Bellman Ford Algorithm Negative Weights in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main purpose of the Bellman Ford algorithm?

The Bellman Ford algorithm finds the shortest path from a single source node to all other nodes in a graph, even if the graph has edges with negative weights.

intermediate

How does Bellman Ford handle negative weight edges differently from Dijkstra's algorithm?

Bellman Ford can handle negative weight edges by relaxing all edges multiple times, while Dijkstra's algorithm assumes all edge weights are non-negative and may fail with negative weights.

intermediate

What does it mean if Bellman Ford detects a negative weight cycle?

It means there is a cycle in the graph where the total sum of edge weights is negative, causing the shortest path to be undefined because you can keep going around the cycle to reduce the path cost infinitely.

beginner

In Bellman Ford, how many times do we relax all edges in the graph?

We relax all edges exactly (V - 1) times, where V is the number of vertices in the graph.

beginner

What is the time complexity of the Bellman Ford algorithm?

The time complexity is O(V * E), where V is the number of vertices and E is the number of edges in the graph.

What does the Bellman Ford algorithm detect besides shortest paths?

A Negative weight cycles

B Maximum flow

C Minimum spanning tree

D Strongly connected components

Bellman Ford can detect negative weight cycles by checking if any edge can still be relaxed after (V - 1) iterations.

How many times are edges relaxed in Bellman Ford for a graph with V vertices?

A V - 1 times

B E times

C V times

D E - 1 times

Edges are relaxed (V - 1) times to ensure shortest paths are found.

Which type of graph edge weights can Bellman Ford handle that Dijkstra cannot?

A Positive weights only

B Zero weights only

C Negative weights

D Non-integer weights

Bellman Ford can handle negative edge weights, unlike Dijkstra.

What happens if a negative weight cycle exists in the graph?

A Bellman Ford runs faster

B Bellman Ford finds shortest paths normally

C Bellman Ford ignores the cycle

D Bellman Ford detects it and reports no solution

Bellman Ford detects negative weight cycles and reports that shortest paths are undefined.

What is the worst-case time complexity of Bellman Ford?

A O(V + E)

B O(V * E)

C O(V^2)

D O(E^2)

Bellman Ford runs in O(V * E) time because it relaxes all edges (V - 1) times.

Explain how the Bellman Ford algorithm finds shortest paths in graphs with negative weights.

Think about how edges are relaxed repeatedly and how cycles affect the result.

Describe the process Bellman Ford uses to detect negative weight cycles.

Focus on what happens after the main relaxation steps.

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