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 it handle graphs with negative weights efficiently?

Scenario Under Consideration

Analyze the time complexity of the following Bellman Ford code snippet.

// Bellman Ford to find shortest paths
// from source in graph with negative weights
for (int i = 1; i <= vertices - 1; i++) {
  foreach (var edge in edges) {
    int u = edge.Source;
    int v = edge.Destination;
    int w = edge.Weight;
    if (dist[u] != int.MaxValue && dist[u] + w < dist[v]) {
      dist[v] = dist[u] + w;
    }
  }
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Relaxing all edges in the graph.
• How many times: The outer loop runs (vertices - 1) times, and inside it, all edges are checked once each time.

How Execution Grows With Input

Each vertex causes a full pass over all edges, so the work grows with both vertices and edges.

Input Size (vertices, edges)	Approx. Operations
10 vertices, 20 edges	~ 9 * 20 = 180 edge relaxations
100 vertices, 500 edges	~ 99 * 500 = 49,500 edge relaxations
1000 vertices, 2000 edges	~ 999 * 2000 = 1,998,000 edge relaxations

Pattern observation: The total work grows roughly by multiplying the number of vertices by the number of edges.

Final Time Complexity

Time Complexity: O(V * E)

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

Common Mistake

[X] Wrong: "Bellman Ford runs as fast as Dijkstra's algorithm, so it should be O(E log V)."

[OK] Correct: Bellman Ford checks all edges multiple times to handle negative weights, so it takes longer than Dijkstra's algorithm which uses a priority queue.

Interview Connect

Understanding Bellman Ford's time complexity helps you explain why it is chosen for graphs with negative weights despite being slower than other algorithms.

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 purpose of the Bellman-Ford algorithm when dealing with graphs that have negative edge weights?

easy

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

B. To detect cycles in an undirected graph

C. To find the minimum spanning tree of a graph

D. To sort vertices in topological order

Solution

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

   The Bellman-Ford algorithm is designed to find shortest paths from a single source vertex to all other vertices in a graph, even if some edges have negative weights.

2. Step 2: Compare with other algorithms

   Unlike Dijkstra's algorithm, Bellman-Ford can handle negative weights and also detect negative weight cycles.

3. Final Answer:

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

4. Quick Check:

   Bellman-Ford finds shortest paths with negative weights [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 detects cycles without finding paths
• Assuming it cannot handle negative weights

2. Which of the following is the correct C# syntax to initialize the distance array for Bellman-Ford with infinite values except the source vertex at index 0 set to 0?

easy

A. int[] dist = new int[n]; for(int i=0; i

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

C. int[] dist = new int[n]; dist.Fill(int.MaxValue); dist[0] = 0;

D. int[] dist = Enumerable.Repeat(int.MaxValue, n).ToArray(); dist[0] = 0;

Solution

1. Step 1: Understand array initialization with default values

   Using Enumerable.Repeat(int.MaxValue, n).ToArray() creates an array filled with int.MaxValue for all elements.

2. Step 2: Set source distance to zero

   After initialization, setting dist[0] = 0 correctly sets the source vertex distance.

3. Final Answer:

   int[] dist = Enumerable.Repeat(int.MaxValue, n).ToArray(); dist[0] = 0; -> Option D

4. Quick Check:

   Correct array initialization with source zero [OK]

Hint: Use Enumerable.Repeat for array filled with max values [OK]

Common Mistakes:

• Using dist.Fill which is not a valid method in C# arrays
• Setting dist[0] before initializing the array
• Not initializing all elements to int.MaxValue

3. Consider the following C# code snippet implementing one iteration of edge relaxation in Bellman-Ford. What will be the output distances array after this iteration?

int n = 4;
int[] dist = {0, int.MaxValue, int.MaxValue, int.MaxValue};
(int u, int v, int w)[] edges = {
  (0, 1, 1),
  (1, 2, -1),
  (2, 3, -1),
  (3, 1, -1)
};

foreach(var edge in edges) {
  if(dist[edge.u] != int.MaxValue && dist[edge.u] + edge.w < dist[edge.v]) {
    dist[edge.v] = dist[edge.u] + edge.w;
  }
}

