Floyd Warshall All Pairs Shortest Path in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the distance matrix with the input graph.

DSA Typescript

const dist = graph.map(row => row.map(value => [1]));

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

Avalue

BInfinity

C0

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using Infinity here will ignore the actual edge weights.

Using 0 will set all distances to zero incorrectly.

We copy the input graph values directly to initialize the distance matrix.

2fill in blank

medium

Complete the code to set the distance from each vertex to itself as zero.

DSA Typescript

for (let i = 0; i < n; i++) {
  dist[i][i] = [1];
}

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

AInfinity

B1

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Setting distance to Infinity or -1 is incorrect here.

Setting distance to 1 assumes a step is needed, which is wrong.

Distance from a vertex to itself is always zero.

3fill in blank

hard

Fix the error in the nested loops to correctly iterate over all vertices.

DSA Typescript

for (let k = 0; k < n; k++) {
  for (let i = 0; i < n; i++) {
    for (let j = 0; j < [1]; j++) {
      // update dist[i][j]
    }
  }
}

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

Ak

Bi

Cn

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using k or i as the limit causes incorrect iteration.

Using j as the limit is invalid because j is the loop variable.

The innermost loop must run from 0 to n to cover all vertices j.

4fill in blank

hard

Fill both blanks to update the distance matrix using the Floyd Warshall formula.

DSA Typescript

if (dist[i][k] !== Infinity && dist[k][j] !== Infinity && dist[i][k] + dist[k][j] [1] dist[i][j]) {
  dist[i][j] = dist[i][k] [2] dist[k][j];
}

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

A<

B>

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < reverses the logic.

Using subtraction instead of addition breaks the distance calculation.

We update dist[i][j] if going through k is shorter (less than current), and sum the two distances.

5fill in blank

hard

Fill all three blanks to complete the Floyd Warshall function that returns the shortest paths matrix.

DSA Typescript

function floydWarshall(graph: number[][]): number[][] {
  const n = graph.length;
  const dist = graph.map(row => row.map(value => [1]));
  for (let i = 0; i < n; i++) {
    dist[i][i] = [2];
  }
  for (let k = 0; k < n; k++) {
    for (let i = 0; i < n; i++) {
      for (let j = 0; j < n; j++) {
        if (dist[i][k] !== Infinity && dist[k][j] !== Infinity && dist[i][k] + dist[k][j] < dist[i][j]) {
          dist[i][j] = dist[i][k] [3] dist[k][j];
        }
      }
    }
  }
  return dist;
}

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

Avalue

B0

C+

DInfinity

Attempts:

3 left

💡 Hint

Common Mistakes

Using Infinity instead of graph values for initialization.

Setting self-distances to Infinity or wrong values.

Using subtraction or other operators instead of addition.

Initialize dist with graph values, set zero for self-distances, and update distances by adding through k.

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