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

Time Complexity: Dijkstra's Algorithm Single Source Shortest Path

O(V^2)

Understanding Time Complexity

We want to understand how the time needed by Dijkstra's algorithm changes as the graph size grows.

Specifically, how the number of nodes and edges affects the work done to find shortest paths.

Scenario Under Consideration

Analyze the time complexity of the following Dijkstra's algorithm code snippet.

void dijkstra(int graph[V][V], int src) {
    int dist[V];
    bool sptSet[V];
    for (int i = 0; i < V; i++) {
        dist[i] = INT_MAX;
        sptSet[i] = false;
    }
    dist[src] = 0;
    for (int count = 0; count < V - 1; count++) {
        int u = minDistance(dist, sptSet);
        sptSet[u] = true;
        for (int v = 0; v < V; v++) {
            if (!sptSet[v] && graph[u][v] > 0 && dist[u] != INT_MAX 
                && dist[u] + graph[u][v] < dist[v]) {
                dist[v] = dist[u] + graph[u][v];
            }
        }
    }
}
    

This code finds shortest paths from one node to all others using a simple array and adjacency matrix.

Identify Repeating Operations

Look at the loops and repeated steps:

• Primary operation: The outer loop runs once for each node (V times).
• Nested operation: Inside, it finds the minimum distance node (scans all V nodes).
• Inner loop: For each chosen node, it checks all neighbors (V nodes) to update distances.
• Dominant work: The nested loops scanning all nodes make the main cost.

How Execution Grows With Input

As the number of nodes (V) grows, the work grows roughly with the square of V.

Input Size (V)	Approx. Operations
10	About 100 (10 x 10) checks
100	About 10,000 (100 x 100) checks
1000	About 1,000,000 (1000 x 1000) checks

Pattern observation: Doubling nodes roughly quadruples the work because of nested loops.

Final Time Complexity

Time Complexity: O(V2)

This means the time needed grows roughly with the square of the number of nodes in the graph.

Common Mistake

[X] Wrong: "Dijkstra's algorithm always runs in linear time because it just visits each node once."

[OK] Correct: The algorithm repeatedly scans all nodes to find the closest unvisited node and updates neighbors, causing nested loops that increase time with the square of nodes.

Interview Connect

Understanding this time complexity helps you explain how your solution scales and why using better data structures can improve performance.

Self-Check

"What if we replace the array with a priority queue (min-heap) to find the minimum distance node? How would the time complexity change?"

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