Bellman Ford Algorithm Negative Weights in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the distance array with infinity except the source vertex.

DSA Java

int[] dist = new int[V];
for (int i = 0; i < V; i++) {
    dist[i] = [1];
}
dist[src] = 0;

Drag options to blanks, or click blank then click option'

AInteger.MAX_VALUE

B0

C-1

DInteger.MIN_VALUE

Attempts:

3 left

💡 Hint

Common Mistakes

Setting distance to 0 for all vertices.

Using Integer.MIN_VALUE which is negative.

Not initializing the array before use.

We initialize all distances to Integer.MAX_VALUE to represent infinity, except the source which is set to 0.

2fill in blank

medium

Complete the code to relax edges inside the main loop of Bellman-Ford.

DSA Java

for (int i = 1; i < V; i++) {
    for (Edge edge : edges) {
        int u = edge.src;
        int v = edge.dest;
        int weight = edge.weight;
        if (dist[u] != Integer.MAX_VALUE && dist[u] + weight [1] dist[v]) {
            dist[v] = dist[u] + weight;
        }
    }
}

Drag options to blanks, or click blank then click option'

A==

B>

C<

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < in the condition.

Not checking if dist[u] is Integer.MAX_VALUE before adding.

Using == or != which are incorrect for relaxation.

We relax edges if the current distance to u plus weight is less than the current distance to v.

3fill in blank

hard

Fix the error in the negative cycle detection condition.

DSA Java

for (Edge edge : edges) {
    int u = edge.src;
    int v = edge.dest;
    int weight = edge.weight;
    if (dist[u] != Integer.MAX_VALUE && dist[u] + weight [1] dist[v]) {
        System.out.println("Graph contains negative weight cycle");
        return;
    }
}

Drag options to blanks, or click blank then click option'

A<

B>

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > which is incorrect for negative cycle detection.

Using == or != which do not detect cycles.

Not checking dist[u] != Integer.MAX_VALUE before comparison.

If after V-1 iterations we can still relax an edge, it means a negative weight cycle exists, so dist[u] + weight < dist[v].

4fill in blank

hard

Fill both blanks to correctly update the distance array and check for overflow.

DSA Java

if (dist[u] != Integer.MAX_VALUE && dist[u] [1] weight < dist[v]) {
    dist[v] = dist[u] [2] weight;
}

Drag options to blanks, or click blank then click option'

A+

B-

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition.

Using > instead of < in the condition.

Not updating dist[v] correctly.

We add weight to dist[u] to get the new distance and check if it is less than dist[v] to update.

5fill in blank

hard

Fill all three blanks to complete the code that prints the shortest distances after Bellman-Ford.

DSA Java

for (int i = 0; i < V; i++) {
    if (dist[i] == [1]) {
        System.out.println("Vertex " + i + ": unreachable");
    } else {
        System.out.println("Vertex " + i + ": " + dist[[2]]);
    }
}

return [3];

Drag options to blanks, or click blank then click option'

AInteger.MAX_VALUE

Bi

Cdist

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong constant for infinity check.

Using wrong index in print statement.

Not returning the distance array.

We check if dist[i] == Integer.MAX_VALUE to represent infinity (unreachable), print unreachable, else print dist[i], then return dist array.

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