Floyd Warshall All Pairs Shortest Path in DSA Javascript - Time & Space Complexity

Time Complexity: Floyd Warshall All Pairs Shortest Path

O(n³)

Understanding Time Complexity

We want to understand how the time needed to find shortest paths between all pairs of points grows as the number of points increases.

How does the running time change when we add more points to the map?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

const floydWarshall = (graph) => {
  const dist = graph.map(row => row.slice());
  const n = graph.length;
  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] + dist[k][j] < dist[i][j]) {
          dist[i][j] = dist[i][k] + dist[k][j];
        }
      }
    }
  }
  return dist;
};
    

This code finds the shortest distance between every pair of points in a graph using three nested loops.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The triple nested loops that check and update distances between all pairs through an intermediate point.
• How many times: Each loop runs n times, so the inner operation runs n x n x n = n³ times.

How Execution Grows With Input

As the number of points (n) increases, the number of operations grows very fast because we check every pair through every possible intermediate point.

Input Size (n)	Approx. Operations
10	1,000
100	1,000,000
1000	1,000,000,000

Pattern observation: When n grows 10 times, operations grow 1000 times, showing a cubic growth pattern.

Final Time Complexity

Time Complexity: O(n³)

This means the time needed grows roughly with the cube of the number of points, so adding more points makes the work much heavier.

Common Mistake

[X] Wrong: "Since there are three loops, the time complexity is just O(n) because loops run one after another."

[OK] Correct: The loops are nested inside each other, so the total work multiplies, not adds, making it O(n³), not O(n).

Interview Connect

Understanding this cubic time complexity helps you explain why Floyd Warshall is good for small graphs but not for very large ones, showing your grasp of algorithm efficiency.

Self-Check

"What if we replaced the triple nested loops with a recursive approach that splits the graph? How might that affect the time complexity?"

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