BFS Breadth First Search on Graph in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of BFS traversal on a simple graph

What is the output of the BFS traversal starting from node 0?

DSA Rust

use std::collections::{VecDeque, HashMap};

fn bfs(graph: &HashMap<i32, Vec<i32>>, start: i32) -> Vec<i32> {
    let mut visited = std::collections::HashSet::new();
    let mut queue = VecDeque::new();
    let mut order = Vec::new();

    visited.insert(start);
    queue.push_back(start);

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

fn main() {
    let mut graph = HashMap::new();
    graph.insert(0, vec![1, 2]);
    graph.insert(1, vec![2]);
    graph.insert(2, vec![0, 3]);
    graph.insert(3, vec![3]);

    let result = bfs(&graph, 0);
    println!("{:?}", result);
}

A [0, 1, 3, 2]

B [0, 1, 2, 3]

C [0, 2, 1, 3]

D [0, 3, 2, 1]

💡 Hint

Remember BFS visits neighbors level by level, starting from the start node.

BFS starts at node 0, visits neighbors 1 and 2 in order, then visits neighbors of 1 (which is 2, already visited), then neighbors of 2 (0 and 3), 0 is visited, so 3 is added last.

🧠 Conceptual

intermediate

BFS queue behavior on a cyclic graph

In BFS, what prevents infinite loops when traversing a graph with cycles?

A Sorting neighbors before visiting

B Limiting the depth of BFS traversal

C Using a visited set to track already visited nodes

D Using a stack instead of a queue

💡 Hint

Think about how BFS avoids revisiting the same node multiple times.

BFS uses a visited set to mark nodes already seen, so it does not enqueue them again, preventing infinite loops in cyclic graphs.

🔧 Debug

advanced

Identify the error in BFS implementation

What error will this BFS code produce when run?

DSA Rust

use std::collections::{VecDeque, HashMap};

fn bfs(graph: &HashMap<i32, Vec<i32>>, start: i32) -> Vec<i32> {
    let mut visited = std::collections::HashSet::new();
    let mut queue = VecDeque::new();
    let mut order = Vec::new();

    queue.push_back(start);

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

fn main() {
    let mut graph = HashMap::new();
    graph.insert(0, vec![1, 2]);
    graph.insert(1, vec![2]);
    graph.insert(2, vec![0, 3]);
    graph.insert(3, vec![3]);

    let result = bfs(&graph, 0);
    println!("{:?}", result);
}

A The BFS will run infinitely due to missing visited insertion for start node

B The code will not compile due to missing semicolon

C The BFS will return an empty vector

D The BFS will panic due to accessing a missing key

💡 Hint

Check if the start node is marked visited before starting the loop.

The start node is never added to the visited set before enqueueing, so it can be re-added multiple times causing infinite loop.

🚀 Application

advanced

Shortest path length using BFS

Given an unweighted graph, which BFS modification correctly computes the shortest path length from start to all nodes?

A Maintain a distance map updated when a node is first visited

B Count the number of nodes visited and divide by total nodes

C Use DFS instead of BFS for shortest path

D Store all paths and pick the shortest after traversal

💡 Hint

Think about how BFS layers correspond to distance from start.

BFS visits nodes in increasing order of distance from start. Recording distance when first visiting a node gives shortest path length.

❓ Predict Output

expert

Output of BFS on disconnected graph

What is the output of BFS starting from node 1 on this disconnected graph?

DSA Rust

use std::collections::{VecDeque, HashMap};

fn bfs(graph: &HashMap<i32, Vec<i32>>, start: i32) -> Vec<i32> {
    let mut visited = std::collections::HashSet::new();
    let mut queue = VecDeque::new();
    let mut order = Vec::new();

    visited.insert(start);
    queue.push_back(start);

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

fn main() {
    let mut graph = HashMap::new();
    graph.insert(0, vec![1]);
    graph.insert(1, vec![]);
    graph.insert(2, vec![3]);
    graph.insert(3, vec![]);

    let result = bfs(&graph, 1);
    println!("{:?}", result);
}

A [0, 1]

B [1, 0]

C [1, 2, 3]

D [1]

💡 Hint

Consider which nodes are reachable from the start node.

Starting from node 1, which has no neighbors, BFS visits only node 1.

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