BFS Breadth First Search on Graph in DSA Rust - Time & Space Complexity

Time Complexity: BFS Breadth First Search on Graph

O(n + m)

Understanding Time Complexity

We want to understand how the time needed to explore a graph grows as the graph gets bigger.

How does BFS handle more nodes and edges in the graph?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn bfs(graph: &Vec, start: usize) {
    let mut visited = vec![false; graph.len()];
    let mut queue = std::collections::VecDeque::new();
    visited[start] = true;
    queue.push_back(start);

    while let Some(node) = queue.pop_front() {
        for &neighbor in &graph[node] {
            if !visited[neighbor] {
                visited[neighbor] = true;
                queue.push_back(neighbor);
            }
        }
    }
}
    

This code visits all nodes reachable from the start node using BFS, exploring neighbors level by level.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop runs as long as there are nodes in the queue, and inside it, the for loop visits all neighbors of the current node.
• How many times: Each node is enqueued and dequeued once, and each edge is checked once during neighbor visits.

How Execution Grows With Input

As the graph grows, BFS visits every node and edge once.

Input Size (n nodes, m edges)	Approx. Operations
10 nodes, 15 edges	About 10 node visits + 15 edge checks = 25 operations
100 nodes, 200 edges	About 100 node visits + 200 edge checks = 300 operations
1000 nodes, 5000 edges	About 1000 node visits + 5000 edge checks = 6000 operations

Pattern observation: The total work grows roughly in proportion to the number of nodes plus edges.

Final Time Complexity

Time Complexity: O(n + m)

This means BFS takes time proportional to the total number of nodes plus edges in the graph.

Common Mistake

[X] Wrong: "BFS always takes O(n²) time because it uses nested loops."

[OK] Correct: BFS visits each node and edge only once, so it does not repeatedly check all pairs. The loops work together to cover all nodes and edges efficiently.

Interview Connect

Understanding BFS time complexity helps you explain how graph searches scale and why BFS is efficient for many problems involving shortest paths or connectivity.

Self-Check

"What if the graph is represented as an adjacency matrix instead of adjacency lists? How would the time complexity change?"

Practice

1. What does BFS (Breadth First Search) primarily use to keep track of nodes to visit next in a graph?

easy

A. A hash map

B. A stack data structure

C. A priority queue

D. A queue data structure

Solution

1. Step 1: Understand BFS traversal method

   BFS explores nodes level by level, visiting all neighbors before moving deeper.

2. Step 2: Identify the data structure used

   To achieve level order traversal, BFS uses a queue to process nodes in the order they are discovered.

3. Final Answer:

   A queue data structure -> Option D

4. Quick Check:

   BFS uses queue = C [OK]

Hint: BFS uses queue to explore neighbors level-wise [OK]

Common Mistakes:

• Confusing BFS with DFS which uses stack
• Thinking BFS uses priority queue
• Assuming BFS uses hash map for traversal order

2. Which of the following Rust code snippets correctly initializes a queue for BFS traversal using the standard library?

easy

A. let mut queue = HashMap::new();

B. let mut queue = Vec::new();

C. let mut queue = VecDeque::new();

D. let mut queue = BinaryHeap::new();

Solution

1. Step 1: Identify the correct queue type in Rust

   Rust's standard library provides VecDeque for efficient queue operations (push_back, pop_front).

2. Step 2: Check each option

   let mut queue = VecDeque::new(); uses VecDeque::new(), which is correct. Vec is a dynamic array, not ideal for queue. HashMap and BinaryHeap are not queues.

3. Final Answer:

   let mut queue = VecDeque::new(); -> Option C

4. Quick Check:

   VecDeque is queue in Rust [OK]

Hint: Use VecDeque for queue in Rust BFS [OK]

Common Mistakes:

• Using Vec which is not efficient for queue
• Using HashMap which is for key-value pairs
• Using BinaryHeap which is a priority queue

3. Given the following Rust code for BFS on a graph represented as adjacency list, what is the order of nodes visited starting from node 0?

use std::collections::VecDeque;

fn bfs(graph: &Vec, start: usize) -> Vec<usize> {
    let mut visited = vec![false; graph.len()];
    let mut queue = VecDeque::new();
    let mut order = Vec::new();

    visited[start] = true;
    queue.push_back(start);

    while let Some(node) = queue.pop_front() {
        order.push(node);
        for &neighbor in &graph[node] {
            if !visited[neighbor] {
                visited[neighbor] = true;
                queue.push_back(neighbor);
            }
        }
    }
    order
}

fn main() {
    let graph = vec![
        vec![1, 2],
        vec![0, 3],
        vec![0, 4],
        vec![1],
        vec![2]
    ];
    let result = bfs(&graph, 0);
    println!("{:?}", result);
}

medium

A. [0, 1, 2, 3, 4]

B. [0, 2, 1, 4, 3]

C. [0, 1, 3, 2, 4]

D. [0, 3, 1, 4, 2]

Solution

1. Step 1: Start BFS from node 0

   Visit node 0, enqueue neighbors 1 and 2.

