Bellman Ford Algorithm Negative Weights in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the distance array with Infinity for all vertices except the source.

DSA Typescript

const distances: number[] = new Array(vertices).fill([1]);
distances[source] = 0;

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

AInfinity

B0

C-Infinity

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Setting all distances to 0, which incorrectly assumes all vertices are reachable immediately.

Using null instead of Infinity, which can cause comparison errors.

We initialize all distances to Infinity to represent that they are initially unreachable, except the source which is set to 0.

2fill in blank

medium

Complete the code to relax edges inside the main loop of Bellman-Ford.

DSA Typescript

for (let i = 0; i < vertices - 1; i++) {
  for (const [u, v, weight] of edges) {
    if (distances[u] !== Infinity && distances[u] + weight [1] distances[v]) {
      distances[v] = distances[u] + weight;
    }
  }
}

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

A>=

B>

C<

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of <, which reverses the relaxation condition.

Using >= or <= which can cause unnecessary updates.

We relax edges when the current known distance to v is greater than the distance to u plus the edge weight.

3fill in blank

hard

Fix the error in the negative cycle detection condition.

DSA Typescript

for (const [u, v, weight] of edges) {
  if (distances[u] !== Infinity && distances[u] + weight [1] distances[v]) {
    return true; // Negative cycle detected
  }
}

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

A>=

B<

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using >= or > which does not correctly detect negative cycles.

Using <= which can cause false positives.

A negative cycle exists if we can still relax an edge, meaning distances[u] + weight is less than distances[v].

4fill in blank

hard

Fill both blanks to complete the function signature and return type for Bellman-Ford.

DSA Typescript

function bellmanFord([1]: number, edges: [number, number, number][]): [2] {
  // function body
}

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

Avertices

Bnumber[]

Cedges

Dboolean

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'edges' as the first parameter name instead of 'vertices'.

Returning a boolean instead of the distances array.

The function takes the number of vertices as 'vertices' and returns an array of distances as 'number[]'.

5fill in blank

hard

Fill all three blanks to complete the edge relaxation inside the Bellman-Ford main loop.

DSA Typescript

for (let i = 0; i < [1] - 1; i++) {
  for (const [u, v, weight] of [2]) {
    if (distances[u] !== Infinity && distances[u] + weight [3] distances[v]) {
      distances[v] = distances[u] + weight;
    }
  }
}

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

Avertices

Bedges

C<

Ddistances

Attempts:

3 left

💡 Hint

Common Mistakes

Using distances instead of edges in the inner loop.

Using > instead of < in the relaxation condition.

The outer loop runs vertices - 1 times, iterating over edges, relaxing when distances[u] + weight is less than distances[v].

Practice

1. What is the main advantage of the Bellman-Ford algorithm compared to Dijkstra's algorithm?

easy

A. It uses less memory.

B. It runs faster on all graphs.

C. It can handle graphs with negative weight edges.

D. It only works on directed acyclic graphs.

Solution

1. Step 1: Understand the limitation of Dijkstra's algorithm

   Dijkstra's algorithm cannot correctly handle graphs with negative weight edges because it assumes that once a node's shortest path is found, it cannot be improved.

2. Step 2: Identify Bellman-Ford's capability

   Bellman-Ford algorithm can handle negative weight edges and detect negative weight cycles, making it more versatile for such graphs.

3. Final Answer:

   It can handle graphs with negative weight edges. -> Option C

4. Quick Check:

   Negative weights handled = A [OK]

Hint: Bellman-Ford works with negative edges, Dijkstra doesn't [OK]

Common Mistakes:

• Thinking Bellman-Ford is always faster
• Believing it uses less memory
• Assuming it only works on acyclic graphs

2. Which of the following is the correct TypeScript syntax to initialize the distance array for Bellman-Ford with infinity except the source node set to 0?

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(0); dist[source] = 0;

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

Solution

1. Step 1: Initialize distances with Infinity

   We start with all distances as Infinity to represent unknown paths, except the source which is 0.

2. Step 2: Correct syntax for array initialization

   Using new Array(n).fill(Infinity) creates an array of size n filled with Infinity, then setting dist[source] = 0 sets the source distance correctly.

