Adjacency List vs Matrix When to Choose Which in DSA C - Build Both Approaches

Adjacency List vs Matrix When to Choose Which

📖 Scenario: Imagine you are helping a city planner who wants to represent roads between different neighborhoods. Some neighborhoods are connected directly by roads, and some are not. You want to help the planner choose the best way to store this road information in a computer.

🎯 Goal: You will create two ways to represent the roads between neighborhoods: one using an adjacency matrix and one using an adjacency list. Then, you will decide which way is better depending on how many roads there are.

📋 What You'll Learn

Create an adjacency matrix for 4 neighborhoods with given roads

Create an adjacency list for the same 4 neighborhoods with the same roads

Count the number of roads using the adjacency matrix

Print the adjacency matrix and adjacency list to compare

💡 Why This Matters

🌍 Real World

City planners, network engineers, and social network analysts use adjacency lists and matrices to represent connections between points or people.

💼 Career

Understanding these representations helps in optimizing storage and speed when working with graphs in software development, data analysis, and network design.

1

Create an adjacency matrix for 4 neighborhoods

Create a 4x4 integer array called adjMatrix initialized with zeros. Then set the following roads (edges) to 1: from neighborhood 0 to 1, from 0 to 2, from 1 to 2, and from 2 to 3.

DSA C

# Create a 4x4 adjacency matrix with zeros
# Set roads as described

Hint

Think of adjMatrix as a table where rows and columns represent neighborhoods. A 1 means a road exists.

2

Create an adjacency list for the same neighborhoods

Create an array of integer pointers called adjList with size 4. Also create an integer array called adjListSizes with size 4 to store the number of roads for each neighborhood. Initialize the adjacency list with the following roads: neighborhood 0 connects to 1 and 2, neighborhood 1 connects to 2, and neighborhood 2 connects to 3.

DSA C

int adjMatrix[4][4] = {0};
adjMatrix[0][1] = 1;
adjMatrix[0][2] = 1;
adjMatrix[1][2] = 1;
adjMatrix[2][3] = 1;

// Create adjacency list arrays
// Your code here

Hint

Each neighborhood has a list of neighbors it connects to. Use arrays for each list and store their sizes.

3

Count the number of roads using the adjacency matrix

Create an integer variable called roadCount and set it to 0. Use a nested for loop with variables i and j to go through adjMatrix. Increase roadCount by 1 every time you find a 1 in adjMatrix[i][j].

DSA C

int adjMatrix[4][4] = {0};
adjMatrix[0][1] = 1;
adjMatrix[0][2] = 1;
adjMatrix[1][2] = 1;
adjMatrix[2][3] = 1;

int neighbors0[] = {1, 2};
int neighbors1[] = {2};
int neighbors2[] = {3};
int neighbors3[] = {};

int* adjList[4] = {neighbors0, neighbors1, neighbors2, neighbors3};
int adjListSizes[4] = {2, 1, 1, 0};

// Count roads using adjMatrix
// Your code here

Hint

Look at each cell in the matrix. If it is 1, it means a road exists, so add to the count.

4

Print the adjacency matrix and adjacency list

Use printf to print the adjacency matrix in a 4x4 grid format. Then print the adjacency list by showing each neighborhood and its connected neighbors. Finally, print the total number of roads stored in roadCount.

DSA C

int adjMatrix[4][4] = {0};
adjMatrix[0][1] = 1;
adjMatrix[0][2] = 1;
adjMatrix[1][2] = 1;
adjMatrix[2][3] = 1;

int neighbors0[] = {1, 2};
int neighbors1[] = {2};
int neighbors2[] = {3};
int neighbors3[] = {};

int* adjList[4] = {neighbors0, neighbors1, neighbors2, neighbors3};
int adjListSizes[4] = {2, 1, 1, 0};

int roadCount = 0;
for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
        if (adjMatrix[i][j] == 1) {
            roadCount++;
        }
    }
}

// Print adjacency matrix and list
// Your code here

Hint

Print the matrix row by row, then print each neighborhood's list of neighbors. Finally, print the road count.

Practice

1. Which data structure is generally better for representing a graph with very few edges compared to the number of vertices?

easy

A. Adjacency List

B. Adjacency Matrix

C. Edge List

D. Incidence Matrix

Solution

1. Step 1: Understand graph density

   A graph with very few edges compared to vertices is called a sparse graph.

2. Step 2: Choose efficient representation for sparse graphs

   Adjacency lists store only existing edges, saving space for sparse graphs, unlike adjacency matrices which use space for all possible edges.

