What is the time complexity of Add Edges to Make Degrees of All Nodes Even?

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

What data structures are used to solve Add Edges to Make Degrees of All Nodes Even?

This problem uses Hash Table, Graph. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Add Edges to Make Degrees of All Nodes Even?

Insight 1: The sum of degrees in any graph is always even. Therefore, the number of nodes with odd degrees must also be even. Insight 2: If there are 0 nodes with odd degrees, we're already done (return true).

Which companies ask LeetCode 2508. Add Edges to Make Degrees of All Nodes Even in interviews?

This problem is frequently asked at Hudson River Trading.

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Add Edges to Make Degrees of All Nodes Even - LeetCode 2508 Solution

Add Edges to Make Degrees of All Nodes Even - Complete Solution Guide

Add Edges to Make Degrees of All Nodes Even is LeetCode problem 2508, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is an undirected graph consisting of n nodes numbered from 1 to n . You are given the integer n and a 2D array edges where edges[i] = [a i , b i ] indicates that there is an edge between nodes a i and b i . The graph can be disconnected. You can add at most two additional edges (possibly none) to this graph so that there are no repeated edges and no self-loops. Return true if it is possible to make the degree of each node in the graph even, otherwise return false . The degree of a node is

Detailed Explanation

The problem asks us to determine if it's possible to make the degree (number of connected edges) of every node in an undirected graph even by adding at most two new edges. The graph can be disconnected initially. We are given the number of nodes 'n' and a list of existing edges 'edges'. We need to return 'true' if it's possible to make all node degrees even by adding 0, 1, or 2 edges, and 'false' otherwise. Constraints include the number of nodes being between 3 and 10^5, and the number of edges being between 2 and 10^5. No self-loops or repeated edges are allowed.

Solution Approach

The solution first constructs an adjacency list representation of the graph and calculates the degree of each node. Then, it identifies the nodes with odd degrees. Based on the number of odd-degree nodes (0, 2, or 4), it checks different conditions to see if adding at most two edges can make all degrees even. The approach leverages the properties of graph degrees and efficiently explores possible edge additions.

Step-by-Step Algorithm

Step 1: Create an adjacency list (adj) and an array (degrees) to store the degree of each node.
Step 2: Populate the adjacency list and degree array based on the given 'edges'. For each edge (u, v), add 'v' to the adjacency list of 'u' and 'u' to the adjacency list of 'v'. Increment the degrees of both 'u' and 'v'.
Step 3: Identify all nodes with odd degrees and store them in a list called 'odd_nodes'.
Step 4: Check the number of odd-degree nodes (k = len(odd_nodes)).
Step 5: If k == 0, return true (all degrees are already even).
Step 6: If k == 2, let the odd-degree nodes be 'u' and 'v'. If there isn't an edge between 'u' and 'v', add an edge between them and return true. Otherwise, check if there is a third node 'w' such that 'w' is not connected to 'u' and 'w' is not connected to 'v'. If such a 'w' exists, add edges (u, w) and (v, w) and return true. If no such w exists, then return false.
Step 7: If k == 4, let the odd-degree nodes be a, b, c, and d. Check if any of the following combinations allow all degrees to become even: (a, b) and (c, d), (a, c) and (b, d), (a, d) and (b, c). For each combination, verify that the edges to be added don't already exist. If any of these combinations work, return true. Otherwise, return false.
Step 8: If k > 4, return false (it's impossible to make all degrees even with at most 2 edges).

Key Insights

Insight 1: The sum of degrees in any graph is always even. Therefore, the number of nodes with odd degrees must also be even.
Insight 2: If there are 0 nodes with odd degrees, we're already done (return true).
Insight 3: If there are 2 nodes with odd degrees, we can simply add an edge between them if one doesn't already exist. If it does exist, we need to find a node that isn't connected to either of these odd degree nodes and connect them via two new edges.
Insight 4: If there are 4 nodes with odd degrees, we can try all possible pairings of these nodes into two edges to see if any combination works.
Insight 5: If there are more than 4 nodes with odd degrees, it's impossible to make all node degrees even by adding at most 2 edges (return false).

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Graph.

Companies

Asked at: Hudson River Trading.

Add Edges to Make Degrees of All Nodes Even | LeetCode 2508

Hard

Hash Table Graph

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

Solution Code

Available in:

import java.util.*;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
    public boolean isPossible(int n, List<List<Integer>> edges) {
        List<Set<Integer>> adj = new ArrayList<>();
        for (int i = 0; i <= n; i++) {
            adj.add(new HashSet<>());
        }

        int[] degrees = new int[n + 1];

        for (List<Integer> edge : edges) {
            int u = edge.get(0);
            int v = edge.get(1);
            adj.get(u).add(v);
            adj.get(v).add(u);
            degrees[u] += 1;
            degrees[v] += 1;
        }

