Floyd Warshall All Pairs Shortest Path in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Floyd Warshall algorithm?

To find the shortest paths between all pairs of vertices in a weighted graph, including graphs with negative edge weights but no negative cycles.

beginner

What is the time complexity of the Floyd Warshall algorithm?

The time complexity is O(V³), where V is the number of vertices in the graph.

intermediate

In Floyd Warshall, what does the matrix entry dist[i][j] represent during the algorithm?

It represents the shortest known distance from vertex i to vertex j using intermediate vertices considered so far.

intermediate

Why can Floyd Warshall handle negative edge weights but not negative cycles?

Because it updates shortest paths by checking all intermediate vertices, but if a negative cycle exists, distances can decrease indefinitely, making shortest paths undefined.

beginner

What is the key idea behind the Floyd Warshall algorithm's triple nested loop?

It tries to improve the shortest path between every pair (i, j) by checking if going through an intermediate vertex k gives a shorter path.

What does the Floyd Warshall algorithm compute?

A Shortest paths between all pairs of vertices

B Shortest path from a single source to all vertices

C Minimum spanning tree

D Topological sort of vertices

Floyd Warshall finds shortest paths between every pair of vertices.

Which of these is a requirement for Floyd Warshall to work correctly?

A No negative cycles in the graph

B Graph must be undirected

C Graph must have only positive weights

D Graph must be a tree

Floyd Warshall can handle negative weights but not negative cycles.

What is the space complexity of Floyd Warshall algorithm for a graph with V vertices?

A O(V³)

B O(V²)

C O(V)

D O(1)

It uses a V x V matrix to store distances, so space is O(V²).

In Floyd Warshall, what does the variable 'k' represent in the triple nested loop?

A The end vertex

B The start vertex

C An intermediate vertex to check for shorter paths

D The weight of an edge

'k' is the intermediate vertex used to update shortest paths.

If dist[i][k] + dist[k][j] < dist[i][j], what should Floyd Warshall do?

A Ignore and keep dist[i][j]

B Remove vertex k

C Set dist[i][j] to zero

D Update dist[i][j] to dist[i][k] + dist[k][j]

The algorithm updates the shortest path if a shorter path through k is found.

Explain how the Floyd Warshall algorithm updates the shortest path matrix step-by-step.

Think about how intermediate vertices help improve paths.

Describe why Floyd Warshall is suitable for graphs with negative edges but not negative cycles.

Consider what happens if you keep going around a negative cycle.

Practice

1. What does the Floyd Warshall algorithm primarily compute in a graph?

easy

A. Shortest paths between all pairs of vertices

B. Minimum spanning tree of the graph

C. Depth-first traversal order

D. Maximum flow between two nodes

Solution

1. Step 1: Understand Floyd Warshall purpose

   The Floyd Warshall algorithm is designed to find shortest paths between every pair of vertices in a weighted graph.

2. Step 2: Compare with other graph algorithms

   Minimum spanning tree, depth-first traversal, and maximum flow solve different problems, not all pairs shortest paths.

3. Final Answer:

   Shortest paths between all pairs of vertices -> Option A

4. Quick Check:

   Floyd Warshall = All pairs shortest paths [OK]

Hint: Floyd Warshall = shortest paths for all vertex pairs [OK]

Common Mistakes:

• Confusing with Dijkstra's single-source shortest path
• Thinking it finds minimum spanning tree
• Mixing with traversal algorithms

2. Which of the following is the correct way to initialize the distance matrix dist for Floyd Warshall in Java for a graph with V vertices?

easy

A. for (int i = 0; i < V; i++) for (int j = 0; j < V; j++) dist[i][j] = (i == j) ? 0 : Integer.MAX_VALUE;

B. for (int i = 0; i < V; i++) dist[i] = new int[V];

C. dist = new int[V][V]; for (int i = 0; i < V; i++) dist[i][i] = 1;

D. for (int i = 0; i < V; i++) for (int j = 0; j < V; j++) dist[i][j] = 0;

Solution

1. Step 1: Understand initialization needs

   Distance from a vertex to itself should be 0, and to others should be infinity (represented by Integer.MAX_VALUE).

2. Step 2: Check each option

   for (int i = 0; i < V; i++) for (int j = 0; j < V; j++) dist[i][j] = (i == j) ? 0 : Integer.MAX_VALUE; correctly sets 0 for same vertices and max value for others. for (int i = 0; i < V; i++) dist[i] = new int[V]; only initializes array but not values. dist = new int[V][V]; for (int i = 0; i < V; i++) dist[i][i] = 1; sets diagonal to 1 incorrectly. for (int i = 0; i < V; i++) for (int j = 0; j < V; j++) dist[i][j] = 0; sets all to 0, which is wrong.

