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

Time Complexity: Graph Terminology Vertices Edges Directed Undirected Weighted

O(E)

Understanding Time Complexity

Graphs are made of points and connections. We want to understand how the work grows when we look at all points and connections.

How does the number of points and connections affect the time to explore or use the graph?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Simple graph traversal example
void printEdges(int vertices, int edges[][2], int edgeCount) {
    for (int i = 0; i < edgeCount; i++) {
        int u = edges[i][0];
        int v = edges[i][1];
        printf("Edge from %d to %d\n", u, v);
    }
}
    

This code prints all edges in a graph represented by an edge list.

Identify Repeating Operations

• Primary operation: Loop over all edges to print each connection.
• How many times: Exactly once for each edge in the graph.

How Execution Grows With Input

As the number of edges grows, the work grows directly with it.

Input Size (edges)	Approx. Operations
10	10 print operations
100	100 print operations
1000	1000 print operations

Pattern observation: The work grows in a straight line with the number of edges.

Final Time Complexity

Time Complexity: O(E)

This means the time to process the graph grows directly with the number of edges.

Common Mistake

[X] Wrong: "The time depends only on the number of vertices, not edges."

[OK] Correct: Edges represent connections, and many operations must look at each edge, so ignoring edges misses the real work.

Interview Connect

Understanding how vertices and edges affect time helps you explain graph problems clearly and choose the right approach confidently.

Self-Check

"What if the graph is directed versus undirected? How would that affect the time complexity of traversing all edges?"

Practice

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

easy

A. A point or node representing an entity in the graph

B. A connection between two nodes

C. A path that starts and ends at the same node

D. A graph with no edges

Solution

1. Step 1: Understand the definition of vertex

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

2. Step 2: Differentiate vertex from edges and other terms

   Edges connect vertices; paths and graph types are different concepts.

3. Final Answer:

   A point or node representing an entity in the graph -> Option A

4. Quick Check:

   Vertex = node/entity [OK]

Hint: Vertices are the points or nodes in a graph [OK]

Common Mistakes:

• Confusing vertices with edges
• Thinking vertices are paths
• Mixing vertex with graph types

2. Which of the following is the correct way to represent a weighted edge from vertex u to vertex v with weight 5 in C struct?

easy

A. struct Edge { int u, v; }; struct Weight { int w; };

B. struct Edge { int u; int v; }; int weight = 5;

C. struct Edge { int u; int v; int weight; };

D. struct Edge { int u; int v; };

Solution

1. Step 1: Identify the correct struct fields for weighted edge

   A weighted edge needs source vertex, destination vertex, and weight as fields.

2. Step 2: Check struct syntax correctness

   struct Edge { int u; int v; int weight; }; correctly declares all three fields in one struct.

3. Final Answer:

   struct Edge { int u; int v; int weight; }; -> Option C

4. Quick Check:

   Weighted edge struct = u, v, weight [OK]

Hint: Weighted edges need u, v, and weight fields in one struct [OK]

Common Mistakes:

• Separating weight outside the edge struct
• Incorrect field order causing confusion
• Missing weight field

3. What is the output of the following code snippet that prints edges of a directed weighted graph?

typedef struct { int u; int v; int weight; } Edge;
Edge edges[] = { {0, 1, 10}, {1, 2, 20}, {2, 0, 30} };
int n = 3;
for (int i = 0; i < n; i++) {
    printf("%d -> %d (weight %d)\n", edges[i].u, edges[i].v, edges[i].weight);
}

medium

A. 0 -> 1 (weight 10)
1 -> 2 (weight 30)
2 -> 0 (weight 20)

B. 0 -> 1 (weight 10)
1 -> 2 (weight 20)
2 -> 0 (weight 30)

C. 0 -> 1 (weight 30)
1 -> 2 (weight 20)
2 -> 0 (weight 10)

D. 1 -> 0 (weight 10)
2 -> 1 (weight 20)
0 -> 2 (weight 30)

Solution

1. Step 1: Read the edges array values

   Edges are: (0,1,10), (1,2,20), (2,0,30).

2. Step 2: Trace the loop printing each edge

   Loop prints each edge in order with format "u -> v (weight w)".

3. Final Answer:

   0 -> 1 (weight 10) 1 -> 2 (weight 20) 2 -> 0 (weight 30) -> Option B

4. Quick Check:

   Print edges as stored = 0 -> 1 (weight 10) 1 -> 2 (weight 20) 2 -> 0 (weight 30) [OK]

Hint: Print edges in array order with their weights [OK]

Common Mistakes:

• Reversing u and v in print
• Mixing up weights between edges
• Printing edges in wrong order

4. Identify the error in this code snippet that tries to create an undirected graph edge list:

typedef struct { int u; int v; } Edge;
Edge edges[2];
edges[0].u = 0; edges[0].v = 1;
edges[1].u = 1; edges[1].v = 0;
// Print edges
for (int i = 0; i <= 2; i++) {
    printf("Edge from %d to %d\n", edges[i].u, edges[i].v);
}

medium

A. Edge direction is incorrect for undirected graph

B. Edges should have weight field

C. Edges array size should be 1, not 2

D. Loop condition should be i < 2, not i <= 2

Solution

1. Step 1: Check array size and loop bounds

   Edges array has size 2, valid indices are 0 and 1.

2. Step 2: Identify loop error

   Loop runs from i=0 to i=2 inclusive, which accesses edges[2] (out of bounds).

3. Final Answer:

   Loop condition should be i < 2, not i <= 2 -> Option D

4. Quick Check:

   Array size 2 means max index 1 [OK]

Hint: Loop index must be less than array size [OK]

Common Mistakes:

• Using <= instead of < in loop
• Assuming edges need weight for undirected
• Confusing edge direction in undirected graphs

5. You have a weighted undirected graph with vertices 0, 1, 2 and edges:
(0,1,4), (1,2,6), (0,2,5)
How many edges will be stored if you represent this graph using an edge list storing each edge only once?

medium

A. 3

B. 6

C. 2

D. 1

Solution

1. Step 1: Understand undirected edge storage

   In undirected graphs, each edge connects two vertices without direction, so store only one entry per edge.

2. Step 2: Count unique edges given

   Edges are (0,1), (1,2), (0,2) -- total 3 unique edges.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Undirected edges stored once = 3 [OK]

Hint: Undirected edges stored once, count unique pairs [OK]

Common Mistakes:

• Counting each edge twice
• Confusing directed with undirected storage
• Ignoring weights when counting edges