Floyd Warshall All Pairs Shortest Path in DSA Typescript - Build from Scratch

Floyd Warshall All Pairs Shortest Path

📖 Scenario: You are working as a city planner. You have a map of cities connected by roads with distances. You want to find the shortest distance between every pair of cities to plan the best routes.

🎯 Goal: Build a program that uses the Floyd Warshall algorithm to find the shortest paths between all pairs of cities in a given map.

📋 What You'll Learn

Create a 2D array called dist representing the distances between cities.

Create a variable numCities to store the number of cities.

Implement the Floyd Warshall algorithm using three nested for loops with variables k, i, and j.

Update the dist array with the shortest distances.

Print the final dist array showing shortest distances between all city pairs.

💡 Why This Matters

🌍 Real World

City planners and GPS systems use all pairs shortest path algorithms to find the best routes between locations.

💼 Career

Understanding Floyd Warshall helps in roles involving route optimization, network analysis, and logistics planning.

1

Create the initial distance matrix

Create a 2D array called dist with these exact values representing distances between 4 cities. Use Infinity where there is no direct road: [[0, 5, Infinity, 10], [Infinity, 0, 3, Infinity], [Infinity, Infinity, 0, 1], [Infinity, Infinity, Infinity, 0]]

DSA Typescript

// Create the 2D array dist with the given distances
// Your code here

Hint

Use const dist: number[][] = [...] to create the matrix exactly as shown.

2

Set the number of cities

Create a variable called numCities and set it to the number of cities in the dist array.

DSA Typescript

const dist: number[][] = [
  [0, 5, Infinity, 10],
  [Infinity, 0, 3, Infinity],
  [Infinity, Infinity, 0, 1],
  [Infinity, Infinity, Infinity, 0]
];
// Create numCities and set it to the length of dist
// Your code here

Hint

Use const numCities: number = dist.length; to get the number of cities.

3

Implement the Floyd Warshall algorithm

Use three nested for loops with variables k, i, and j from 0 to numCities. Inside the innermost loop, update dist[i][j] to the minimum of its current value and dist[i][k] + dist[k][j].

DSA Typescript

const dist: number[][] = [
  [0, 5, Infinity, 10],
  [Infinity, 0, 3, Infinity],
  [Infinity, Infinity, 0, 1],
  [Infinity, Infinity, Infinity, 0]
];

const numCities: number = dist.length;

// Implement Floyd Warshall algorithm using for loops
// Your code here

Hint

Use three nested loops and update dist[i][j] if a shorter path is found through k.

4

Print the shortest distance matrix

Print the dist array after running the Floyd Warshall algorithm. Use console.log(dist) to display the final shortest distances between all pairs of cities.

DSA Typescript

const dist: number[][] = [
  [0, 5, Infinity, 10],
  [Infinity, 0, 3, Infinity],
  [Infinity, Infinity, 0, 1],
  [Infinity, Infinity, Infinity, 0]
];

const numCities: number = dist.length;

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

// Print the dist array
// Your code here

Hint

Use console.log(dist) to print the final matrix.

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