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 analyze how the Bellman Ford algorithm's running time grows with the size of the graph.

We want to know how the number of edges and vertices affects the work done.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void bellmanFord(int V, int E, int edges[][3], int src, int dist[]) {
    for (int i = 0; i < V; i++)
        dist[i] = INT_MAX;
    dist[src] = 0;

    for (int i = 1; i <= V - 1; i++) {
        for (int j = 0; j < E; j++) {
            int u = edges[j][0];
            int v = edges[j][1];
            int w = edges[j][2];
            if (dist[u] != INT_MAX && dist[u] + w < dist[v])
                dist[v] = dist[u] + w;
        }
    }
}
    

This code finds shortest paths from a source vertex to all others, even with negative weights.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The inner loop over all edges that relaxes distances.
• How many times: This inner loop runs for each vertex minus one times (V-1 times).

How Execution Grows With Input

The algorithm repeats checking all edges many times, so work grows with both vertices and edges.

Input Size (V, E)	Approx. Operations
V=10, E=20	About 9 * 20 = 180
V=100, E=500	About 99 * 500 = 49,500
V=1000, E=2000	About 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 multiplying the number of vertices and edges in the graph.

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, so the time depends on edges, not just vertices.

Interview Connect

Understanding this complexity helps you explain how Bellman Ford handles negative weights efficiently compared to other algorithms.

Self-Check

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

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: Recognize Bellman-Ford's capability

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

3. Final Answer:

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

4. Quick Check:

   Negative weights handled = A [OK]

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

Common Mistakes:

• Thinking Bellman-Ford is always faster
• Assuming it uses less memory
• Believing it 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. int dist[V]; for(int i=0; i

B. int dist[V]; for(int i=0; i

C. int dist[V]; for(int i=0; i

D. int dist[V]; for(int i=0; i

Solution

1. Step 1: Understand distance initialization

   Distances to all vertices except the source should be set to a very large number (infinity), here represented by INT_MAX, to indicate they are initially unreachable.

2. Step 2: Set source distance to zero

   The distance to the source vertex itself must be zero because the shortest path from source to source is zero.

3. Final Answer:

   int dist[V]; for(int i=0; i -> Option B

4. Quick Check:

   Initialize all to max, source to 0 = C [OK]

Hint: Set all distances to INT_MAX except source = 0 [OK]

Common Mistakes:

• Initializing all distances to zero
• Using -1 for unreachable distances
• Setting all distances to source index

3. What will be the output distances array after running Bellman-Ford on this graph with edges:
edges = [(0,1,4), (0,2,5), (1,2,-3), (2,3,4)] starting from source 0?

medium

A. [0, 4, 1, 5]

B. [0, 4, 1, 9]

C. [0, 4, 5, 9]

D. [0, 4, -3, 4]

Solution

1. Step 1: Initialize distances

   dist = [0, INF, INF, INF]

2. Step 2: Relax edges repeatedly

   First iteration:
   dist[1] = min(INF, 0+4) = 4
   dist[2] = min(INF, 0+5) = 5
   dist[2] = min(5, 4 + (-3)) = 1
   dist[3] = min(INF, 1 + 4) = 5
   Second iteration:
   dist[1] stays 4
   dist[2] stays 1
   dist[3] stays 5
   No further updates.

3. Final distances:

   [0, 4, 1, 5]

4. Final Answer:

   [0, 4, 1, 5] -> Option A

5. Quick Check:

   Relax edges to update distances = A [OK]

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

Common Mistakes:

• Not relaxing edges enough times
• Ignoring negative edge effect
• Wrong initial distances

4. Identify the error in this Bellman-Ford code snippet in C:

for (int i = 0; i < V; i++) {
  for (int j = 0; j < E; j++) {
    int u = edges[j].src;
    int v = edges[j].dest;
    int w = edges[j].weight;
    if (dist[u] + w < dist[v]) {
      dist[v] = dist[u] + w;
    }
  }
}

medium

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

B. The inner loop should run V times, not E times.

C. The distance array is not initialized.

D. The weight comparison should be dist[v] + w < dist[u].

Solution

1. Step 1: Check loop iteration count

   Bellman-Ford requires relaxing edges exactly V-1 times to find shortest paths correctly.

2. Step 2: Identify the incorrect loop count

   The outer loop runs V times here, which is one extra iteration and unnecessary.

3. Final Answer:

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

4. Quick Check:

   Outer loop runs V-1 times = D [OK]

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

Common Mistakes:

• Running outer loop V times instead of V-1
• Confusing inner loop count with vertex count
• Incorrect weight comparison direction

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 source 0?

hard

A. It will run infinitely without stopping.

B. It will find shortest paths without any issues.

C. It will detect a negative weight cycle.

D. It will give incorrect distances but no error.

Solution

1. Step 1: Analyze 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 by checking for further relaxation after V-1 iterations. Here, it will detect the negative cycle.

3. Final Answer:

   It will detect a negative weight cycle. -> Option C

4. Quick Check:

   Negative cycle detected = A [OK]

Hint: Negative cycle if distances keep decreasing after V-1 relaxations [OK]

Common Mistakes:

• Assuming algorithm runs infinitely
• Believing it finds shortest paths despite negative cycle
• Ignoring cycle detection step