Dijkstra's Algorithm Single Source Shortest Path in DSA Javascript - 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.

function dijkstra(graph, start) {
  const distances = {};
  const visited = new Set();
  const pq = new MinPriorityQueue();

  for (const node in graph) distances[node] = Infinity;
  distances[start] = 0;
  pq.enqueue(start, 0);

  while (!pq.isEmpty()) {
    const { element: current } = pq.dequeue();
    if (visited.has(current)) continue;
    visited.add(current);

    for (const neighbor in graph[current]) {
      const newDist = distances[current] + graph[current][neighbor];
      if (newDist < distances[neighbor]) {
        distances[neighbor] = newDist;
        pq.enqueue(neighbor, newDist);
      }
    }
  }
  return distances;
}
    

This code finds the shortest path from one node to all others using a priority queue and updates distances step by step.

Identify Repeating Operations

Look for loops and repeated steps that take most time.

• Primary operation: The while loop runs until all nodes are visited, and inside it, the for loop checks neighbors of the current node.
• How many times: The while loop runs up to V times (number of nodes), and the inner for loop runs for all edges connected to nodes, totaling E times (number of edges).

How Execution Grows With Input

As the graph grows, the number of nodes (V) and edges (E) increase, affecting the work done.

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

Pattern observation: The work grows roughly with the number of nodes plus edges, so bigger graphs mean more steps.

Final Time Complexity

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

This means the time grows with nodes and edges, multiplied by the cost of managing the priority queue, which depends on nodes.

Common Mistake

[X] Wrong: "Dijkstra's algorithm always runs in simple linear time O(V + E)."

[OK] Correct: Managing the priority queue adds a log factor, so the actual time depends on log V too, making it slower than just linear.

Interview Connect

Understanding this time complexity helps you explain how efficient your solution is and why certain data structures matter.

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 sort vertices in topological order

B. To find the maximum flow in a network

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 algorithm's goal

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

2. Step 2: Compare with other graph algorithms

   Other algorithms like max flow or cycle detection serve different purposes, so they do not match Dijkstra's goal.

3. Final Answer:

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

4. Quick Check:

   Shortest path from one source = To find the shortest path from a single source to all other vertices [OK]

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

Common Mistakes:

• Confusing Dijkstra with max flow 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 JavaScript?

easy

A. const dist = new Array(n).fill(Infinity); dist[source] = 0;

B. const dist = Array(n).fill(0); dist[source] = Infinity;

C. const dist = new Array(n).fill(null); dist[source] = 0;

D. const dist = new Array(n).fill(-1); dist[source] = 0;

Solution

1. Step 1: Understand distance initialization

   Distances to all nodes except the source should start as Infinity to represent unknown paths, and the source distance is 0.

2. Step 2: Check each option

   const dist = new Array(n).fill(Infinity); dist[source] = 0; correctly sets all distances to Infinity and source to 0. Others either set wrong initial values or invert source distance.

3. Final Answer:

   const dist = new Array(n).fill(Infinity); dist[source] = 0; -> Option A

4. Quick Check:

   Distance array starts with Infinity except source = const dist = new Array(n).fill(Infinity); dist[source] = 0; [OK]

Hint: Start distances as Infinity, source as 0 [OK]

Common Mistakes:

• Setting source distance to Infinity
• Using 0 for all distances initially
• Using null or -1 incorrectly

3. Consider the following JavaScript code snippet implementing part of Dijkstra's Algorithm. What will be the output of the distance array after running this code?

const graph = {
  0: [{node:1, weight:4}, {node:2, weight:1}],
  1: [{node:3, weight:1}],
  2: [{node:1, weight:2}, {node:3, weight:5}],
  3: []
};
const dist = [0, Infinity, Infinity, Infinity];
const visited = new Set();

function dijkstra() {
  while (visited.size < 4) {
    let u = -1;
    let minDist = Infinity;
    for (let i = 0; i < 4; i++) {
      if (!visited.has(i) && dist[i] < minDist) {
        minDist = dist[i];
        u = i;
      }
    }
    visited.add(u);
    for (const neighbor of graph[u]) {
      if (dist[u] + neighbor.weight < dist[neighbor.node]) {
        dist[neighbor.node] = dist[u] + neighbor.weight;
      }
    }
  }
}
dijkstra();
console.log(dist);

medium

A. [0, 4, 1, 6]

B. [0, 3, 1, 4]

C. [0, 2, 1, 3]

D. [0, 5, 1, 7]

Solution

1. Step 1: Trace initial distances and first iteration

   Start dist = [0, Infinity, Infinity, Infinity]. Pick node 0 (dist=0). Update neighbors: node 1 = 4, node 2 = 1.

2. Step 2: Next iterations and updates

   Pick node 2 (dist=1). Update node 1: min(4, 1+2=3) = 3; node 3: min(Infinity, 1+5=6) = 6. Pick node 1 (dist=3). Update node 3: min(6, 3+1=4) = 4. Pick node 3 (dist=4). No neighbors.

3. Final Answer:

   [0, 3, 1, 4] -> Option B

4. Quick Check:

   Distances after updates = [0, 3, 1, 4] [OK]

Hint: Pick unvisited node with smallest dist, update neighbors [OK]

Common Mistakes:

• Not updating distances correctly
• Choosing wrong next node
• Ignoring neighbor updates

4. The following code snippet is intended to implement Dijkstra's Algorithm but contains a bug. What is the bug?

const dist = new Array(n).fill(Infinity);
dist[source] = 0;
const visited = new Set();

while (visited.size < n) {
  let u = -1;
  let minDist = Infinity;
  for (let i = 0; i < n; i++) {
    if (dist[i] < minDist) {
      minDist = dist[i];
      u = i;
    }
  }
  visited.add(u);
  for (const neighbor of graph[u]) {
    if (dist[u] + neighbor.weight < dist[neighbor.node]) {
      dist[neighbor.node] = dist[u] + neighbor.weight;
    }
  }
}

medium

A. The visited set is not updated inside the loop

B. The distance array is not initialized correctly

C. The code does not check if node i is already visited before selecting u

D. The neighbor weights are not added to dist[u]

Solution

1. Step 1: Analyze node selection loop

   The loop selects node u with smallest dist[i] but does not check if i is already visited, so it may pick visited nodes again.

2. Step 2: Understand impact of missing visited check

   Without skipping visited nodes, the algorithm can loop incorrectly or update distances wrongly, breaking correctness.

3. Final Answer:

   The code does not check if node i is already visited before selecting u -> Option C

4. Quick Check:

   Must skip visited nodes when picking u = The code does not check if node i is already visited before selecting u [OK]

Hint: Always skip visited nodes when choosing next node [OK]

Common Mistakes:

• Forgetting to check visited nodes in selection
• Incorrect distance initialization
• Not updating visited set

5. You have a weighted graph with 5 nodes (0 to 4). The edges are:

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

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

hard

A. 6

B. 7

C. 8

D. 5

Solution

1. Step 1: Trace shortest paths from node 0

   Start dist: 0=0, others=Infinity. From 0: dist[1]=2, dist[2]=4.

2. Step 2: Update distances via neighbors

   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.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Shortest distance to node 4 = 6 [OK]

Hint: Check all paths, pick minimum sum of weights [OK]

Common Mistakes:

• Ignoring shorter path through node 2
• Adding wrong edge weights
• Not updating distances properly