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

Recall & Review

beginner

What problem does the Floyd Warshall algorithm solve?

It finds the shortest paths between all pairs of nodes in a weighted graph, including graphs with negative edge weights but no negative cycles.

intermediate

What is the main idea behind the Floyd Warshall algorithm?

It uses dynamic programming to update shortest paths by considering each node as an intermediate point, improving paths step-by-step until all shortest paths are found.

intermediate

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

It represents the shortest known distance from node i to node j considering only intermediate nodes from the set {1, 2, ..., k} at step k.

beginner

What is the time complexity of the Floyd Warshall algorithm?

The time complexity is O(n³), where n is the number of nodes in the graph, because it uses three nested loops over all nodes.

advanced

Can Floyd Warshall detect negative weight cycles? If yes, how?

Yes, by checking if any dist[i][i] becomes negative after the algorithm finishes, indicating a negative cycle reachable from node i.

What does Floyd Warshall algorithm compute?

A Shortest paths between all pairs of nodes

B Shortest path from a single source to all nodes

C Minimum spanning tree

D Topological sort

Floyd Warshall finds shortest paths between every pair of nodes.

Which of these is true about Floyd Warshall algorithm?

A It can handle negative edges but no negative cycles

B It works only on graphs without negative edges

C It finds shortest paths using BFS

D It uses a priority queue

Floyd Warshall can handle negative edges but fails if negative cycles exist.

What is the main data structure used in Floyd Warshall?

A Adjacency list

B Queue

C Stack

D Distance matrix

Floyd Warshall uses a matrix to store shortest distances between nodes.

What does a negative value in dist[i][i] after Floyd Warshall indicate?

A No path exists

B Graph is disconnected

C Negative weight cycle reachable from node i

D Algorithm failed

Negative dist[i][i] means a negative cycle involving node i.

What is the time complexity of Floyd Warshall algorithm?

A O(n log n)

B O(n³)

C O(n²)

D O(m + n)

Floyd Warshall uses three nested loops over n nodes, so O(n³).

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

Think about dynamic programming and intermediate nodes.

Describe how Floyd Warshall algorithm can detect negative weight cycles in a graph.

Focus on the meaning of dist[i][i] after algorithm finishes.

Practice

1. What does Floyd-Warshall compute that single-source shortest path algorithms like Dijkstra do not?

easy

A. Shortest path from one source to all vertices

B. Shortest paths between ALL pairs of vertices in one run

C. Minimum spanning tree of the graph

D. Topological order of vertices

Solution

1. Step 1: Single-source vs all-pairs

   Dijkstra/Bellman-Ford compute shortest paths from ONE source to all others. To get all-pairs with Dijkstra, run it V times = O(V(E + V log V)).

2. Step 2: Floyd-Warshall

   Floyd-Warshall computes shortest paths between every pair (i,j) in a single O(V³) pass. Simpler to implement for dense graphs.

3. Final Answer:

   All-pairs shortest paths in one pass. -> Option B

4. Quick Check:

   Floyd-Warshall = all-pairs. Dijkstra = single-source. [OK]

Hint: Floyd-Warshall = all-pairs shortest path. O(V³) time, O(V²) space. [OK]

Common Mistakes:

• Confusing with Dijkstra (single-source)
• Thinking it finds MST
• Confusing with topological sort

2. What is the Floyd-Warshall recurrence? dist[i][j] after considering vertex k as intermediate is:

medium

A. dist[i][j] = dist[i][k] + dist[k][j]

B. dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])

C. dist[i][j] = max(dist[i][j], dist[i][k] + dist[k][j])

D. dist[i][j] = dist[i][k] * dist[k][j]

Solution

1. Step 1: Core idea

   For each intermediate vertex k: is the path i→k→j shorter than the current best i→j?

2. Step 2: Recurrence

   dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]). Keep the minimum of: current best OR going through k.

3. Final Answer:

   min(dist[i][j], dist[i][k] + dist[k][j]) -> Option B

4. Quick Check:

   Always take the minimum - relaxation. [OK]

Hint: Floyd-Warshall: dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]). Relax via each intermediate k. [OK]

Common Mistakes:

• Using + without min (overwrites current best)
• Using max instead of min
• Multiplying instead of adding edge weights

3. In TypeScript, how do you initialize dist[i][i] and dist[i][j] (no direct edge) for Floyd-Warshall?

medium

A. dist[i][i] = 0, dist[i][j] = -1

B. dist[i][i] = 0, dist[i][j] = Infinity

C. dist[i][i] = 1, dist[i][j] = 0

D. dist[i][i] = Infinity, dist[i][j] = Infinity

Solution

1. Step 1: Self-distance

   dist[i][i] = 0 - zero cost to stay at the same vertex.

2. Step 2: No edge

   dist[i][j] = Infinity when no direct edge exists - represents unreachable initially. In TypeScript: Infinity works correctly with Math.min().

3. Final Answer:

   dist[i][i]=0, no edge=Infinity -> Option B

4. Quick Check:

   Self=0, no edge=Infinity. Infinity + x = Infinity in JS - works with min(). [OK]

Hint: Init: self=0, no edge=Infinity. TypeScript Infinity works correctly with Math.min(). [OK]

Common Mistakes:

• Using -1 for no edge (breaks min comparison)
• Setting self-distance to Infinity
• Using 0 for no edge (treats as zero-weight edge)

4. What is the correct loop order for Floyd-Warshall in TypeScript?

hard

A. for i → for j → for k

B. for k → for i → for j

C. for i → for k → for j

D. for j → for k → for i

Solution

1. Step 1: Why k is outermost

   k is the intermediate vertex being 'unlocked' at each outer iteration. When we process intermediate k, all pairs i→j must be relaxed using k.

2. Step 2: Correctness

   If k is innermost, we might use a later-unlocked vertex when relaxing - violating the DP ordering. k must be outermost so all (i,j) pairs are fully relaxed before moving to k+1.

3. Final Answer:

   k outermost, then i, then j. -> Option B

4. Quick Check:

   Floyd-Warshall loop order: k→i→j. k must be outermost. [OK]

Hint: Loop order: k (intermediate) → i (source) → j (dest). k must be outermost - critical for correctness. [OK]

Common Mistakes:

• Putting i outermost (most common mistake)
• Putting j outermost
• Any order other than k→i→j gives wrong results

5. How do you detect a negative cycle using Floyd-Warshall after completion?

hard

A. Check if any dist[i][j] === -Infinity

B. Check if any dist[i][i] < 0 - a vertex with negative self-distance indicates a negative cycle

C. Check if the total sum of all dist values is negative

D. Run the algorithm a second time and compare results

Solution

1. Step 1: Negative cycle effect

   If a negative cycle exists reachable from i back to i, the shortest path dist[i][i] will become negative (going around the cycle reduces cost below 0).

2. Step 2: Detection

   After the algorithm completes, check: for any i, if dist[i][i] < 0, a negative cycle exists involving i.

3. Final Answer:

   Check dist[i][i] < 0 for any i. -> Option B

4. Quick Check:

   Negative cycle: dist[i][i] < 0 after Floyd-Warshall. [OK]

Hint: Negative cycle detection: dist[i][i] < 0 after Floyd-Warshall. Self-distance going negative = cycle. [OK]

Common Mistakes:

• Checking dist[i][j] for non-self pairs
• Summing all distances
• Running algorithm twice