DFS Depth First Search on Graph in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start DFS from the given node.

DSA Rust

fn dfs(graph: &Vec<Vec<usize>>, node: usize, visited: &mut Vec<bool>) {
    if visited[node] {
        return;
    }
    visited[node] = true;
    for &neighbor in &graph[node] {
        dfs(graph, [1], visited);
    }
}

Drag options to blanks, or click blank then click option'

Avisited.len()

Bnode

C0

Dneighbor

Attempts:

3 left

💡 Hint

Common Mistakes

Calling dfs on the current node again causes infinite recursion.

Using an index out of bounds instead of the neighbor.

The DFS function should recursively call itself on each neighbor of the current node to explore the graph deeply.

2fill in blank

medium

Complete the code to mark the current node as visited.

DSA Rust

fn dfs(graph: &Vec<Vec<usize>>, node: usize, visited: &mut Vec<bool>) {
    if visited[node] {
        return;
    }
    visited[[1]] = true;
    for &neighbor in &graph[node] {
        dfs(graph, neighbor, visited);
    }
}

Drag options to blanks, or click blank then click option'

A0

Bneighbor

Cnode

Dvisited.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Marking neighbors as visited before visiting them.

Using an incorrect index causing panic.

We mark the current node as visited to avoid revisiting it and causing infinite loops.

3fill in blank

hard

Fix the error in the DFS function call to avoid infinite recursion.

DSA Rust

fn dfs(graph: &Vec<Vec<usize>>, node: usize, visited: &mut Vec<bool>) {
    if visited[node] {
        return;
    }
    visited[node] = true;
    for &neighbor in &graph[node] {
        dfs(graph, [1], visited);  // Fix this line
    }
}

Drag options to blanks, or click blank then click option'

A0

Bneighbor

Cvisited.len()

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Calling dfs on the current node causes infinite recursion.

Using an invalid index causing panic.

The recursive call should be on the neighbor node to explore new nodes, not the current node again.

4fill in blank

hard

Fill both blanks to create a DFS function that prints nodes in order of visit.

DSA Rust

fn dfs(graph: &Vec<Vec<usize>>, node: usize, visited: &mut Vec<bool>) {
    if visited[node] {
        return;
    }
    visited[node] = true;
    println!("Visited: {}", [1]);
    for &neighbor in &graph[[2]] {
        dfs(graph, neighbor, visited);
    }
}

Drag options to blanks, or click blank then click option'

Anode

Bneighbor

C0

Dvisited.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Printing neighbor instead of current node.

Using wrong index to access neighbors.

We print the current node and iterate over its neighbors to continue DFS.

5fill in blank

hard

Fill all three blanks to implement DFS that collects visited nodes in order.

DSA Rust

fn dfs_collect(graph: &Vec<Vec<usize>>, node: usize, visited: &mut Vec<bool>, result: &mut Vec<usize>) {
    if visited[[1]] {
        return;
    }
    visited[node] = true;
    result.[2](node);
    for &neighbor in &graph[[3]] {
        dfs_collect(graph, neighbor, visited, result);
    }
}

Drag options to blanks, or click blank then click option'

Anode

Bpush

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push to add nodes.

Checking visited status of wrong index.

Accessing neighbors with wrong index.

We check if the current node is visited, mark it, add it to result using push, and explore neighbors.

Practice

1. What is the main idea behind Depth First Search (DFS) on a graph?

easy

A. Explore as far as possible along each branch before backtracking

B. Visit all nodes at the current depth before moving deeper

C. Visit nodes randomly without any order

D. Only visit nodes with the highest value first

Solution

1. Step 1: Understand DFS traversal

   DFS explores a graph by going deep into one branch before exploring others.

2. Step 2: Compare with other traversal methods

   BFS visits nodes level by level, but DFS goes deep first.

3. Final Answer:

   Explore as far as possible along each branch before backtracking -> Option A

4. Quick Check:

   DFS = deep exploration first [OK]

Hint: DFS means go deep first, then backtrack [OK]

Common Mistakes:

• Confusing DFS with BFS
• Thinking DFS visits nodes level-wise
• Assuming DFS visits nodes randomly

2. Which of the following is the correct Rust function signature for a recursive DFS function on a graph represented as adjacency list?

easy

A. fn dfs(graph: &mut Vec, node: usize, visited: Vec<bool>)

B. fn dfs(graph: Vec, node: usize, visited: Vec<bool>)

C. fn dfs(graph: &Vec, node: usize, visited: &mut Vec<bool>)

D. fn dfs(graph: Vec, node: usize, visited: &mut Vec<bool>)

Solution

1. Step 1: Understand parameter passing in Rust for DFS

   The graph should be passed by reference to avoid copying, and visited vector should be mutable reference to mark nodes visited.

