How do you solve Add Edges to Make Degrees of All Nodes Even on LeetCode?

Add Edges to Make Degrees of All Nodes Even (LeetCode #2508) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The degree of a node is even if it has an even number of edges connected to it.

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

The optimal solution for Add Edges to Make Degrees of All Nodes Even runs in O(n) time and uses O(n) space. This complexity is due to counting degrees and checking pairs of odd nodes, which is linear in relation to the number of edges.

What is the brute force solution for Add Edges to Make Degrees of All Nodes Even?

The brute force approach for Add Edges to Make Degrees of All Nodes Even is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks all possible pairs of nodes to see if adding an edge between them can make all node degrees even. This is simple but inefficient for large graphs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Add Edges to Make Degrees of All Nodes Even?

Add Edges to Make Degrees of All Nodes Even uses the following concepts: Hash Table, Graph Theory. The recommended patterns to study are: Graph Theory, Set Operations.

What are the interview tips for Add Edges to Make Degrees of All Nodes Even?

Always clarify the constraints and edge cases before diving into coding. Think about the properties of the graph and how they relate to the problem. Practice explaining your thought process as you code to simulate the interview environment.

What are common mistakes when solving Add Edges to Make Degrees of All Nodes Even?

Overlooking the condition that only 0, 2, or 4 odd degree nodes can exist for a valid solution. Failing to check for duplicate edges when considering new edges.

#2508

Add Edges to Make Degrees of All Nodes Even

Hard

Hash TableGraph TheoryGraph TheorySet Operations

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the fact that the number of nodes with odd degrees must be 0, 2, or 4 for it to be possible to make all degrees even. This allows us to directly check combinations without brute-forcing.

⚙️

Algorithm

5 steps

1Step 1: Count the degree of each node from the given edges.
2Step 2: Identify nodes with odd degrees.
3Step 3: If there are 0 odd nodes, return true.
4Step 4: If there are 2 odd nodes, check if they can be connected without creating a duplicate edge.
5Step 5: If there are 4 odd nodes, check if any two can be connected without creating a duplicate edge.

solution.py22 lines

1# Full working Python code
2from collections import defaultdict
3
4def canMakeDegreesEven(n, edges):
5    degree = [0] * (n + 1)
6    edgeSet = set()
7    for a, b in edges:
8        degree[a] += 1
9        degree[b] += 1
10        edgeSet.add((a, b))
11        edgeSet.add((b, a))
12    odd_nodes = [i for i in range(1, n + 1) if degree[i] % 2 != 0]
13    if len(odd_nodes) == 0:
14        return True
15    if len(odd_nodes) == 2:
16        return (odd_nodes[0], odd_nodes[1]) not in edgeSet
17    if len(odd_nodes) == 4:
18        for i in range(4):
19            for j in range(i + 1, 4):
20                if (odd_nodes[i], odd_nodes[j]) not in edgeSet:
21                    return True
22    return False

ℹ

Complexity note: This complexity is due to counting degrees and checking pairs of odd nodes, which is linear in relation to the number of edges.

1The degree of a node is even if it has an even number of edges connected to it.
2Adding an edge between two nodes affects the degree of both nodes.

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