3. Final Answer:

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

4. Quick Check:

   Initialize with Infinity except source 0 = C [OK]

Hint: Fill array with Infinity, then set source to 0 [OK]

Common Mistakes:

• Filling with 0 instead of Infinity
• Setting source distance to Infinity
• Using -Infinity which is incorrect

3. Consider the following TypeScript code snippet implementing one iteration of edge relaxation in Bellman-Ford:

for (const [u, v, w] of edges) {
  if (dist[u] !== Infinity && dist[u] + w < dist[v]) {
    dist[v] = dist[u] + w;
  }
}
console.log(dist);

If edges = [[0,1,4],[1,2,-2],[0,2,5]] and initial dist = [0, Infinity, Infinity], what will be the output after this iteration?

medium

A. [0, 4, 5]

B. [0, 5, 3]

C. [0, Infinity, Infinity]

D. [0, 4, 2]

Solution

1. Step 1: Relax edge (0,1,4)

   dist[0] = 0, 0 + 4 = 4 < dist[1] (Infinity), so dist[1] = 4.

2. Step 2: Relax edge (1,2,-2)

   dist[1] = 4, 4 + (-2) = 2 < dist[2] (Infinity), so dist[2] = 2.

3. Step 3: Relax edge (0,2,5)

   dist[0] = 0, 0 + 5 = 5 < dist[2] (2)? No, so no update.

4. Final Answer:

   [0, 4, 2] -> Option D

5. Quick Check:

   Relax edges updates dist to [0,4,2] [OK]

Hint: Relax edges in order, update dist if smaller [OK]

Common Mistakes:

• Not updating dist[2] after second edge
• Updating dist[2] incorrectly on last edge
• Ignoring initial Infinity checks

4. The following TypeScript code attempts to detect a negative weight cycle after Bellman-Ford iterations:

for (const [u, v, w] of edges) {
  if (dist[u] + w < dist[v]) {
    console.log('Negative cycle detected');
  }
}

What is the bug in this code snippet?

medium

A. It does not check if dist[u] is Infinity before comparison.

B. It should update dist[v] inside the loop.

C. It should use <= instead of < in the condition.

D. It should run inside the main relaxation loop.

Solution

1. Step 1: Understand the condition for negative cycle detection

   After all relaxation rounds, if any edge can still relax dist[v], a negative cycle exists.

2. Step 2: Identify missing Infinity check

   If dist[u] is Infinity (unreachable), dist[u] + w is invalid and can cause wrong detection or runtime error.

3. Final Answer:

   It does not check if dist[u] is Infinity before comparison. -> Option A

4. Quick Check:

   Check dist[u] !== Infinity before comparing [OK]

Hint: Always check dist[u] !== Infinity before edge relaxation [OK]

Common Mistakes:

• Forgetting to check dist[u] Infinity
• Trying to update dist[v] in detection phase
• Misplacing detection inside main loop

5. You have a graph with edges: [[0,1,1],[1,2,-1],[2,3,-1],[3,1,-1]]. Using Bellman-Ford from source 0, what will be the result after all iterations?

hard

A. Algorithm will run infinitely without detection.

B. Negative cycle detected because the cycle 1->2->3->1 sums to -3.

C. Shortest distances: [0, 1, 2, 3]

D. Shortest distances: [0, 1, 0, -1]

Solution

1. Step 1: Identify the cycle and its weight

   The cycle 1->2->3->1 has weights 1->2 (-1), 2->3 (-1), 3->1 (-1), total -3, which is negative.

2. Step 2: Bellman-Ford detects negative cycles

   After |V|-1 iterations, if any edge can still relax distances, it means a negative cycle exists.

3. Step 3: Result of running Bellman-Ford

   The algorithm will detect the negative cycle and report it instead of giving shortest distances.

4. Final Answer:

   Negative cycle detected because the cycle 1->2->3->1 sums to -3. -> Option B

5. Quick Check:

   Negative cycle sum < 0 means detection = D [OK]

Hint: Sum cycle weights; negative sum means cycle detected [OK]

Common Mistakes:

• Ignoring negative cycle and reporting distances
• Assuming algorithm runs infinitely
• Miscomputing cycle weight