Floyd Warshall All Pairs Shortest Path in DSA Rust - 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 work change when we add more points to the map?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn floyd_warshall(graph: &mut Vec<Vec<i32>>) {
    let n = graph.len();
    for k in 0..n {
        for i in 0..n {
            for j in 0..n {
                if graph[i][k] != i32::MAX && graph[k][j] != i32::MAX && graph[i][k] + graph[k][j] < graph[i][j] {
                    graph[i][j] = graph[i][k] + graph[k][j];
                }
            }
        }
    }
}

This code updates the shortest distances between every pair of points by checking if going through an intermediate point is shorter.

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 each intermediate point.
• How many times: Each loop runs from 0 to n-1, so the innermost operation runs about n x n x n = n³ times.

How Execution Grows With Input

As the number of points (n) grows, the number of checks grows much faster because we check every pair through every intermediate point.

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

Pattern observation: The work grows by the cube of n, so doubling n makes the work about eight times bigger.

Final Time Complexity

Time Complexity: O(n³)

This means the time needed grows very quickly as the number of points increases, because we check all pairs through all intermediate points.

Common Mistake

[X] Wrong: "Since we only have three loops, the time complexity should be O(n)."

[OK] Correct: Each loop runs n times, so the total work multiplies, making it n x n x n = n³, not just n.

Interview Connect

Understanding this cubic growth 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 used a graph with fewer edges and skipped some checks? How would the time complexity change?"

Practice

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

easy

A. To perform a depth-first search traversal

B. To find the minimum spanning tree of a graph

C. To detect cycles in a directed graph

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

Solution

1. Step 1: Understand Floyd Warshall algorithm goal

   The Floyd Warshall algorithm is designed to find shortest paths between every pair of vertices in a weighted graph, including negative weights but no negative cycles.

2. Step 2: Compare with other graph algorithms

   Minimum spanning tree algorithms find a tree connecting all vertices with minimum total edge weight, cycle detection finds cycles, and DFS is a traversal method. These are different from Floyd Warshall's all pairs shortest path goal.

3. Final Answer:

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

4. Quick Check:

   Floyd Warshall = All pairs shortest paths [OK]

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

Common Mistakes:

• Confusing with minimum spanning tree algorithms
• Thinking it only finds shortest path from one source
• Mixing with cycle detection or traversal algorithms

2. Which of the following Rust code snippets correctly initializes the distance matrix for Floyd Warshall with 3 vertices and no edges (infinity represented by i32::MAX)?

easy

A. let mut dist = vec![vec![0; 3]; 3]; for i in 0..3 { dist[i][i] = i32::MAX; }

B. let mut dist = vec![vec![i32::MAX; 3]; 3]; for i in 0..3 { dist[i][i] = 0; }

C. let mut dist = vec![vec![i32::MAX; 3]; 3]; for i in 0..3 { dist[i][i] = i32::MAX; }

D. let mut dist = vec![vec![0; 3]; 3]; for i in 0..3 { dist[i][i] = 0; }

Solution

1. Step 1: Initialize matrix with infinity

   We start with all distances as infinity (i32::MAX) because no edges exist initially.

2. Step 2: Set distance from vertex to itself as zero

   Distance from a vertex to itself is zero, so dist[i][i] = 0 for all i.

3. Final Answer:

   let mut dist = vec![vec![i32::MAX; 3]; 3]; for i in 0..3 { dist[i][i] = 0; } -> Option B

4. Quick Check:

   Init with i32::MAX, diagonal zero [OK]

Hint: Set diagonal to zero, others to infinity for dist matrix [OK]

Common Mistakes:

• Setting diagonal to infinity instead of zero
• Initializing with zeros everywhere
• Mixing up i32::MAX and 0 values

3. Given the following Rust code snippet implementing Floyd Warshall on a graph with 3 vertices and edges: 0->1 (4), 1->2 (3), 0->2 (10), what is the shortest distance from vertex 0 to vertex 2 after running the algorithm?

let n = 3;
let mut dist = vec![vec![i32::MAX; n]; n];
for i in 0..n { dist[i][i] = 0; }
dist[0][1] = 4;
dist[1][2] = 3;
dist[0][2] = 10;

for k in 0..n {
    for i in 0..n {
        for j in 0..n {
            if dist[i][k] != i32::MAX && dist[k][j] != i32::MAX {
                dist[i][j] = dist[i][j].min(dist[i][k] + dist[k][j]);
            }
        }
    }
}

println!("{}", dist[0][2]);

medium

A. 7

B. 10

C. 4

D. 3

Solution

1. Step 1: Initial distances

   dist[0][2] starts at 10, dist[0][1] = 4, dist[1][2] = 3.

2. Step 2: Update via intermediate vertex 1

   Check if path 0->1->2 is shorter: 4 + 3 = 7, which is less than 10, so update dist[0][2] to 7.

3. Final Answer:

   7 -> Option A

4. Quick Check:

   0->1->2 = 7 < 10 [OK]

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

Common Mistakes:

• Not updating dist[0][2] via vertex 1
• Confusing initial edge weights with final shortest paths
• Ignoring i32::MAX checks causing overflow

4. Identify the bug in this Rust Floyd Warshall implementation snippet that causes incorrect shortest path results:

for k in 0..n {
    for i in 0..n {
        for j in 0..n {
            if dist[i][k] + dist[k][j] < dist[i][j] {
                dist[i][j] = dist[i][k] + dist[k][j];
            }
        }
    }
}

medium

A. Using '<' instead of '<=' in comparison

B. Incorrect loop order; should be i, j, k

C. Missing check for i32::MAX before addition causing overflow

D. Not initializing diagonal elements to zero

Solution

1. Step 1: Analyze addition of distances

   Adding dist[i][k] + dist[k][j] without checking if either is i32::MAX can cause integer overflow.

2. Step 2: Importance of checks before addition

   Proper Floyd Warshall requires checking dist[i][k] != i32::MAX and dist[k][j] != i32::MAX before adding to avoid overflow and incorrect updates.

3. Final Answer:

   Missing check for i32::MAX before addition causing overflow -> Option C

4. Quick Check:

   Check for infinity before addition to avoid overflow [OK]

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

Common Mistakes:

• Ignoring overflow from adding i32::MAX
• Changing loop order unnecessarily
• Confusing comparison operators
• Forgetting to initialize diagonal distances

5. You have a graph with 4 vertices and edges: 0->1 (5), 1->2 (2), 2->3 (1), and 0->3 (20). Using Floyd Warshall, what is the shortest distance from vertex 0 to vertex 3 after the algorithm completes?

hard

A. 8

B. 20

C. 5

D. 10

Solution

1. Step 1: Initial distances

   Direct edge 0->3 is 20. Other edges: 0->1=5, 1->2=2, 2->3=1.

2. Step 2: Calculate path 0->1->2->3

   Sum is 5 + 2 + 1 = 8, which is less than 20, so shortest path updates to 8.

3. Final Answer:

   8 -> Option A

4. Quick Check:

   0->1->2->3 = 8 < 20 [OK]

Hint: Sum edges along path to find shorter distance than direct edge [OK]

Common Mistakes:

• Choosing direct edge weight without checking intermediate paths
• Adding edges incorrectly
• Ignoring path through multiple vertices