Bellman Ford Algorithm Negative Weights in DSA Java - 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 (Edge edge : edges) {
    if (dist[edge.src] + edge.weight < dist[edge.dest]) {
      dist[edge.dest] = dist[edge.src] + edge.weight;
    }
  }
}
    

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

Identify Repeating Operations

Look for loops or repeated steps that take most time.

• Primary operation: Checking and updating distances for each edge (edge relaxation).
• How many times: The outer loop runs V-1 times, and the inner loop runs once for each edge E.

How Execution Grows With Input

As the number of vertices (V) and edges (E) grow, the total work grows by multiplying these two.

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

Pattern observation: The work grows roughly by multiplying vertices and edges, so doubling both roughly quadruples the work.

Final Time Complexity

Time Complexity: O(V * E)

This means the time grows by the number of vertices times the number of edges, so bigger graphs take more time.

Common Mistake

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

[OK] Correct: The edge loop runs many times (V-1), so total work depends on both vertices and edges, not just edges.

Interview Connect

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

This skill shows you can analyze algorithms beyond just coding them.

Self-Check

"What if we stopped the algorithm early when no distances change in a full pass? 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 can handle graphs with negative weight edges.

B. It runs faster on all graphs.

C. It uses less memory.

D. It only works on directed acyclic graphs.

Solution

1. Step 1: Understand Dijkstra's limitation

   Dijkstra's algorithm cannot handle negative weight edges correctly because it assumes once a node is visited, its shortest path is finalized.

2. Step 2: Bellman-Ford's capability

   Bellman-Ford algorithm relaxes edges multiple times and can detect negative weight edges, allowing it to work correctly even with negative weights.

3. Final Answer:

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

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 only works on acyclic graphs
• Confusing memory usage advantages

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

easy

A. int[] dist = new int[V]; Arrays.fill(dist, -1); dist[src] = 0;

B. int[] dist = new int[V]; dist[src] = Integer.MAX_VALUE;

C. int[] dist = new int[V]; Arrays.fill(dist, Integer.MAX_VALUE); dist[src] = 0;

D. int[] dist = new int[V]; Arrays.fill(dist, 0);

Solution

1. Step 1: Initialize distances to max value

   We set all distances to Integer.MAX_VALUE to represent infinity initially.

2. Step 2: Set source distance to zero

   The source vertex distance is set to 0 because distance from source to itself is zero.

3. Final Answer:

   int[] dist = new int[V]; Arrays.fill(dist, Integer.MAX_VALUE); dist[src] = 0; -> Option C

4. Quick Check:

   Distance init with max + source zero = D [OK]

Hint: Initialize all distances to max, source to zero [OK]

Common Mistakes:

• Setting source distance to max instead of zero
• Filling distances with zero instead of max
• Using -1 which is invalid for distances

3. Consider the following graph edges and weights:

int[][] edges = {
  {0, 1, 4},
  {0, 2, 5},
  {1, 2, -3},
  {2, 3, 4}
};
int V = 4; int src = 0;

What is the shortest distance array after running Bellman-Ford from source 0?

medium

A. [0, 4, 5, 9]

B. [0, 4, 1, 5]

C. [0, 4, 1, 9]

D. [0, 0, -3, 1]

Solution

1. Step 1: Initialize distances

   dist = [0, INF, INF, INF]

2. Step 2: Relax edges V-1 times

   Iteration 1: - Edge 0->1: dist[1] = min(INF, 0+4) = 4 - Edge 0->2: dist[2] = min(INF, 0+5) = 5 - Edge 1->2: dist[2] = min(5, 4-3) = 1 - Edge 2->3: dist[3] = min(INF, 1+4) = 5 Iteration 2 and 3 do not improve distances further.

3. Final Answer:

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

4. Quick Check:

   Relax edges 3 times, final dist = [0,4,1,5] [OK]

Hint: Relax edges V-1 times, update distances each time [OK]

Common Mistakes:

• Not updating dist[2] after edge 1->2 relaxation
• Ignoring negative edge effect
• Stopping after first iteration

4. Given this snippet of Bellman-Ford code, what is the error?

for (int i = 0; i < V; i++) {
  for (int[] edge : edges) {
    int u = edge[0], v = edge[1], w = edge[2];
    if (dist[u] != Integer.MAX_VALUE && 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 condition dist[u] != Integer.MAX_VALUE is incorrect.

C. The edge array indexing is wrong.

D. The distance update should be dist[u] - w instead of dist[u] + w.

Solution

1. Step 1: Check loop iteration count

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

2. Step 2: Identify loop error

   The code runs the outer loop V times, which is one extra iteration and can cause incorrect behavior 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: Run outer loop V-1 times for correct relaxation [OK]

Common Mistakes:

• Running loop V times instead of V-1
• Misunderstanding distance initialization condition
• Incorrect edge array indexing

5. You run Bellman-Ford on a graph with edges:

int[][] edges = {
  {0, 1, 1},
  {1, 2, -1},
  {2, 3, -1},
  {3, 1, -1}
};
int V = 4; int src = 0;

What will the algorithm detect after V-1 iterations?

hard

A. Algorithm will throw an exception due to negative weights.

B. No negative weight cycle; shortest paths found.

C. Distances will all be zero.

D. A negative weight cycle exists reachable from the source.

Solution

1. Step 1: Identify cycle in graph

   Edges 1->2, 2->3, 3->1 form a cycle with total weight -3 (negative).

2. Step 2: Bellman-Ford detects negative cycles

   After V-1 iterations, the algorithm checks for further relaxation. If possible, it means a negative weight cycle exists reachable from source.

3. Final Answer:

   A negative weight cycle exists reachable from the source. -> Option D

4. Quick Check:

   Negative cycle detected = C [OK]

Hint: Check for extra relaxation after V-1 iterations to find negative cycles [OK]

Common Mistakes:

• Assuming no negative cycle despite cycle with negative sum
• Expecting exception on negative weights
• Thinking distances become zero automatically