Bellman Ford Algorithm Negative Weights in DSA Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What problem does the Bellman Ford algorithm solve?

It finds the shortest path from a single source vertex to all other vertices in a weighted graph, even if some edges have negative weights.

beginner

Why can Dijkstra's algorithm fail with negative edge weights?

Because it assumes that once a vertex's shortest path is found, it cannot be improved. Negative weights can reduce path costs later, breaking this assumption.

intermediate

How many times does Bellman Ford relax all edges in the graph?

It relaxes all edges exactly |V| - 1 times, where |V| is the number of vertices in the graph.

intermediate

What does Bellman Ford do to detect negative weight cycles?

After |V| - 1 relaxations, it checks all edges one more time. If any edge can still be relaxed, it means a negative weight cycle exists.

beginner

What is the time complexity of the Bellman Ford algorithm?

O(|V| * |E|), where |V| is the number of vertices and |E| is the number of edges.

What is the main advantage of Bellman Ford over Dijkstra's algorithm?

A It only works on unweighted graphs

B It works with graphs that have negative edge weights

C It uses less memory

D It is faster for all graphs

Bellman Ford can handle negative edge weights, unlike Dijkstra's algorithm.

How many times does Bellman Ford relax edges in the worst case?

A |V| times

B |E| times

C |V| - 1 times

D Once

Bellman Ford relaxes all edges |V| - 1 times to find shortest paths.

What does it mean if Bellman Ford can still relax an edge after |V| - 1 iterations?

A There is a negative weight cycle

B Graph is disconnected

C All shortest paths are found

D Graph has no edges

Further relaxation indicates a negative weight cycle in the graph.

Which data structure is primarily used to represent the graph in Bellman Ford?

A Edge list

B Adjacency list

C Adjacency matrix

D Stack

Bellman Ford typically uses an edge list to relax edges efficiently.

What is the time complexity of Bellman Ford algorithm?

A O(|V| + |E|)

B O(|E|^2)

C O(|V|^2)

D O(|V| * |E|)

Bellman Ford runs in O(|V| * |E|) time.

Explain how Bellman Ford algorithm detects negative weight cycles.

Think about what happens if distances keep improving after expected iterations.

Describe the step-by-step process of the Bellman Ford algorithm for finding shortest paths.

Start from setting initial distances and repeat edge relaxation.

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