Complete Solution Guide

Count Unreachable Pairs of Nodes in an Undirected Graph is LeetCode problem 2316, 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

You are given an integer n . There is an undirected graph with n nodes, numbered from 0 to n - 1 . You are given a 2D integer array edges where edges[i] = [a i , b i ] denotes that there exists an undirected edge connecting nodes a i and b i . Return the number of pairs of different nodes that are unreachable from each other . Example 1: Input: n = 3, edges = [[0,1],[0,2],[1,2]] Output: 0 Explanation: There are no pairs of nodes that are unreachable from each other. Therefore, we return 0. Examp

Detailed Explanation

The problem asks us to find the number of pairs of nodes in an undirected graph that are unreachable from each other. We are given the number of nodes `n` and a list of edges `edges`. Two nodes are considered unreachable if there is no path connecting them in the graph. The goal is to count all such pairs of nodes.

Solution Approach

The solution uses the Union-Find algorithm to find connected components in the graph. The algorithm iterates through the edges and merges the nodes connected by each edge into the same connected component. After processing all edges, it iterates through all nodes and calculates the size of each connected component. Then it computes the number of unreachable pairs based on the sizes of the connected components.

Step-by-Step Algorithm

Step 1: Initialize a `parent` array where `parent[i]` represents the parent of node `i`. Initially, each node is its own parent (i.e., each node is in its own connected component). Initialize a `size` array where `size[i]` represents the size of the connected component that node `i` belongs to. Initially, each component has a size of 1.
Step 2: Implement the `find(i)` function, which finds the root (representative) of the connected component that node `i` belongs to. Path compression is used to optimize the find operation by updating the parent of each node along the path to point directly to the root.
Step 3: Implement the `union(i, j)` function, which merges the connected components that nodes `i` and `j` belong to. Union by size is used to optimize the union operation by attaching the smaller tree to the larger tree. This helps to keep the trees relatively flat, which reduces the time complexity of the `find` operation.
Step 4: Iterate through the `edges` array. For each edge `(u, v)`, call the `union(u, v)` function to merge the connected components that nodes `u` and `v` belong to.
Step 5: Initialize `unreachable_pairs` to 0 and `nodes_so_far` to 0.
Step 6: Iterate through all nodes from 0 to `n-1`. For each node `i`, check if `parent[i] == i`. If it is, then node `i` is the root of a connected component.
Step 7: If node `i` is a root, calculate `component_size = size[i]`. Add `component_size * nodes_so_far` to `unreachable_pairs`. Then, add `component_size` to `nodes_so_far`.
Step 8: Return `unreachable_pairs`.

Key Insights

Insight 1: We can use the Union-Find algorithm to efficiently determine the connected components in the graph.
Insight 2: Once we know the size of each connected component, we can calculate the number of unreachable pairs by multiplying the size of each component with the sum of the sizes of all the previous components.
Insight 3: Using path compression and union by size in the Union-Find implementation optimizes the algorithm's performance.

Complexity Analysis

Time Complexity: O(n + e)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Commvault.

Count Unreachable Pairs of Nodes in an Undirected Graph | LeetCode 2316

Medium

Depth-First Search Breadth-First Search Union Find Graph

Time: O(n + e)Space: O(n)

Solution Code

Available in:

class Solution {
    public long countPairs(int n, int[][] edges) {
        int[] parent = new int[n];
        int[] size = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
            size[i] = 1;
        }

        function find = (i) -> {
            if (parent[i] == i) {
                return i;
            }
            parent[i] = find.apply(parent[i]);
            return parent[i];
        };

        interface function {
            int apply(int i);
        }

        BiConsumer union = (i, j) -> {
            int root_i = find.apply(i);
            int root_j = find.apply(j);
            if (root_i != root_j) {
                if (size[root_i] < size[root_j]) {
                    int temp = root_i;
                    root_i = root_j;
                    root_j = temp;
                }
                parent[root_j] = root_i;
                size[root_i] += size[root_j];
            }
        };

        interface BiConsumer {
            void accept(int i, int j);
        }


        for (int[] edge : edges) {
            union.accept(edge[0], edge[1]);
        }

        long unreachable_pairs = 0;
        long nodes_so_far = 0;
        for (int i = 0; i < n; i++) {
            if (parent[i] == i) {
                long component_size = size[i];
                unreachable_pairs += component_size * nodes_so_far;
                nodes_so_far += component_size;
            }
        }

        return unreachable_pairs;
    }
}

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Problem Info

Description

Problem 2316: Count Unreachable Pairs of Nodes in an Undirected Graph - Detailed explanation and solution approaches.

