Bellman Ford Algorithm Negative Weights in DSA Typescript - Time & Space Complexity

Time Complexity: Bellman Ford Algorithm Negative Weights

O(V * E)

Understanding Time Complexity

We want to understand how the Bellman Ford algorithm's running time changes as the graph size grows.

Specifically, how does the number of edges and vertices affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function bellmanFord(edges: [number, number, number][], V: number, src: number): number[] {
  const dist = Array(V).fill(Infinity);
  dist[src] = 0;
  for (let i = 0; i < V - 1; i++) {
    for (const [u, v, w] of edges) {
      if (dist[u] !== Infinity && dist[u] + w < dist[v]) {
        dist[v] = dist[u] + w;
      }
    }
  }
  return dist;
}
    

This code finds shortest paths from a source to all vertices, even with negative weights, by relaxing edges repeatedly.

Identify Repeating Operations

• Primary operation: The inner loop that checks and updates distances for all edges.
• How many times: This inner loop runs once for every edge, repeated (V - 1) times where V is the number of vertices.

How Execution Grows With Input

As the number of vertices (V) and edges (E) grow, the total operations grow roughly by E times V.

Input Size (V, E)	Approx. Operations
V=10, E=20	About 10 * 20 = 200
V=100, E=500	About 100 * 500 = 50,000
V=1000, E=2000	About 1000 * 2000 = 2,000,000

Pattern observation: Operations grow linearly with edges and vertices multiplied together.

Final Time Complexity

Time Complexity: O(V * E)

This means the time needed grows proportionally to the number of vertices times the number of edges.

Common Mistake

[X] Wrong: "The algorithm runs in O(E) time because it just loops over edges."

[OK] Correct: The edge loop runs multiple times--once for each vertex minus one--so total work multiplies by V.

Interview Connect

Understanding this time complexity helps you explain why Bellman Ford is slower than other shortest path methods but necessary for graphs with negative weights.

Self-Check

"What if we stopped the algorithm early when no distances change in an iteration? How would that affect the time complexity?"

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