Bellman Ford Algorithm Negative Weights in DSA C++ - 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 Bellman Ford code snippet.

for (int i = 1; i <= V - 1; i++) {
  for (auto edge : edges) {
    int u = edge.u, v = edge.v, w = edge.w;
    if (dist[u] != INT_MAX && dist[u] + w < dist[v]) {
      dist[v] = dist[u] + w;
    }
  }
}
    

This code relaxes all edges V-1 times to find shortest paths, even with negative weights.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Relaxing each edge by checking and updating distances.
• How many times: The inner loop runs for all edges, repeated V-1 times (outer loop).

How Execution Grows With Input

Each vertex causes all edges to be checked once, so work grows with both vertices and edges.

Input Size (V, E)	Approx. Operations
V=10, E=20	9 * 20 = 180
V=100, E=200	99 * 200 = 19,800
V=1000, E=2000	999 * 2000 = 1,998,000

Pattern observation: Operations grow roughly by multiplying vertices and edges.

Final Time Complexity

Time Complexity: O(V * E)

This means the time grows by the number of vertices times the number of edges in the graph.

Common Mistake

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

[OK] Correct: The edges are checked multiple times--once for each vertex minus one--so the total work depends on both vertices and edges.

Interview Connect

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

Self-Check

"What if we stop the algorithm early when no distance updates happen 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 only works on directed acyclic graphs.

B. It runs faster on all graphs.

C. It uses less memory.

D. It can handle graphs with negative weight edges.

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 D

4. Quick Check:

   Negative weights handled = D [OK]

Hint: Bellman Ford works with negative edges, Dijkstra does not [OK]

Common Mistakes:

• Thinking Bellman Ford is always faster than Dijkstra
• Believing Bellman Ford cannot detect negative cycles
• Assuming Bellman Ford only works on acyclic graphs

2. Which of the following is the correct way to initialize the distance array in Bellman Ford algorithm in C++?

easy

A. vector<int> dist(V, 0); dist[src] = INT_MAX;

B. vector<int> dist(V, INT_MAX); dist[src] = 0;

C. vector<int> dist(V, -1); dist[src] = 0;

D. vector<int> dist(V, INT_MIN); dist[src] = 0;

Solution

1. Step 1: Understand distance initialization

   Distances to all vertices except the source should be set to a very large number (infinity equivalent), here INT_MAX, to represent unknown distance.

2. Step 2: Set source distance to zero

   The source vertex distance is zero because the distance from source to itself is zero.

3. Final Answer:

   vector<int> dist(V, INT_MAX); dist[src] = 0; -> Option B

4. Quick Check:

   Initialize with INT_MAX and source zero = B [OK]

Hint: Initialize all distances to INT_MAX except source zero [OK]

Common Mistakes:

• Setting all distances to zero
• Using INT_MIN which is incorrect for distances
• Not setting source distance to zero

3. Consider the following graph edges and weights:

Edges: (0->1, 4), (0->2, 5), (1->2, -3), (2->3, 4), (3->1, 6)

What is the shortest distance from vertex 0 to vertex 3 after running Bellman Ford?

medium

A. 5

B. 9

C. 6

D. 7

Solution

1. Step 1: Initialize distances

   dist[0]=0, dist[1]=∞, dist[2]=∞, dist[3]=∞.

2. Step 2: Relax edges repeatedly

   After first iteration: dist[1]=4, dist[2]=5, dist[3]=∞.
   Relax (1->2): dist[2]=4 + (-3) = 1.
   Relax (2->3): dist[3]=1 + 4 = 5.
   Relax (3->1): dist[1]=min(4, 5 + 6=11) = 4.
   Second iteration: No further improvements.
   Final dist[3] = 5.

3. Step 3: Check for negative cycles

   No negative cycle detected.

4. Final Answer:

   5 -> Option A

5. Quick Check:

   Shortest path dist to 3 = 5 [OK]

Hint: Relax edges V-1 times, track minimum distances [OK]

Common Mistakes:

• Not updating distances after negative edge relaxation
• Confusing edge directions
• Stopping relaxation too early

4. The following C++ code snippet attempts to implement Bellman Ford but has a bug:

for (int i = 0; i < V; i++) {
  for (auto edge : edges) {
    if (dist[edge.u] + edge.w < dist[edge.v]) {
      dist[edge.v] = dist[edge.u] + edge.w;
    }
  }
}

What is the bug?

medium

A. The outer loop should run V-1 times, not V times.

B. The condition should use <= instead of <.

C. The distance array is not initialized.

D. Edges should be relaxed in reverse order.

Solution

1. Step 1: Understand Bellman Ford iteration count

   Bellman Ford requires relaxing edges exactly V-1 times to find shortest paths without negative cycles.

2. Step 2: Identify the loop error

   The code runs the outer loop V times, which is one extra iteration and may cause incorrect detection of negative cycles or unnecessary work.

3. Final Answer:

   The outer loop should run V-1 times, not V times. -> Option A

4. Quick Check:

   Outer loop count = V-1 [OK]

Hint: Relax edges exactly V-1 times for correct results [OK]

Common Mistakes:

• Running outer loop V times instead of V-1
• Changing comparison operator incorrectly
• Ignoring distance initialization

5. Given a graph with edges:

(0->1, 1), (1->2, -1), (2->3, -1), (3->1, -1)

What will Bellman Ford detect when run from vertex 0?

hard

A. Graph is disconnected.

B. Shortest distances without any issues.

C. A negative weight cycle reachable from source.

D. Algorithm will run infinitely.

Solution

1. Step 1: Identify the cycle in the graph

   The edges (1->2), (2->3), and (3->1) form a cycle with total weight -1 + -1 + -1 = -3, which is negative.

2. Step 2: Understand Bellman Ford's negative cycle detection

   Bellman Ford detects negative weight cycles reachable from the source by checking if any edge can still be relaxed after V-1 iterations.

3. Step 3: Conclusion

   Since the negative cycle is reachable from vertex 0, Bellman Ford will detect it and report its presence.

4. Final Answer:

   A negative weight cycle reachable from source. -> Option C

5. Quick Check:

   Negative cycle detected = A [OK]

Hint: Negative cycle if edges relax after V-1 iterations [OK]

Common Mistakes:

• Ignoring negative cycles reachable from source
• Assuming algorithm runs infinitely
• Thinking graph is disconnected