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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

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

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

Avalue

Bvalue === 0 ? Infinity : 0

Cvalue === 0 ? 0 : Infinity

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 for no edge instead of Infinity.

Not handling zero distance for same node.

Copy the graph values to dist. The graph has 0 for same node distance and Infinity for no direct edge.

2fill in blank

medium

Complete the code to update the distance using the intermediate node k.

DSA Javascript

if (dist[i][k] !== Infinity && dist[k][j] !== Infinity) {
  dist[i][j] = Math.min(dist[i][j], dist[i][k] [1] dist[k][j]);
}

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

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

Not checking for Infinity before adding.

We add distances from i to k and k to j to check if path is shorter.

3fill in blank

hard

Fix the error in the nested loops for Floyd Warshall algorithm.

DSA Javascript

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'

An

Bk

Ci

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong loop variable as limit.

Not iterating over all nodes j.

The inner loop must run from 0 to n to cover all nodes j.

4fill in blank

hard

Fill both blanks to complete the condition that checks if a shorter path exists.

DSA Javascript

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

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

Ak

Bi

Cj

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong indices in condition.

Not checking both parts of the path.

We check distances from i to k and k to j to update dist[i][j].

5fill in blank

hard

Fill all three blanks to create the final nested loops for Floyd Warshall algorithm.

DSA Javascript

for (let [1] = 0; [1] < n; [1]++) {
  for (let [2] = 0; [2] < n; [2]++) {
    for (let [3] = 0; [3] < n; [3]++) {
      if (dist[[2]][[1]] !== Infinity && dist[[1]][[3]] !== Infinity) {
        dist[[2]][[3]] = Math.min(dist[[2]][[3]], dist[[2]][[1]] + dist[[1]][[3]]);
      }
    }
  }
}

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

Ak

Bi

Cj

Dm

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing variable names or using undefined variables.

Incorrect loop order.

Standard variable names: k for intermediate, i and j for nodes.

Practice

1. What is the main purpose of the Floyd Warshall algorithm in graph theory?

easy

A. To detect cycles in a graph

B. To find the shortest path from a single source to all other vertices

C. To find the shortest paths between all pairs of vertices in a weighted graph

D. To sort vertices in topological order

Solution

1. Step 1: Understand Floyd Warshall's goal

   The Floyd Warshall algorithm calculates shortest paths between every pair of vertices in a weighted graph.

2. Step 2: Differentiate from other algorithms

   Unlike Dijkstra's algorithm which finds shortest paths from one source, Floyd Warshall finds paths for all pairs.

3. Final Answer:

   To find the shortest paths between all pairs of vertices in a weighted graph -> Option C

4. Quick Check:

   All pairs shortest path = Floyd Warshall [OK]

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

Common Mistakes:

• Confusing with Dijkstra's single source shortest path
• Thinking it detects cycles
• Assuming it sorts vertices

2. Which of the following is the correct way to initialize the distance matrix dist for Floyd Warshall in JavaScript given a graph adjacency matrix graph?

easy

A. const dist = graph.map(row => row.slice());

B. const dist = graph.slice();

C. const dist = graph.map(row => row);

D. const dist = [...graph];

Solution

1. Step 1: Understand deep copy necessity

   We need a deep copy of the 2D array so changes to dist do not affect graph.

2. Step 2: Check each option for deep copy

   const dist = graph.map(row => row.slice()); uses map and slice to copy each row array, creating a deep copy. Others create shallow copies.

3. Final Answer:

   const dist = graph.map(row => row.slice()); -> Option A

4. Quick Check:

   Deep copy with map and slice = const dist = graph.map(row => row.slice()); [OK]

Hint: Use map with slice for deep copy of 2D arrays [OK]

Common Mistakes:

• Using shallow copy causing original graph to change
• Using spread operator which copies only top-level array
• Using map without slice copying references

3. What will be the output distance matrix after running Floyd Warshall on the following graph adjacency matrix where Infinity means no direct edge?

const graph = [
  [0, 3, Infinity],
  [2, 0, 5],
  [Infinity, 1, 0]
];

medium

A. [[0, 3, 7], [2, 0, 5], [3, 1, 0]]

B. [[0, 3, 5], [2, 0, 5], [Infinity, 1, 0]]

C. [[0, 3, Infinity], [2, 0, 5], [3, 1, 0]]

D. [[0, 3, 8], [2, 0, 5], [3, 1, 0]]

Solution

1. Step 1: Initialize dist matrix same as graph

   dist = [[0,3,∞],[2,0,5],[∞,1,0]]

2. Step 2: Update dist using Floyd Warshall iterations

   Check paths through each vertex k=0,1,2:
   k=0: dist[2][0] = min(∞, dist[2][0]+dist[0][0])=∞ no change
   k=1: dist[0][2] = min(∞, dist[0][1]+dist[1][2])=3+5=8
   dist[2][0] = min(∞, dist[2][1]+dist[1][0])=1+2=3
   k=2: no shorter paths found
   Final dist = [[0,3,8],[2,0,5],[3,1,0]]

3. Final Answer:

   [[0, 3, 8], [2, 0, 5], [3, 1, 0]] -> Option D

4. Quick Check:

   Updated shortest paths matrix = [[0, 3, 8], [2, 0, 5], [3, 1, 0]] [OK]

Hint: Check all intermediate vertices to update shortest paths [OK]

Common Mistakes:

• Not updating paths through intermediate vertices
• Confusing Infinity with zero
• Missing updates for indirect paths

4. Identify the error in this Floyd Warshall implementation snippet:

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

medium

A. Missing check if dist[i][k] or dist[k][j] is Infinity before addition

B. The loops should be reversed: i, j, then k

C. No error, the code is correct

D. The condition should use <= instead of <

Solution

1. Step 1: Analyze addition of distances

   Adding dist[i][k] + dist[k][j] without checking if either is Infinity can cause incorrect results or overflow.

2. Step 2: Identify missing Infinity checks

   Before adding, must ensure dist[i][k] !== Infinity and dist[k][j] !== Infinity to avoid invalid sums.

3. Final Answer:

   Missing check if dist[i][k] or dist[k][j] is Infinity before addition -> Option A

4. Quick Check:

   Check Infinity before sum to avoid errors [OK]

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

Common Mistakes:

• Ignoring Infinity causing wrong updates
• Changing loop order unnecessarily
• Using <= instead of < causing redundant updates

5. Given a weighted directed graph with possible negative edges but no negative cycles, how does Floyd Warshall detect the presence of a negative cycle after running the algorithm?

hard

A. By checking if any dist[i][j] is Infinity

B. By checking if any diagonal element dist[i][i] becomes negative

C. By verifying if the sum of all distances is zero

D. By checking if any dist[i][j] is negative for i != j

Solution

1. Step 1: Understand negative cycle effect on dist matrix

   Negative cycles cause the shortest path from a vertex to itself to become negative.

2. Step 2: Check diagonal elements after Floyd Warshall

   If any dist[i][i] < 0, it means a negative cycle exists involving vertex i.

3. Final Answer:

   By checking if any diagonal element dist[i][i] becomes negative -> Option B

4. Quick Check:

   Negative cycle detected if dist[i][i] < 0 [OK]

Hint: Check diagonal dist[i][i] for negativity to find negative cycles [OK]

Common Mistakes:

• Checking off-diagonal elements for negative cycles
• Confusing Infinity with negative cycles
• Assuming sum of distances indicates cycles