Topics

Depth-First Search Breadth-First Search Union Find Graph

Companies

Commvault

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2316

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: Initialize a `parent` array where `parent[i]` represents the parent of node `i`. Initially, each node is its own parent (i.e., each node is in its own connected component). Initialize a `size` array where `size[i]` represents the size of the connected component that node `i` belongs to. Initially, each component has a size of 1.
2
Step 2: Implement the `find(i)` function, which finds the root (representative) of the connected component that node `i` belongs to. Path compression is used to optimize the find operation by updating the parent of each node along the path to point directly to the root.
3
Step 3: Implement the `union(i, j)` function, which merges the connected components that nodes `i` and `j` belong to. Union by size is used to optimize the union operation by attaching the smaller tree to the larger tree. This helps to keep the trees relatively flat, which reduces the time complexity of the `find` operation.
4
Step 4: Iterate through the `edges` array. For each edge `(u, v)`, call the `union(u, v)` function to merge the connected components that nodes `u` and `v` belong to.
5
Step 5: Initialize `unreachable_pairs` to 0 and `nodes_so_far` to 0.
6
Step 6: Iterate through all nodes from 0 to `n-1`. For each node `i`, check if `parent[i] == i`. If it is, then node `i` is the root of a connected component.
7
Step 7: If node `i` is a root, calculate `component_size = size[i]`. Add `component_size * nodes_so_far` to `unreachable_pairs`. Then, add `component_size` to `nodes_so_far`.
8
Step 8: Return `unreachable_pairs`.

💡

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 + e)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the Union-Find operations and the iterations through nodes and edges. The space complexity is determined by the size of the `parent` and `size` arrays.

📖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) or Breadth-First Search (BFS)

We can use DFS or BFS to find the connected components. For each node, we can start a DFS or BFS and visit all reachable nodes. Then, we can count the size of each component and calculate the number of unreachable pairs.

Time: O(n + e), where n is the number of nodes and e is the number of edges.Space: O(n), due to the visited array and the recursion stack or queue.

Trade-offs: DFS/BFS might be slightly less efficient in practice due to the overhead of recursive calls or queue operations compared to the optimized Union-Find implementation, particularly for large graphs. However, DFS/BFS might be more intuitive to implement for some.

✨

Key Insights

1Insight 1: We can use the Union-Find algorithm to efficiently determine the connected components in the graph.
2Insight 2: Once we know the size of each connected component, we can calculate the number of unreachable pairs by multiplying the size of each component with the sum of the sizes of all the previous components.
3Insight 3: Using path compression and union by size in the Union-Find implementation optimizes the algorithm's performance.

📚

Key Terms Explained

For Students

Graph

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

Edge Cases

Edge case 1: An empty graph (edges is empty). The algorithm correctly handles this case because the Union-Find data structure is initialized such that each node is initially its own component. The outer loop will still correctly calculate the result.
Edge case 2: A fully connected graph. In this case, all nodes will be in the same connected component. The algorithm correctly calculates that there are 0 unreachable pairs.
Edge case 3: A graph with multiple disconnected components. The algorithm correctly identifies each component and uses the size of each to calculate the number of unreachable pairs.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using path compression and union by size in Union-Find, leading to higher time complexity and potential TLE (Time Limit Exceeded) errors for large graphs.

✗Common mistake 2: Incorrectly calculating the number of unreachable pairs by not considering the sizes of the connected components or by double-counting pairs.

✗Common mistake 3: Failing to handle the case where the graph is disconnected, assuming all nodes are reachable from each other.

✅

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. This algorithm can be used to determine the number of pairs of computers in a network that cannot communicate with each other due to network partitioning.

🏦Real-world scenario 2: Social network analysis. It can be used to identify disconnected groups of people in a social network, where people within each group are connected to each other but not to people in other groups.

🎮Real-world scenario 3: Recommendation systems. It can be used to identify groups of users with similar preferences who are not yet connected in the system, allowing for targeted recommendations to bridge these gaps.

📱Real-world scenario 4: Analyzing road networks: Determining segments of a road network that are unreachable from other segments due to road closures or natural disasters.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: November 4, 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

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

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count Unreachable Pairs of Nodes in an Undirected Graph"?

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?