Floyd Warshall All Pairs Shortest Path in DSA Typescript

Mental Model

Find the shortest paths between every pair of points by checking if going through a middle point is better than going directly.

Analogy: Imagine you want to find the quickest way to travel between every pair of cities. Sometimes going through a third city is faster than going straight. You check all possible middle cities to find the best routes.

Graph with 4 nodes:
  (0)---5---(1)
   | \       |
   10  3     1
   |     \   |
  (3)---2---(2)

Distance matrix before:
[0, 5, 3, 10]
[∞, 0, 1, ∞]
[∞, ∞, 0, 2]
[∞, ∞, ∞, 0]

Dry Run Walkthrough

Input: Graph with 4 nodes and edges: 0->1=5, 0->3=10, 0->2=3, 1->2=1, 2->3=2; find shortest paths between all pairs

Goal: Calculate shortest distances between every pair of nodes using Floyd Warshall algorithm

Step 1: Initialize distance matrix with direct edges and 0 for self loops

[[0, 5, 3, 10],
 [∞, 0, 1, ∞],
 [∞, ∞, 0, 2],
 [∞, ∞, ∞, 0]]

Why: Start with known direct distances and zero for same node

Step 2: Use node 0 as intermediate to update distances

[[0, 5, 3, 10],
 [∞, 0, 1, ∞],
 [∞, ∞, 0, 2],
 [∞, ∞, ∞, 0]]

Why: No shorter paths found through node 0 yet

Step 3: Use node 1 as intermediate to update distances

[[0, 5, 3, 10],
 [∞, 0, 1, ∞],
 [∞, ∞, 0, 2],
 [∞, ∞, ∞, 0]]

Why: 0->1->2: 5 + 1 = 6 > 3 (direct), so no update. No shorter paths found through node 1.

Step 4: Use node 2 as intermediate to update distances

[[0, 5, 3, 5],
 [∞, 0, 1, 3],
 [∞, ∞, 0, 2],
 [∞, ∞, ∞, 0]]

Why: 0->2->3 is shorter (3 + 2 = 5) than 0->3 direct (10), update 0->3 to 5; 1->2->3 is shorter (1 + 2 = 3) than ∞, update 1->3 to 3

Step 5: Use node 3 as intermediate to update distances

[[0, 5, 3, 5],
 [∞, 0, 1, 3],
 [∞, ∞, 0, 2],
 [∞, ∞, ∞, 0]]

Why: No shorter paths found through node 3

Result:

[[0, 5, 3, 5],
 [∞, 0, 1, 3],
 [∞, ∞, 0, 2],
 [∞, ∞, ∞, 0]]

Annotated Code

DSA Typescript

class FloydWarshall {
  static INF = Number.MAX_SAFE_INTEGER;

  static floydWarshall(graph: number[][]): number[][] {
    const n = graph.length;
    const dist = Array.from({ length: n }, (_, i) =>
      Array.from({ length: n }, (_, j) => graph[i][j])
    );

    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] !== FloydWarshall.INF && dist[k][j] !== FloydWarshall.INF) {
            if (dist[i][k] + dist[k][j] < dist[i][j]) {
              dist[i][j] = dist[i][k] + dist[k][j];
            }
          }
        }
      }
    }
    return dist;
  }
}

// Driver code
const INF = FloydWarshall.INF;
const graph = [
  [0, 5, 3, 10],
  [INF, 0, 1, INF],
  [INF, INF, 0, 2],
  [INF, INF, INF, 0],
];

const result = FloydWarshall.floydWarshall(graph);
for (const row of result) {
  console.log(row.map(x => (x === INF ? '∞' : x)).join(' '));
}

for (let k = 0; k < n; k++) {

Use each node as intermediate to check for shorter paths

if (dist[i][k] !== FloydWarshall.INF && dist[k][j] !== FloydWarshall.INF) {

Check if path through k is possible

if (dist[i][k] + dist[k][j] < dist[i][j]) {

Update distance if going through k is shorter

dist[i][j] = dist[i][k] + dist[k][j];

Set new shorter distance

Output

0 5 3 5

∞ 0 1 3

∞ ∞ 0 2

∞ ∞ ∞ 0

Complexity Analysis

Time: O(n^3) because we check all pairs (i, j) for every intermediate node k

Space: O(n^2) because we store distances between all pairs in a matrix

vs Alternative: Compared to running Dijkstra from each node (O(n * (E + V log V))), Floyd Warshall is simpler but slower for sparse graphs

Edge Cases

Graph with no edges except self loops

Distances remain zero on diagonal and infinity elsewhere

DSA Typescript

if (dist[i][k] !== FloydWarshall.INF && dist[k][j] !== FloydWarshall.INF) {

Graph with negative edge weights but no negative cycles

Algorithm correctly finds shortest paths including negative edges

DSA Typescript

if (dist[i][k] + dist[k][j] < dist[i][j]) {

Graph with negative cycle

Algorithm does not detect negative cycles; distances may become incorrect or negative infinity

DSA Typescript

No explicit guard; Floyd Warshall does not handle negative cycle detection here

When to Use This Pattern

When you need shortest paths between all pairs of points and the graph may have negative edges but no negative cycles, use Floyd Warshall to systematically check all intermediate nodes.

Common Mistakes

Mistake: Not checking if intermediate paths exist before adding distances, causing overflow or wrong results
Fix: Add condition to check dist[i][k] and dist[k][j] are not infinity before summing

Mistake: Updating distances without comparing if new path is shorter
Fix: Only update if dist[i][k] + dist[k][j] < dist[i][j]

Summary

Finds shortest paths between all pairs of nodes by considering each node as an intermediate step.

Use when you need all pairs shortest paths and the graph may have negative edges but no negative cycles.

The key insight is that any shortest path can be built by combining shorter paths through intermediate nodes.