2. Step 2: Visit nodes in queue order

   Dequeue 1, visit it, enqueue neighbor 3 (0 already visited). Dequeue 2, visit it, enqueue neighbor 4.

3. Step 3: Continue visiting remaining nodes

   Dequeue 3, visit it (neighbor 1 visited). Dequeue 4, visit it (neighbor 2 visited).

4. Final Answer:

   [0, 1, 2, 3, 4] -> Option A

5. Quick Check:

   BFS order from 0 = [0,1,2,3,4] [OK]

Hint: BFS visits neighbors level-wise in order discovered [OK]

Common Mistakes:

• Mixing BFS with DFS order
• Visiting neighbors out of order
• Forgetting to mark visited nodes

4. The following Rust BFS code has a bug causing it to panic at runtime. What is the error?

use std::collections::VecDeque;

fn bfs(graph: &Vec, start: usize) -> Vec<usize> {
    let mut visited = vec![false; graph.len()];
    let mut queue = VecDeque::new();
    let mut order = Vec::new();

    visited[start] = true;
    queue.push_back(start);

    while let Some(node) = queue.pop_front() {
        order.push(node);
        for &neighbor in &graph[node] {
            if !visited[neighbor] {
                queue.push_back(neighbor);
            }
        }
    }
    order
}

fn main() {
    let graph = vec![
        vec![1, 2],
        vec![0, 3],
        vec![0, 4],
        vec![1],
        vec![2]
    ];
    let result = bfs(&graph, 0);
    println!("{:?}", result);
}

medium

A. Missing visited[neighbor] = true; causes infinite loop

B. Queue is not initialized properly

C. Graph adjacency list is empty

D. Start node index is out of bounds

Solution

1. Step 1: Analyze BFS loop and visited marking

   The code pushes neighbors to queue but never marks them visited.

2. Step 2: Understand consequence of missing visited update

   Without marking visited, neighbors get re-added repeatedly causing infinite loop or panic.

3. Final Answer:

   Missing visited[neighbor] = true; causes infinite loop -> Option A

4. Quick Check:

   Mark visited before enqueue = prevent infinite loop [OK]

Hint: Always mark visited before enqueueing neighbors [OK]

Common Mistakes:

• Forgetting to mark nodes visited before enqueue
• Assuming queue initialization causes panic
• Ignoring start node validity

5. You want to find the shortest path length from node 0 to node 4 in an unweighted graph using BFS. Which modification to the BFS code below will correctly compute the shortest distance?

use std::collections::VecDeque;

fn shortest_path_length(graph: &Vec, start: usize, target: usize) -> Option<usize> {
    let mut visited = vec![false; graph.len()];
    let mut distance = vec![0; graph.len()];
    let mut queue = VecDeque::new();

    visited[start] = true;
    queue.push_back(start);

    while let Some(node) = queue.pop_front() {
        if node == target {
            return Some(distance[node]);
        }
        for &neighbor in &graph[node] {
            if !visited[neighbor] {
                visited[neighbor] = true;
                queue.push_back(neighbor);
                // What to add here?
            }
        }
    }
    None
}

fn main() {
    let graph = vec![
        vec![1, 2],
        vec![0, 3],
        vec![0, 4],
        vec![1],
        vec![2]
    ];
    let result = shortest_path_length(&graph, 0, 4);
    println!("{:?}", result);
}

hard

A. distance[neighbor] = 1;

B. distance[neighbor] = distance[node] + 1;

C. distance[neighbor] += 1;

D. distance[neighbor] = distance[node];

Solution

1. Step 1: Understand BFS distance tracking

   Distance to neighbor is distance to current node plus one edge.

2. Step 2: Choose correct distance update

   Assign distance[neighbor] = distance[node] + 1 to track shortest path length.

3. Final Answer:

   distance[neighbor] = distance[node] + 1; -> Option B

4. Quick Check:

   Distance increments by 1 per edge [OK]

Hint: Set neighbor distance = current node distance + 1 [OK]

Common Mistakes:

• Not updating distance correctly
• Using += 1 without initialization
• Setting distance to constant 1 always