3. Final Answer:

   for (int i = 0; i < V; i++) for (int j = 0; j < V; j++) dist[i][j] = (i == j) ? 0 : Integer.MAX_VALUE; -> Option A

4. Quick Check:

   Initialize dist[i][i] = 0, others = infinity [OK]

Hint: Set zero for same nodes, max value for others [OK]

Common Mistakes:

• Not setting distance to self as zero
• Setting all distances to zero
• Forgetting to initialize the matrix fully

3. What will be the output distance matrix after running Floyd Warshall on this graph with 3 vertices and edges: 0->1=4, 1->2=3, 0->2=10?

int V = 3;
int[][] dist = {
  {0, 4, 10},
  {Integer.MAX_VALUE, 0, 3},
  {Integer.MAX_VALUE, Integer.MAX_VALUE, 0}
};
// Floyd Warshall algorithm applied here

medium

A. [[0, 4, 3], [∞, 0, 3], [∞, ∞, 0]]

B. [[0, 4, 10], [∞, 0, 3], [∞, ∞, 0]]

C. [[0, 7, 10], [∞, 0, 3], [∞, ∞, 0]]

D. [[0, 4, 7], [∞, 0, 3], [∞, ∞, 0]]

Solution

1. Step 1: Initial distances

   Start with dist matrix: 0->1=4, 1->2=3, 0->2=10, others infinity or 0 on diagonal.

2. Step 2: Update using intermediate vertices

   Check if path 0->1->2 is shorter than 0->2 directly: 4 + 3 = 7 < 10, so update dist[0][2] = 7.

3. Final Answer:

   [[0, 4, 7], [∞, 0, 3], [∞, ∞, 0]] -> Option D

4. Quick Check:

   0->2 updated to 7 via 1 [OK]

Hint: Check if indirect path is shorter than direct [OK]

Common Mistakes:

• Not updating indirect shorter paths
• Confusing infinity with zero
• Ignoring intermediate vertex updates

4. Identify the bug in this Floyd Warshall code snippet:

for (int k = 0; k < V; k++) {
  for (int i = 0; i < V; i++) {
    for (int j = 0; j < V; j++) {
      if (dist[i][k] + dist[k][j] < dist[i][j]) {
        dist[i][j] = dist[i][k] + dist[k][j];
      }
    }
  }
}

Assume dist uses Integer.MAX_VALUE for infinity.

medium

A. Missing initialization of dist matrix

B. Loop variables are in wrong order

C. No check for Integer.MAX_VALUE before addition causing overflow

D. Should use '>' instead of '<' in if condition

Solution

1. Step 1: Understand addition with infinity

   Adding Integer.MAX_VALUE + any positive number causes overflow and incorrect results.

2. Step 2: Check code for safety

   Code lacks checks to ensure dist[i][k] and dist[k][j] are not Integer.MAX_VALUE before adding.

3. Final Answer:

   No check for Integer.MAX_VALUE before addition causing overflow -> Option C

4. Quick Check:

   Check for infinity before addition to avoid overflow [OK]

Hint: Always check for infinity before adding distances [OK]

Common Mistakes:

• Ignoring overflow from Integer.MAX_VALUE + number
• Assuming loops order causes error
• Confusing comparison operator direction

5. Given a graph with 4 vertices and edges:
0->1=5, 0->3=10, 1->2=3, 2->3=1
Use Floyd Warshall to find the shortest distance from vertex 0 to 3 after all updates.
What is the shortest distance from 0 to 3?

hard

A. 10

B. 9

C. 8

D. 7

Solution

1. Step 1: Initial distances

   Direct edges: 0->1=5, 0->3=10, 1->2=3, 2->3=1. Other distances are infinity or 0 on diagonal.

2. Step 2: Check indirect paths

   Path 0->1->2->3 distance = 5 + 3 + 1 = 9, which is less than direct 0->3=10, so update shortest distance to 9.

3. Final Answer:

   9 -> Option B

4. Quick Check:

   0->1->2->3 = 9 < 10 [OK]

Hint: Sum edges along path to find shortest distance [OK]

Common Mistakes:

• Ignoring multi-step paths
• Choosing direct edge without checking alternatives
• Adding edges incorrectly