Bellman-Ford algorithm in Data Structures Theory - Time & Space Complexity

Time Complexity: Bellman-Ford algorithm

O(V * E)

Understanding Time Complexity

We want to understand how the time needed by the Bellman-Ford algorithm grows as the graph size increases.

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

Scenario Under Consideration

Analyze the time complexity of the Bellman-Ford algorithm below.

function BellmanFord(graph, source):
  distance = array of size V with all values infinity
  distance[source] = 0
  for i in 1 to V-1:
    for each edge (u, v) in graph.edges:
      if distance[u] + weight(u, v) < distance[v]:
        distance[v] = distance[u] + weight(u, v)
  return distance
    

This code finds shortest paths from one vertex to all others by relaxing edges repeatedly.

Identify Repeating Operations

Look at the loops and repeated steps:

• Primary operation: Checking and updating distances for each edge.
• How many times: The inner loop runs once for every edge, repeated V-1 times (where V is number of vertices).

How Execution Grows With Input

The work grows with both the number of vertices (V) and edges (E).

Input Size (V, E)	Approx. Operations
V=10, E=20	About 9 * 20 = 180 checks
V=100, E=500	About 99 * 500 = 49,500 checks
V=1000, E=5000	About 999 * 5000 = 4,995,000 checks

Pattern observation: The total work increases roughly by multiplying vertices and edges.

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(V^2) time because it has two loops over vertices."

[OK] Correct: The inner loop runs over edges, not vertices. The number of edges can be very different from vertices, so the complexity depends on both V and E, not just V squared.

Interview Connect

Understanding this complexity helps you explain how graph size affects performance and why Bellman-Ford is slower than some other shortest path algorithms.

Self-Check

"What if the graph is dense, meaning E is close to V squared? How would that affect the time complexity?"

Practice

1. What is the primary purpose of the Bellman-Ford algorithm in graph theory?

easy

A. To sort vertices in topological order

B. To find the minimum spanning tree of a graph

C. To detect cycles in an undirected graph

D. To find the shortest path from a single source to all vertices, even with negative edge weights

Solution

1. Step 1: Understand Bellman-Ford's goal

   The Bellman-Ford algorithm is designed to find shortest paths from one source vertex to all other vertices in a graph.

2. Step 2: Recognize its unique feature

   Unlike some algorithms, Bellman-Ford can handle graphs with negative edge weights and detect negative weight cycles.

3. Final Answer:

   To find the shortest path from a single source to all vertices, even with negative edge weights -> Option D

4. Quick Check:

   Bellman-Ford finds shortest paths with negative edges = A [OK]

Hint: Bellman-Ford handles negative edges and finds shortest paths [OK]

Common Mistakes:

• Confusing Bellman-Ford with algorithms for minimum spanning trees
• Thinking it only works with positive edge weights
• Assuming it sorts vertices or detects cycles in undirected graphs

2. Which of the following is the correct step repeated in the Bellman-Ford algorithm?

easy

A. Relax all edges |V|-1 times

B. Relax all edges once

C. Relax edges only if weights are positive

D. Relax edges in topological order

Solution

1. Step 1: Recall the relaxation process

   Bellman-Ford repeatedly relaxes all edges to update shortest path estimates.

2. Step 2: Identify the number of repetitions

   It performs this relaxation exactly |V|-1 times, where |V| is the number of vertices.

3. Final Answer:

   Relax all edges |V|-1 times -> Option A

4. Quick Check:

   Relax edges |V|-1 times = D [OK]

Hint: Relax edges |V|-1 times to find shortest paths [OK]

Common Mistakes:

• Relaxing edges only once
• Relaxing edges only if weights are positive
• Assuming topological order is required

3. Consider the graph with edges and weights:
A -> B (4), A -> C (5), B -> C (-3), C -> D (2)
Using Bellman-Ford from vertex A, what is the shortest distance to vertex C?

medium

A. 5

B. 1

C. -3

D. 4

Solution

1. Step 1: Initialize distances from A

   Distance to A=0, B=∞, C=∞, D=∞ initially.

2. Step 2: Relax edges repeatedly

   First relaxation: A->B=4, A->C=5. Then B->C updates C to 4 + (-3) = 1. C->D updates D to 1 + 2 = 3.

3. Final Answer:

   1 -> Option B

4. Quick Check:

   Shortest path A->B->C = 4 + (-3) = 1 [OK]

Hint: Check paths through neighbors to update shortest distances [OK]

Common Mistakes:

• Ignoring negative edge from B to C
• Taking direct edge A->C without relaxation
• Confusing edge weights with distances

4. You run Bellman-Ford on a graph and after |V|-1 iterations, you find an edge u -> v where distance[u] + weight(u,v) < distance[v]. What does this indicate?

medium

A. The graph contains a negative weight cycle reachable from the source

B. The algorithm has completed successfully with correct distances

C. There is a disconnected vertex in the graph

D. The graph is undirected

Solution

1. Step 1: Understand the relaxation check after iterations

   After |V|-1 relaxations, no edge should improve distances if no negative cycles exist.

2. Step 2: Interpret the condition

   If an edge can still relax a distance, it means a negative weight cycle exists reachable from the source.

3. Final Answer:

   The graph contains a negative weight cycle reachable from the source -> Option A

4. Quick Check:

   Extra relaxation means negative cycle = C [OK]

Hint: Extra relaxation after |V|-1 means negative cycle [OK]

Common Mistakes:

• Thinking algorithm finished correctly despite extra relaxation
• Confusing negative cycle with disconnected vertices
• Assuming graph type (directed/undirected) affects this check

5. Given a graph with vertices V = {S, A, B, C} and edges with weights:
S->A (1), A->B (-1), B->C (-1), C->A (-1)
What will Bellman-Ford detect when run from source S?

hard

A. It will find shortest paths without issues

B. It will report no path to vertex C

C. It will detect a negative weight cycle involving vertices A, B, and C

D. It will fail due to disconnected graph

Solution

1. Step 1: Analyze the cycle formed by A, B, and C

   The edges A->B (-1), B->C (-1), and C->A (-1) form a cycle with total weight -3, which is negative.

2. Step 2: Understand Bellman-Ford behavior on negative cycles

   Bellman-Ford detects negative weight cycles reachable from the source, which prevents shortest path calculation.

3. Final Answer:

   It will detect a negative weight cycle involving vertices A, B, and C -> Option C

4. Quick Check:

   Negative cycle A-B-C with sum -3 detected = B [OK]

Hint: Check cycles' total weight; negative sum means cycle detected [OK]

Common Mistakes:

• Assuming shortest paths exist despite negative cycle
• Ignoring the cycle formed by A, B, and C
• Thinking graph is disconnected when it is not