Dijkstra's Algorithm Single Source Shortest Path in DSA C - Build from Scratch

Dijkstra's Algorithm Single Source Shortest Path

📖 Scenario: You are working as a city planner. You have a map of intersections connected by roads with distances. You want to find the shortest path from a starting intersection to all other intersections.

🎯 Goal: Build a program that uses Dijkstra's algorithm to find the shortest distance from a given start intersection to every other intersection in the city.

📋 What You'll Learn

Create a graph using an adjacency matrix with exact distances

Set a starting intersection index

Implement Dijkstra's algorithm to find shortest paths

Print the shortest distances from the start to all intersections

💡 Why This Matters

🌍 Real World

City planners and GPS systems use shortest path algorithms like Dijkstra's to find quickest routes between locations.

💼 Career

Understanding Dijkstra's algorithm is essential for software engineers working on navigation, mapping, and network routing applications.

1

Create the graph as an adjacency matrix

Create a 2D integer array called graph with 4 rows and 4 columns representing the distances between intersections. Use these exact values: row 0: 0, 5, 0, 10; row 1: 5, 0, 3, 0; row 2: 0, 3, 0, 1; row 3: 10, 0, 1, 0.

DSA C

# Create the adjacency matrix graph with the given distances
// Your code here

Hint

Use a 2D array with the exact distances for each row as shown.

2

Set the starting intersection index

Create an integer variable called start and set it to 0 to represent the starting intersection.

DSA C

#include 

int graph[4][4] = {
    {0, 5, 0, 10},
    {5, 0, 3, 0},
    {0, 3, 0, 1},
    {10, 0, 1, 0}
};

int main() {
    // Set the start intersection index
    // Your code here
    return 0;
}

Hint

Declare an integer variable named start and assign it the value 0.

3

Implement Dijkstra's algorithm to find shortest paths

Write the core logic of Dijkstra's algorithm inside main. Use arrays dist[4] to store shortest distances and visited[4] to track visited intersections. Initialize dist with a large number (e.g., 10000) except for dist[start] which is 0. Loop 4 times to pick the unvisited intersection with the smallest distance, mark it visited, and update distances to neighbors if shorter paths are found.

DSA C

#include 

int graph[4][4] = {
    {0, 5, 0, 10},
    {5, 0, 3, 0},
    {0, 3, 0, 1},
    {10, 0, 1, 0}
};

int main() {
    int start = 0;
    int dist[4];
    int visited[4] = {0};
    int i, j;

    // Initialize dist array
    // Your code here

    // Dijkstra's algorithm main loop
    // Your code here

    return 0;
}

Hint

Initialize distances to a large number except the start. Then repeat picking the closest unvisited node and update neighbors.

4

Print the shortest distances from the start

Add a for loop to print the shortest distances from start to each intersection using printf. Print exactly this format for each intersection i: Distance from 0 to i is dist[i] with a newline.

DSA C

Hint

Use a for loop and printf to show each distance in the exact format.

Practice

1. What is the main purpose of Dijkstra's Algorithm in graph theory?

easy

A. To sort the vertices of a graph topologically

B. To find the minimum spanning tree of a graph

C. To detect cycles in a graph

D. To find the shortest path from a single source to all other vertices

Solution

1. Step 1: Understand the goal of Dijkstra's Algorithm

   Dijkstra's Algorithm is designed to find the shortest distance from one starting point (source) to every other point in a graph with non-negative edge weights.

2. Step 2: Compare with other graph algorithms

   Minimum spanning tree algorithms find a tree connecting all vertices with minimum total edge weight, cycle detection finds loops, and topological sort orders vertices in directed acyclic graphs. These are different goals.

3. Final Answer:

   To find the shortest path from a single source to all other vertices -> Option D

4. Quick Check:

   Dijkstra's Algorithm = shortest paths from source [OK]

Hint: Dijkstra finds shortest paths from one start point [OK]

Common Mistakes:

• Confusing with minimum spanning tree algorithms
• Thinking it detects cycles
• Assuming it sorts vertices

2. Which of the following is the correct way to initialize the distance array in Dijkstra's Algorithm in C?

easy

A. int dist[V]; for(int i=0; i

B. int dist[V] = {INT_MAX};

C. int dist[V] = {0};

D. int dist[V] = NULL;

Solution

1. Step 1: Understand distance initialization

   Distances should start as infinity (or a very large number) except the source vertex, to represent unknown shortest paths.

