Dijkstra's Algorithm Single Source Shortest Path in DSA C# - Time & Space Complexity

Time Complexity: Dijkstra's Algorithm Single Source Shortest Path

O((V + E) log V)

Understanding Time Complexity

We want to understand how the time needed to find shortest paths grows as the graph gets bigger.

How does the algorithm's work increase when there are more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Simplified Dijkstra's Algorithm using adjacency list and priority queue
int V = graph.Length; // number of vertices
int[] dist = new int[V];
for (int i = 0; i < V; i++) dist[i] = int.MaxValue;
dist[source] = 0;
var pq = new PriorityQueue<int, int>();
pq.Enqueue(source, 0);

while (pq.Count > 0) {
    var u = pq.Dequeue();
    foreach (var (v, w) in graph[u]) {
        if (dist[u] + w < dist[v]) {
            dist[v] = dist[u] + w;
            pq.Enqueue(v, dist[v]);
        }
    }
}
    

This code finds the shortest distance from one node to all others using a priority queue and adjacency list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop runs as long as there are nodes in the priority queue.
• How many times: Each edge is checked at most once when relaxing distances, so roughly once per edge.
• Inside the while loop, the foreach loop goes through all neighbors of the current node.
• Priority queue operations (enqueue and dequeue) happen for nodes when their distances improve.

How Execution Grows With Input

As the number of nodes (V) and edges (E) grow, the operations increase based on how many edges and nodes are processed.

Input Size (V, E)	Approx. Operations
10 nodes, 20 edges	About 20 edge checks and 10 node queue operations
100 nodes, 500 edges	About 500 edge checks and 100 node queue operations
1000 nodes, 5000 edges	About 5000 edge checks and 1000 node queue operations

Pattern observation: The work grows roughly with the number of edges and nodes combined, not just nodes alone.

Final Time Complexity

Time Complexity: O((V + E) log V)

This means the time grows with the number of nodes plus edges, multiplied by the cost of priority queue operations.

Common Mistake

[X] Wrong: "Dijkstra's algorithm always runs in O(V²) time regardless of data structures."

[OK] Correct: Using a priority queue and adjacency list reduces the time to O((V + E) log V), which is much faster especially for sparse graphs.

Interview Connect

Understanding this time complexity helps you explain how your solution scales and why choosing the right data structures matters.

Self-Check

"What if we used a simple array instead of a priority queue? How would the time complexity change?"

Practice

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

easy

A. To find the minimum spanning tree of a graph

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

C. To detect cycles in a graph

D. To sort the vertices in topological order

Solution

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

   Dijkstra's Algorithm is designed to find the shortest paths from one starting point (source) to all other points in a weighted graph.

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 a DAG. These are different goals.

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Confusing shortest path with minimum spanning tree
• Thinking it detects cycles
• Mixing it with topological sorting

2. Which of the following C# code snippets correctly initializes the distance array for Dijkstra's Algorithm for a graph with 5 vertices?

easy

A. int[] dist = new int[5]; for(int i = 0; i < 5; i++) dist[i] = -1;

B. int[] dist = new int[5] {0, 0, 0, 0, 0};

C. int[] dist = new int[5]; dist.Fill(0);

D. int[] dist = new int[5]; for(int i = 0; i < 5; i++) dist[i] = int.MaxValue;

Solution

1. Step 1: Understand initialization for Dijkstra's distances

   Distances are set to a very large number (infinity) initially, except the source vertex which is set to 0 later.

2. Step 2: Check each option

   int[] dist = new int[5]; for(int i = 0; i < 5; i++) dist[i] = int.MaxValue; sets all distances to int.MaxValue, which is correct. int[] dist = new int[5] {0, 0, 0, 0, 0}; sets all to 0, which is wrong. int[] dist = new int[5]; dist.Fill(0); uses a non-existent Fill method. int[] dist = new int[5]; for(int i = 0; i < 5; i++) dist[i] = -1; sets distances to -1, which is invalid for distances.

3. Final Answer:

   int[] dist = new int[5]; for(int i = 0; i < 5; i++) dist[i] = int.MaxValue; -> Option D

4. Quick Check:

   Initialize distances to max value [OK]

Hint: Use int.MaxValue to represent infinity distances [OK]

Common Mistakes:

• Setting all distances to zero initially
• Using negative values for distances
• Using non-existent array methods

3. Consider the following C# code snippet implementing Dijkstra's Algorithm on a graph with 4 vertices (0 to 3). What is the shortest distance from vertex 0 to vertex 3?