3. Final Answer:

   Adjacency List -> Option A

4. Quick Check:

   Sparse graph = Adjacency List [OK]

Hint: Sparse graph? Use adjacency list for space efficiency [OK]

Common Mistakes:

• Choosing adjacency matrix for sparse graphs wastes space
• Confusing edge list with adjacency matrix
• Thinking adjacency matrix is always better

2. Which of the following is the correct way to declare an adjacency matrix for a graph with 4 vertices in C?

easy

A. int graph[4][4][4];

B. int graph[4];

C. int graph[][] = new int[4][4];

D. int graph[4][4];

Solution

1. Step 1: Recall C array declaration syntax

   In C, a 2D array for 4 vertices is declared as int graph[4][4];

2. Step 2: Check options for correctness

   int graph[4][4]; matches correct syntax; int graph[4]; is 1D array; int graph[][] = new int[4][4]; uses Java syntax; int graph[4][4][4]; is 3D array.

3. Final Answer:

   int graph[4][4]; -> Option D

4. Quick Check:

   C 2D array syntax = int graph[4][4]; [OK]

Hint: C 2D arrays use int name[rows][cols]; syntax [OK]

Common Mistakes:

• Using Java or other language syntax in C
• Declaring wrong array dimensions
• Confusing 1D and 2D arrays

3. Given the following adjacency matrix for a graph with 3 vertices:

int graph[3][3] = {
  {0, 1, 0},
  {1, 0, 1},
  {0, 1, 0}
};

What is the output of the code snippet that prints all neighbors of vertex 1?

for (int i = 0; i < 3; i++) {
  if (graph[1][i] == 1) {
    printf("%d ", i);
  }
}

medium

A. 0 2

B. 1 2

C. 0 1

D. 1 0 2

Solution

1. Step 1: Analyze adjacency matrix row for vertex 1

   Row 1 is {1, 0, 1}, meaning edges to vertices 0 and 2.

2. Step 2: Trace loop printing neighbors

   Loop checks graph[1][i] == 1; prints i when true. i=0 and i=2 satisfy condition.

3. Final Answer:

   0 2 -> Option A

4. Quick Check:

   Neighbors of vertex 1 = 0 and 2 [OK]

Hint: Check matrix row for 1s to find neighbors [OK]

Common Mistakes:

• Confusing row and column indices
• Printing vertex itself as neighbor
• Misreading matrix values

4. The following code attempts to create an adjacency list for a graph with 3 vertices but has an error:

int* graph[3];
for (int i = 0; i < 3; i++) {
  graph[i] = malloc(sizeof(int));
  graph[i][0] = i;
}

What is the main issue?

medium

A. Array size should be 4 instead of 3

B. Using malloc instead of calloc

C. Memory allocated is too small for multiple neighbors

D. graph should be declared as int graph[3][3]

Solution

1. Step 1: Check memory allocation size

   malloc(sizeof(int)) allocates space for only one int, but adjacency list needs space for multiple neighbors.

2. Step 2: Understand adjacency list needs dynamic size

   Allocating only one int is insufficient; should allocate enough space or use dynamic structures like linked lists.

3. Final Answer:

   Memory allocated is too small for multiple neighbors -> Option C

4. Quick Check:

   Insufficient malloc size = Error [OK]

Hint: Allocate enough memory for all neighbors in adjacency list [OK]

Common Mistakes:

• Confusing malloc and calloc
• Wrong array declaration type
• Assuming fixed size adjacency list

5. You have a graph with 1000 vertices and about 10,000 edges. You want to quickly check if an edge exists between any two vertices. Which representation is best and why?

hard

A. Adjacency List, because it uses less memory for sparse graphs

B. Adjacency Matrix, because it allows O(1) edge existence check

C. Edge List, because it is easiest to implement

D. Incidence Matrix, because it stores edges explicitly

Solution

1. Step 1: Analyze graph size and density

   1000 vertices and 10,000 edges means the graph is moderately dense (max edges ~499,500).

2. Step 2: Consider edge existence check speed

   Adjacency matrix allows checking edge existence in O(1) time by direct index access.

3. Step 3: Compare memory vs speed tradeoff

   Though adjacency matrix uses more memory, fast edge check is priority here.

4. Final Answer:

   Adjacency Matrix, because it allows O(1) edge existence check -> Option B

5. Quick Check:

   Fast edge check = Adjacency Matrix [OK]

Hint: Need fast edge check? Use adjacency matrix despite memory [OK]

Common Mistakes:

• Choosing adjacency list only for memory without speed need
• Ignoring graph density impact
• Confusing edge list with adjacency matrix