Graph Terminology Vertices Edges Directed Undirected Weighted in DSA Java - Time & Space Complexity

Time Complexity: Graph Terminology Vertices Edges Directed Undirected Weighted

O(V)

Understanding Time Complexity

Graphs are made of points and connections. We want to understand how the time to work with these parts grows as the graph gets bigger.

How does the number of points and connections affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Simple graph representation
class Graph {
  int vertices; // number of points
  LinkedList<Integer>[] edges; // connections list

  Graph(int v) {
    vertices = v;
    edges = new LinkedList[v];
    for (int i = 0; i < v; i++) {
      edges[i] = new LinkedList<Integer>();
    }
  }

  void addEdge(int from, int to) {
    edges[from].add(to); // add connection
  }
}
    

This code creates a graph with points and connections using lists. It adds a connection from one point to another.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop to initialize the list of connections for each point.
• How many times: Once for each point (vertices times).

How Execution Grows With Input

As the number of points grows, the initialization work grows in a straight line. Adding one connection is a quick step.

Input Size (vertices)	Approx. Operations
10	10 loops to create lists
100	100 loops to create lists
1000	1000 loops to create lists

Pattern observation: The work grows directly with the number of points.

Final Time Complexity

Time Complexity: O(V)

This means the time to set up the graph grows directly with the number of points.

Common Mistake

[X] Wrong: "Adding a connection takes time depending on all points or connections."

[OK] Correct: Adding a connection is just putting one item in a list, which is quick and does not depend on the total size.

Interview Connect

Understanding how graph size affects work helps you explain and write efficient graph code confidently in interviews.

Self-Check

"What if we used a matrix instead of lists to store connections? How would the time complexity change?"

Practice

1. Which of the following best describes a vertex in a graph?

easy

A. A point or node where edges meet

B. A line connecting two vertices

C. A graph with no edges

D. A graph that has weights on edges

Solution

1. Step 1: Understand the definition of vertex

   A vertex is a fundamental unit in a graph representing a point or node.

2. Step 2: Differentiate vertex from edges and graph types

   Edges connect vertices; weights and graph types describe edges or the whole graph, not vertices.

3. Final Answer:

   A point or node where edges meet -> Option A

4. Quick Check:

   Vertex = point/node [OK]

Hint: Vertices are points/nodes, edges are connections [OK]

Common Mistakes:

• Confusing edges with vertices
• Thinking weights define vertices
• Mixing graph types with vertex definition

2. Which of the following is the correct way to declare an undirected weighted graph using adjacency list in Java?

easy

A. Map graph = new HashMap<>();

B. Map graph = new TreeMap<>();

C. Map>> graph = new TreeSet<>();

D. Map>> graph = new HashMap<>();

Solution

1. Step 1: Identify data structure for weighted edges

   Weighted edges need to store both neighbor and weight, so Pair is used.

2. Step 2: Choose correct map and list types

   HashMap with List of pairs is standard for adjacency list of weighted graphs.

3. Final Answer:

   Map>> graph = new HashMap<>(); -> Option D

4. Quick Check:

   Weighted graph uses Pair for edges [OK]

Hint: Weighted edges need pairs: neighbor and weight [OK]

Common Mistakes:

• Using List<Integer> for weighted edges
• Using TreeSet instead of List for adjacency
• Confusing Map types for graph

3. Consider the following Java code snippet for a directed graph:

Map graph = new HashMap<>();
graph.put(1, Arrays.asList(2, 3));
graph.put(2, Arrays.asList(3));
graph.put(3, Arrays.asList());
System.out.println(graph.get(1));

What will be the output?

medium

A. null

B. [1, 2, 3]

C. [2, 3]

D. []

Solution

1. Step 1: Check what is stored for key 1

   Key 1 maps to list containing 2 and 3.

2. Step 2: Print the list for key 1

   Printing graph.get(1) outputs the list [2, 3].

3. Final Answer:

   [2, 3] -> Option C

4. Quick Check:

   graph.get(1) = [2, 3] [OK]

Hint: Print adjacency list for key to see neighbors [OK]

Common Mistakes:

• Expecting all vertices in output
• Confusing keys and values
• Assuming empty list prints null

4. The following code attempts to add an edge to an undirected graph represented by adjacency list. What is the bug?

void addEdge(Map graph, int u, int v) {
    graph.get(u).add(v);
}

medium

A. It adds edge only from u to v, missing v to u

B. It uses get() instead of put() to add edge

C. It should use a Set instead of List for adjacency

D. It should clear the adjacency list before adding

Solution

1. Step 1: Understand undirected graph edge addition

   Undirected edges must be added both ways: u to v and v to u.

2. Step 2: Analyze given code

   Code adds edge only from u to v, missing the reverse edge.

3. Final Answer:

   It adds edge only from u to v, missing v to u -> Option A

4. Quick Check:

   Undirected edges need both directions [OK]

Hint: Undirected edges require adding both directions [OK]

Common Mistakes:

• Adding edge only one way in undirected graph
• Confusing get() and put() usage
• Clearing adjacency list mistakenly

5. You have a weighted directed graph represented as:

Map>> graph = new HashMap<>();
graph.put(1, Arrays.asList(new Pair<>(2, 5), new Pair<>(3, 10)));
graph.put(2, Arrays.asList(new Pair<>(3, 2)));
graph.put(3, Arrays.asList());

How would you find the total weight of edges going out from vertex 1?

hard

A. Sum weights 10 + 2 = 12

B. Sum weights 5 + 10 = 15

C. Sum weights 5 + 2 = 7

D. Count edges: 2

Solution

1. Step 1: Identify edges from vertex 1

   Edges from 1 go to 2 with weight 5, and to 3 with weight 10.

2. Step 2: Sum the weights of these edges

   Total weight = 5 + 10 = 15.

3. Final Answer:

   Sum weights 5 + 10 = 15 -> Option B

4. Quick Check:

   Total outgoing weight from 1 = 15 [OK]

Hint: Add weights of all edges from the vertex [OK]

Common Mistakes:

• Adding weights from wrong vertices
• Counting edges instead of summing weights
• Mixing weights of different edges