        List<Integer> odd_nodes = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            if (degrees[i] % 2 != 0) {
                odd_nodes.add(i);
            }
        }

        int k = odd_nodes.size();

        if (k == 0) {
            return true;
        }

        if (k == 2) {
            int u = odd_nodes.get(0);
            int v = odd_nodes.get(1);
            if (!adj.get(u).contains(v)) {
                return true;
            }

            for (int w = 1; w <= n; w++) {
                if (w != u && w != v) {
                    if (!adj.get(u).contains(w) && !adj.get(v).contains(w)) {
                        return true;
                    }
                }
            }

            return false;
        }

        if (k == 4) {
            int a = odd_nodes.get(0);
            int b = odd_nodes.get(1);
            int c = odd_nodes.get(2);
            int d = odd_nodes.get(3);

            if (!adj.get(a).contains(b) && !adj.get(c).contains(d)) {
                return true;
            }

            if (!adj.get(a).contains(c) && !adj.get(b).contains(d)) {
                return true;
            }

            if (!adj.get(a).contains(d) && !adj.get(b).contains(c)) {
                return true;
            }

            return false;
        }

        return false;
    }
}

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Problem Info

Description

Problem 2508: Add Edges to Make Degrees of All Nodes Even - Detailed explanation and solution approaches.

Topics

Hash Table Graph

Companies

Hudson River Trading

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2508

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: Create an adjacency list (adj) and an array (degrees) to store the degree of each node.
2
Step 2: Populate the adjacency list and degree array based on the given 'edges'. For each edge (u, v), add 'v' to the adjacency list of 'u' and 'u' to the adjacency list of 'v'. Increment the degrees of both 'u' and 'v'.
3
Step 3: Identify all nodes with odd degrees and store them in a list called 'odd_nodes'.
4
Step 4: Check the number of odd-degree nodes (k = len(odd_nodes)).
5
Step 5: If k == 0, return true (all degrees are already even).
6
Step 6: If k == 2, let the odd-degree nodes be 'u' and 'v'. If there isn't an edge between 'u' and 'v', add an edge between them and return true. Otherwise, check if there is a third node 'w' such that 'w' is not connected to 'u' and 'w' is not connected to 'v'. If such a 'w' exists, add edges (u, w) and (v, w) and return true. If no such w exists, then return false.
7
Step 7: If k == 4, let the odd-degree nodes be a, b, c, and d. Check if any of the following combinations allow all degrees to become even: (a, b) and (c, d), (a, c) and (b, d), (a, d) and (b, c). For each combination, verify that the edges to be added don't already exist. If any of these combinations work, return true. Otherwise, return false.
8
Step 8: If k > 4, return false (it's impossible to make all degrees even with at most 2 edges).

💡

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 determined by the most time-consuming operations: constructing the adjacency list and iterating through the nodes to find those with odd degrees and performing the checks in cases where there are 2 or 4 odd degree nodes. The space complexity arises from storing the graph's adjacency list and the degree array.

📖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)

While DFS can be used for analyzing connectivity, it doesn't directly help in solving this problem. The core of the problem relies on the degrees of the nodes rather than connectivity, therefore, a DFS based approach would not be beneficial.

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

Trade-offs: DFS is not a useful alternative approach in this specific problem.

✨

Key Insights

1Insight 1: The sum of degrees in any graph is always even. Therefore, the number of nodes with odd degrees must also be even.
2Insight 2: If there are 0 nodes with odd degrees, we're already done (return true).
3Insight 3: If there are 2 nodes with odd degrees, we can simply add an edge between them if one doesn't already exist. If it does exist, we need to find a node that isn't connected to either of these odd degree nodes and connect them via two new edges.
4Insight 4: If there are 4 nodes with odd degrees, we can try all possible pairings of these nodes into two edges to see if any combination works.
5Insight 5: If there are more than 4 nodes with odd degrees, it's impossible to make all node degrees even by adding at most 2 edges (return false).

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Graph

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

Edge Cases

Edge case 1: The graph is already valid (all degrees are even). The solution handles this correctly by returning true when k == 0.
Edge case 2: The graph is disconnected. The solution correctly calculates the degrees of nodes regardless of connectivity. The checks for adding edges consider the entire node set, not just a connected component.
Edge case 3: When k = 2, the edge between the two odd degree nodes already exists. The code correctly identifies and handles this case by checking for another node to connect them to.
Edge case 4: When k = 4, none of the possible pairings are valid due to edges already existing. The code correctly returns false in this situation.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if the edge being added already exists. This would lead to incorrect results, as adding a duplicate edge is not allowed.

✗Common mistake 2: Not considering all possible pairings of odd-degree nodes when k = 4. Failing to explore all possibilities would lead to a wrong result.

✗Common mistake 3: Incorrectly calculating the degree of a node, either by not updating it correctly when adding edges, or by using an inefficient data structure for storing edges.

✗Common mistake 4: Assuming the graph is connected and failing to handle disconnected graphs correctly.

✅

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 design. Imagine a network where each device should have an even number of connections for stability. This algorithm could be used to determine if adding a few links can balance the connections.

🏦Real-world scenario 2: Data integrity. Consider a system where data chunks are linked together, and each chunk should have an even number of links for redundancy. This algorithm could verify if the system can be corrected with a few additional links.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate 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 16, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

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

@TechLeadJul 7

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I represent a graph in code?

Use adjacency list (HashMap or array of lists) for sparse graphs - it's space-efficient O(V+E). Use adjacency matrix for dense graphs or when you need O(1) edge lookup. For weighted graphs, store (neighbor, weight) pairs in adjacency lists.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Add Edges to Make Degrees of All Nodes Even"?

For this hard-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether graph techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?