2. Step 2: Check each option

   fn dfs(graph: &Vec, node: usize, visited: &mut Vec<bool>) passes graph by reference and visited as mutable reference, which is correct for DFS recursion.

3. Final Answer:

   fn dfs(graph: &Vec, node: usize, visited: &mut Vec<bool>) -> Option C

4. Quick Check:

   Graph by ref + visited mutable ref = correct DFS signature [OK]

Hint: Pass graph by ref and visited as mutable ref [OK]

Common Mistakes:

• Passing graph or visited by value causing copies
• Not using mutable reference for visited
• Using mutable reference for graph unnecessarily

3. What is the output of the following Rust DFS traversal starting from node 0?

fn dfs(graph: &Vec, node: usize, visited: &mut Vec<bool>, result: &mut Vec<usize>) {
    visited[node] = true;
    result.push(node);
    for &neighbor in &graph[node] {
        if !visited[neighbor] {
            dfs(graph, neighbor, visited, result);
        }
    }
}

fn main() {
    let graph = vec![
        vec![1, 2],
        vec![0, 3],
        vec![0],
        vec![1]
    ];
    let mut visited = vec![false; 4];
    let mut result = Vec::new();
    dfs(&graph, 0, &mut visited, &mut result);
    println!("{:?}", result);
}

medium

A. [0, 2, 1, 3]

B. [0, 1, 3, 2]

C. [0, 1, 2, 3]

D. [1, 0, 2, 3]

Solution

1. Step 1: Trace DFS starting at node 0

   Start at 0, mark visited, add 0. Neighbors: 1, 2. Visit 1 first.

2. Step 2: Continue DFS from node 1

   Visit 1, add 1. Neighbors: 0 (visited), 3 (not visited). Visit 3 next.

3. Step 3: Visit node 3

   Visit 3, add 3. Neighbor: 1 (visited). Backtrack to 0 and visit 2.

4. Step 4: Visit node 2

   Visit 2, add 2. Neighbor: 0 (visited). DFS complete.

5. Final Answer:

   [0, 1, 3, 2] -> Option B

6. Quick Check:

   DFS order from 0 = [0,1,3,2] [OK]

Hint: Follow neighbors in order, mark visited, backtrack [OK]

Common Mistakes:

• Visiting neighbors out of order
• Not marking visited nodes
• Confusing BFS order with DFS

4. Identify the bug in this DFS Rust code snippet and choose the fix:

fn dfs(graph: &Vec, node: usize, visited: &mut Vec<bool>) {
    visited[node] = true;
    for neighbor in &graph[node] {
        if visited[*neighbor] {
            dfs(graph, *neighbor, visited);
        }
    }
}

medium

A. Change if visited[*neighbor] to if !visited[*neighbor]

B. Change visited[node] = true to visited[node] = false

C. Remove the for loop and call dfs only once

D. Change neighbor to node inside the loop

Solution

1. Step 1: Analyze the condition for recursive DFS call

   The code calls dfs only if visited[*neighbor] is true, meaning already visited nodes are revisited, causing infinite recursion.

2. Step 2: Correct the condition to avoid revisiting

   Change condition to if !visited[*neighbor] to visit only unvisited neighbors.

3. Final Answer:

   Change if visited[*neighbor] to if !visited[*neighbor] -> Option A

4. Quick Check:

   Visit unvisited neighbors only [OK]

Hint: Call DFS only on unvisited neighbors [OK]

Common Mistakes:

• Calling DFS on visited nodes causing infinite loops
• Incorrectly marking nodes as unvisited
• Removing necessary loops

5. Given a graph with possible cycles, how can you modify DFS in Rust to detect if the graph contains a cycle?

hard

A. Use BFS instead of DFS to detect cycles

B. Only mark nodes visited after all neighbors are visited

C. Ignore visited nodes and continue traversal

D. Use a recursion stack (call stack) boolean vector to track nodes in the current path and detect back edges

Solution

1. Step 1: Understand cycle detection in DFS

   To detect cycles, track nodes currently in recursion stack to find back edges indicating cycles.

2. Step 2: Implement recursion stack tracking

   Use a boolean vector to mark nodes in current DFS path; if a neighbor is in this stack, a cycle exists.

3. Step 3: Other options are incorrect

   Marking visited after neighbors or using BFS won't detect cycles reliably; ignoring visited nodes causes infinite loops.

4. Final Answer:

   Use a recursion stack (call stack) boolean vector to track nodes in the current path and detect back edges -> Option D

5. Quick Check:

   Recursion stack tracks cycles in DFS [OK]

Hint: Track recursion stack to find cycles in DFS [OK]

Common Mistakes:

• Not tracking recursion stack causing missed cycles
• Using BFS which needs extra logic for cycles
• Ignoring visited nodes causing infinite loops