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

int V = 4;
int[] dist = new int[V];
bool[] sptSet = new bool[V];

for (int i = 0; i < V; i++) dist[i] = int.MaxValue;
dist[0] = 0;

for (int count = 0; count < V - 1; count++) {
  int u = -1;
  int minDist = int.MaxValue;
  for (int v = 0; v < V; v++) {
    if (!sptSet[v] && dist[v] < minDist) {
      minDist = dist[v];
      u = v;
    }
  }
  sptSet[u] = true;

  for (int v = 0; v < V; v++) {
    if (!sptSet[v] && graph[u, v] != 0 && dist[u] != int.MaxValue && dist[u] + graph[u, v] < dist[v]) {
      dist[v] = dist[u] + graph[u, v];
    }
  }
}

Console.WriteLine(dist[3]);

medium

A. 6

B. 7

C. 9

D. 10

Solution

1. Step 1: Trace the algorithm's distance updates

   Start dist = [0, ∞, ∞, ∞]. - Pick vertex 0: update dist[1] = 1, dist[2] = 4. - Pick vertex 1 (dist=1): update dist[2] = min(4, 1+2=3) = 3, dist[3] = 1+6=7. - Pick vertex 2 (dist=3): update dist[3] = min(7, 3+3=6) = 6. - Pick vertex 3 (dist=6): done.

2. Step 2: Final shortest distance to vertex 3

   After all updates, dist[3] = 6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Shortest distance to vertex 3 = 6 [OK]

Hint: Update distances only if new path is shorter [OK]

Common Mistakes:

• Not updating distances when a shorter path is found
• Confusing vertex indices
• Ignoring zero-weight edges as no edge

4. The following C# code snippet for Dijkstra's Algorithm has a bug. What is the error?

int V = 3;
int[] dist = new int[V];
bool[] visited = new bool[V];

for (int i = 0; i < V; i++)
  dist[i] = int.MaxValue;
dist[0] = 0;

for (int count = 0; count < V; count++) {
  int u = 0;
  for (int v = 0; v < V; v++) {
    if (!visited[v] && dist[v] < dist[u])
      u = v;
  }
  visited[u] = true;
  // Update distances here (omitted)
}

medium

A. Variable u is initialized to 0 causing incorrect minimum selection

B. dist array is not initialized properly

C. visited array is not updated

D. Loop runs one time too many

Solution

1. Step 1: Analyze initialization of u in inner loop

   Variable u is set to 0 before checking for minimum dist among unvisited vertices. If dist[0] is int.MaxValue, comparison fails and minimum may not be found correctly.

2. Step 2: Correct approach for minimum selection

   u should be initialized to -1 or a sentinel, and updated only when a better candidate is found to avoid wrong minimum selection.

3. Final Answer:

   Variable u is initialized to 0 causing incorrect minimum selection -> Option A

4. Quick Check:

   Initialize u properly to find minimum dist [OK]

Hint: Initialize min index to -1 before searching minimum [OK]

Common Mistakes:

• Initializing min index to 0 without checking if visited
• Not updating visited array
• Incorrect loop count

5. You have a weighted graph with 5 vertices and edges:

0 -> 1 (2), 0 -> 2 (4),
1 -> 2 (1), 1 -> 3 (7),
2 -> 4 (3), 3 -> 4 (1)

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

hard

A. 7

B. 8

C. 6

D. 9

Solution

1. Step 1: Trace shortest paths from vertex 0

   Start dist = [0, ∞, ∞, ∞, ∞]. - From 0: dist[1] = 2, dist[2] = 4. - From 1: dist[2] = min(4, 2+1=3) = 3, dist[3] = 2+7=9. - From 2: dist[4] = 3+3=6. - From 3: dist[4] = min(6, 9+1=10) = 6.

2. Step 2: Final shortest distance to vertex 4

   The shortest distance to vertex 4 is 6 via path 0->1->2->4.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Shortest path distance to 4 = 6 [OK]

Hint: Check all paths and pick minimum sum of edges [OK]

Common Mistakes:

• Ignoring shorter indirect paths
• Adding edge weights incorrectly
• Assuming direct edge is always shortest