Console.WriteLine(string.Join(", ", dist));

medium

A. 0, int.MaxValue, int.MaxValue, int.MaxValue

B. 0, 1, int.MaxValue, int.MaxValue

C. 0, -2, 0, -1

D. 0, 1, 0, -1

Solution

1. Step 1: Relax edge (0,1,1)

   dist[0] = 0, so dist[1] = min(int.MaxValue, 0+1) = 1

2. Step 2: Relax edge (1,2,-1)

   dist[1] = 1, so dist[2] = min(int.MaxValue, 1-1) = 0

3. Step 3: Relax edge (2,3,-1)

   dist[2] = 0, so dist[3] = min(int.MaxValue, 0-1) = -1

4. Step 4: Relax edge (3,1,-1)

   dist[3] = -1, so dist[1] = min(1, -1-1) = -2

5. Final Answer:

   0, -2, 0, -1 -> Option C

6. Quick Check:

   Distances updated correctly after one iteration [OK]

Hint: Relax edges in order, update distances if smaller [OK]

Common Mistakes:

• Not updating dist[1] after last edge relaxation
• Ignoring negative weights in relaxation
• Assuming no change if dist[v] is int.MaxValue

4. The following C# code snippet is intended to detect a negative weight cycle using Bellman-Ford. Identify the error:

bool HasNegativeCycle(int n, (int u, int v, int w)[] edges, int source) {
  int[] dist = Enumerable.Repeat(int.MaxValue, n).ToArray();
  dist[source] = 0;
  for(int i = 1; i <= n; i++) {
    foreach(var edge in edges) {
      if(dist[edge.u] != int.MaxValue && dist[edge.u] + edge.w < dist[edge.v]) {
        dist[edge.v] = dist[edge.u] + edge.w;
        if(i == n) return true;
      }
    }
  }
  return false;
}

medium

A. The loop should run from 1 to n-1, not 1 to n

B. The distance array should be initialized with zeros

C. The source vertex distance should be int.MaxValue, not 0

D. The return true should be outside the inner foreach loop

Solution

1. Step 1: Understand Bellman-Ford iteration count

   Bellman-Ford relaxes edges exactly n-1 times to find shortest paths without negative cycles.

2. Step 2: Detect negative cycle on nth iteration

   If any edge can still be relaxed on the nth iteration, a negative cycle exists.

3. Step 3: Identify loop range error

   The code runs the loop from 1 to n inclusive. Bellman-Ford should relax edges exactly n-1 times, then check separately on the nth iteration.

4. Final Answer:

   The loop should run from 1 to n-1, not 1 to n -> Option A

5. Quick Check:

   Relax edges n-1 times before checking cycle [OK]

Hint: Relax edges n-1 times, then check for cycle [OK]

Common Mistakes:

• Running relaxation loop n times instead of n-1
• Initializing distances incorrectly
• Returning true inside inner loop prematurely

5. You have a graph with 5 vertices and edges with some negative weights but no negative cycles. How many times must the Bellman-Ford algorithm relax all edges to guarantee correct shortest paths?

hard

A. 6 times

B. 4 times

C. 3 times

D. 5 times

Solution

1. Step 1: Recall Bellman-Ford relaxation count rule

   Bellman-Ford requires relaxing all edges exactly n-1 times, where n is the number of vertices.

2. Step 2: Apply to given graph

   For 5 vertices, the algorithm must relax edges 5-1 = 4 times to guarantee shortest paths.

3. Final Answer:

   4 times -> Option B

4. Quick Check:

   Relax edges n-1 times for n vertices [OK]

Hint: Relax edges n-1 times for n vertices [OK]

Common Mistakes:

• Relaxing edges n times instead of n-1
• Confusing number of edges with number of vertices
• Assuming negative weights require more iterations