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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the distance to the source node as zero.

DSA Javascript

distances[source] = [1];

Drag options to blanks, or click blank then click option'

A-1

BInfinity

C0

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Setting the distance to Infinity instead of zero.

Using null or negative values for initial distance.

The distance from the source node to itself is always zero.

2fill in blank

medium

Complete the code to pick the unvisited node with the smallest distance.

DSA Javascript

let currentNode = unvisitedNodes.reduce((minNode, node) => distances[node] < distances[minNode] ? node : minNode, [1]);

Drag options to blanks, or click blank then click option'

Aundefined

BunvisitedNodes[0]

Cnull

DInfinity

Attempts:

3 left

💡 Hint

Common Mistakes

Using source, which may not be unvisited after the first iteration.

Using null or undefined which cause errors in comparison.

Using Infinity which is not a node.

We use the first node in unvisitedNodes as the initial value to correctly find the minimum among unvisited nodes.

3fill in blank

hard

Fix the error in updating the distance if a shorter path is found.

DSA Javascript

if (distances[currentNode] + graph[currentNode][neighbor] [1] distances[neighbor]) {
  distances[neighbor] = distances[currentNode] + graph[currentNode][neighbor];
  previous[neighbor] = currentNode;
}

Drag options to blanks, or click blank then click option'

A!=

B==

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using greater than operator which updates on longer paths.

Using equality or inequality which do not check for shorter paths.

We update the distance only if the new path is shorter (less than current known distance).

4fill in blank

hard

Fill both blanks to correctly initialize distances and previous nodes for all nodes.

DSA Javascript

for (const node of nodes) {
  distances[node] = [1];
  previous[node] = [2];
}

Drag options to blanks, or click blank then click option'

AInfinity

Bnull

C0

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Setting distances to zero which is only for source node.

Using undefined instead of null for previous nodes.

Distances start as Infinity because we don't know paths yet. Previous nodes start as null because no paths are known.

5fill in blank

hard

Fill all three blanks to build the shortest path by backtracking from the target node.

DSA Javascript

const path = [];
let current = [1];
while (current !== [2]) {
  path.unshift(current);
  current = [3][current];
}

Drag options to blanks, or click blank then click option'

Atarget

Bnull

Cprevious

Dsource

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from source instead of target.

Checking for undefined instead of null.

Using distances instead of previous to backtrack.

We start from the target node, backtrack until previous is null, using the previous map to find the path.

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