2. Step 2: Check each option

   int dist[V] = {0}; initializes all distances to 0, which is incorrect. int dist[V]; for(int i=0; i

3. Final Answer:

   int dist[V]; for(int i=0; i -> Option A

4. Quick Check:

   Initialize distances to INT_MAX with loop [OK]

Hint: Use loop to set all distances to INT_MAX [OK]

Common Mistakes:

• Using {INT_MAX} initializes only first element
• Setting all distances to 0
• Assigning NULL to int array

3. What will be the output distances array after running Dijkstra's Algorithm from source vertex 0 on the following graph represented by adjacency matrix?

int graph[4][4] = {
  {0, 1, 4, 0},
  {1, 0, 2, 6},
  {4, 2, 0, 3},
  {0, 6, 3, 0}
};

medium

A. [0, 1, 4, 7]

B. [0, 1, 3, 5]

C. [0, 1, 3, 6]

D. [0, 2, 4, 6]

Solution

1. Step 1: Initialize distances and start from vertex 0

   Distances: dist[0]=0, dist[1]=∞, dist[2]=∞, dist[3]=∞. Update neighbors of 0: dist[1]=1, dist[2]=4.

2. Step 2: Pick vertex with smallest tentative distance and update neighbors

   Pick vertex 1 (dist=1). Update dist[2] = min(4, 1+2=3) = 3, dist[3] = min(∞, 1+6=7) = 7. Next pick vertex 2 (dist=3). Update dist[3] = min(7, 3+3=6) = 6. Finally pick vertex 3 (dist=6).

3. Final Answer:

   [0, 1, 3, 6] -> Option C

4. Quick Check:

   Distances after relaxation = [0,1,3,6] [OK]

Hint: Relax edges stepwise, pick smallest distance vertex next [OK]

Common Mistakes:

• Not updating distances after relaxation
• Choosing wrong next vertex
• Ignoring shorter paths through intermediate vertices

4. Identify the error in the following snippet of Dijkstra's Algorithm in C:

for (int count = 0; count < V-1; count++) {
  int u = minDistance(dist, sptSet);
  sptSet[u] = 1;
  for (int v = 0; v < V; v++) {
    if (!sptSet[v] && graph[u][v] && dist[u] != INT_MAX && dist[u] + graph[u][v] < dist[v]) {
      dist[v] = dist[u] + graph[u][v];
    }
  }
}

medium

A. The loop should run from 1 to V instead of 0 to V-1

B. The condition 'graph[u][v]' should be 'graph[u][v] != 0' to check edge existence

C. The sptSet array should be initialized with 1 instead of 0

D. The dist[u] check should be 'dist[u] == INT_MAX' instead of '!='

Solution

1. Step 1: Analyze edge existence condition

   The condition 'graph[u][v]' relies on implicit truthiness of non-zero values. In C, this works but is less clear. Using 'graph[u][v] != 0' explicitly checks for edge presence and improves readability.

2. Step 2: Check other conditions

   Loop range from 0 to V-1 is correct for V vertices. sptSet should start with 0 (false). The dist[u] check must be '!= INT_MAX' to ensure u is reachable before relaxing edges.

3. Final Answer:

   The condition 'graph[u][v]' should be 'graph[u][v] != 0' to check edge existence -> Option B

4. Quick Check:

   Explicit edge check improves clarity [OK]

Hint: Use explicit '!= 0' to check edges in adjacency matrix [OK]

Common Mistakes:

• Assuming loop range is wrong
• Initializing sptSet incorrectly
• Reversing dist[u] check logic

5. Given a weighted graph with vertices 0 to 5 and edges:

0-1: 7, 0-2: 9, 0-5: 14, 1-2: 10, 1-3: 15, 2-3: 11, 2-5: 2, 3-4: 6, 4-5: 9

Using Dijkstra's Algorithm from source 0, what is the shortest distance to vertex 4?

hard

A. 20

B. 26

C. 21

D. 19

Solution

1. Step 1: Initialize distances and start from vertex 0

   dist[0]=0, others = ∞. Update neighbors: dist[1]=7, dist[2]=9, dist[5]=14.

2. Step 2: Pick vertex with smallest dist and update neighbors

   Pick 1 (7): update dist[3] = min(∞, 7+15=22) = 22. Pick 2 (9): update dist[3] = min(22, 9+11=20) = 20, dist[5] = min(14, 9+2=11) = 11. Pick 5 (11): update dist[4] = min(∞, 11+9=20) = 20. Pick 3 (20): update dist[4] = min(20, 20+6=26) = 20. Pick 4 (20).

3. Final Answer:

   20 -> Option A

4. Quick Check:

   Shortest path to 4 = 20 [OK]

Hint: Relax edges carefully, update shortest distances stepwise [OK]

Common Mistakes:

• Not updating dist[5] to smaller value
• Ignoring shorter path through vertex 2
• Adding edges incorrectly