Bridges in Graph Tarjan's Algorithm in DSA Rust - Time & Space Complexity

Time Complexity: Bridges in Graph Tarjan's Algorithm

O(n + m)

Understanding Time Complexity

We want to understand how the time needed to find bridges in a graph grows as the graph gets bigger.

Specifically, how the algorithm's steps increase with more nodes and edges.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn dfs(
    u: usize,
    parent: usize,
    time: &mut usize,
    graph: &Vec<Vec<usize>>,
    visited: &mut Vec<bool>,
    disc: &mut Vec<usize>,
    low: &mut Vec<usize>,
    bridges: &mut Vec<(usize, usize)>
) {
    visited[u] = true;
    *time += 1;
    disc[u] = *time;
    low[u] = *time;

    for &v in &graph[u] {
        if v == parent { continue; }
        if !visited[v] {
            dfs(v, u, time, graph, visited, disc, low, bridges);
            low[u] = low[u].min(low[v]);
            if low[v] > disc[u] {
                bridges.push((u, v));
            }
        } else {
            low[u] = low[u].min(disc[v]);
        }
    }
}

This code runs a depth-first search to find all bridges in an undirected graph using Tarjan's algorithm.

Identify Repeating Operations

• Primary operation: Recursive DFS visits each node and explores its neighbors.
• How many times: Each node is visited once, and each edge is checked once or twice.

How Execution Grows With Input

As the number of nodes (n) and edges (m) grow, the algorithm visits each node once and checks each edge once.

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

Pattern observation: The operations grow roughly in proportion to nodes plus edges, so doubling nodes and edges roughly doubles the work.

Final Time Complexity

Time Complexity: O(n + m)

This means the time needed grows linearly with the total number of nodes and edges in the graph.

Common Mistake

[X] Wrong: "The algorithm checks every possible pair of nodes, so it must be O(n²)."

[OK] Correct: The algorithm only visits each node and edge once, not all pairs, so it runs much faster than O(n²).

Interview Connect

Understanding this linear time complexity helps you explain how efficient graph algorithms work in practice.

It shows you can handle large graphs without slowdowns, a key skill in many coding challenges.

Self-Check

"What if the graph was directed instead of undirected? How would the time complexity change?"

Practice

1. What is the main purpose of Tarjan's algorithm when finding bridges in a graph?

easy

A. To detect cycles in a graph

B. To find the shortest path between two nodes

C. To find edges that, if removed, increase the number of connected components

D. To sort nodes in topological order

Solution

1. Step 1: Understand what a bridge is in a graph

   A bridge is an edge that, when removed, increases the number of connected components in the graph.

2. Step 2: Identify Tarjan's algorithm goal

   Tarjan's algorithm finds such edges efficiently by using DFS and tracking discovery and low times.

3. Final Answer:

   To find edges that, if removed, increase the number of connected components -> Option C

4. Quick Check:

   Bridges = edges increasing components [OK]

Hint: Bridges break connectivity when removed [OK]

Common Mistakes:

• Confusing bridges with cycles
• Thinking it finds shortest paths
• Mixing bridges with articulation points

2. Which of the following Rust code snippets correctly declares a vector to store adjacency lists for a graph with 5 nodes?

easy

A. let graph: Vec = vec![vec![0]; 5];

B. let graph: Vec<usize> = vec![0; 5];

C. let graph: Vec = vec![vec![]; 4];

D. let graph: Vec = vec![vec![]; 5];

Solution

1. Step 1: Understand adjacency list representation

   Each node has a vector of neighbors, so graph is a vector of vectors.

2. Step 2: Check vector initialization for 5 nodes

   vec![vec![]; 5] creates 5 empty vectors, one per node.

3. Final Answer:

   let graph: Vec = vec![vec![]; 5]; -> Option D

4. Quick Check:

   Adjacency list for 5 nodes = vec![vec![]; 5] [OK]

Hint: Use vec![vec![]; n] for adjacency lists [OK]

Common Mistakes:

• Using wrong vector type (Vec<usize> instead of Vec)
• Initializing with wrong size
• Filling inner vectors with zeros instead of empty

3. Given the following Rust code snippet implementing part of Tarjan's algorithm, what will be printed as bridges for the graph edges?

let edges = vec![(0,1), (1,2), (2,0), (1,3)];
// Graph: triangle 0-1-2 and a bridge edge 1-3
// After running Tarjan's algorithm, which edges are bridges?
println!("Bridges: {:?}", vec![(1,3)]);

medium

A. Bridges: [(1, 3)]

B. Bridges: [(0, 1), (1, 2), (2, 0)]

C. Bridges: []

D. Bridges: [(0, 3)]

Solution

1. Step 1: Analyze the graph structure

   The edges (0,1), (1,2), (2,0) form a cycle (triangle), so none are bridges.

2. Step 2: Identify the bridge edge

   The edge (1,3) connects node 3 to the triangle and is a bridge because removing it disconnects node 3.

3. Final Answer:

   Bridges: [(1, 3)] -> Option A

4. Quick Check:

   Only edge (1,3) breaks connectivity [OK]

Hint: Edges in cycles are not bridges [OK]

Common Mistakes:

• Marking cycle edges as bridges
• Ignoring leaf node connections
• Confusing bridge edges with articulation points

4. Identify the bug in this Rust snippet of Tarjan's algorithm for finding bridges:

fn dfs(u: usize, parent: usize, graph: &Vec, disc: &mut Vec<usize>, low: &mut Vec<usize>, time: &mut usize, bridges: &mut Vec<(usize, usize)>) {
    disc[u] = *time;
    low[u] = *time;
    *time += 1;
    for &v in &graph[u] {
        if v == parent {
            continue;
        }
        if disc[v] == 0 {
            dfs(v, u, graph, disc, low, time, bridges);
            low[u] = low[u].min(low[v]);
            if low[v] > disc[u] {
                bridges.push((u, v));
            }
        } else {
            low[u] = low[u].min(disc[v]);
        }
    }
}

medium

A. disc and low arrays are not initialized with zeros

B. The parent check should be if v != parent instead of v == parent

C. The time variable should be passed by value, not mutable reference

D. The condition low[v] > disc[u] should be low[v] < disc[u]

Solution

1. Step 1: Check initialization of disc and low arrays

   disc and low must be initialized with zeros to detect unvisited nodes (disc[v] == 0).

2. Step 2: Verify other logic correctness

   Parent check and time passing are correct; condition low[v] > disc[u] is correct for bridges.

3. Final Answer:

   disc and low arrays are not initialized with zeros -> Option A

4. Quick Check:

   Uninitialized disc causes wrong visited checks [OK]

Hint: Initialize disc and low with zeros before DFS [OK]

Common Mistakes:

• Not initializing disc and low arrays
• Wrong parent check condition
• Passing time by value instead of mutable reference

5. You have a graph with 6 nodes and edges: (0,1), (1,2), (2,0), (1,3), (3,4), (4,5), (5,3). Using Tarjan's algorithm, which edges are bridges in this graph?

hard

A. [(3, 4), (4, 5), (5, 3)]

B. [(1, 3)]

C. [(0, 1), (1, 2), (2, 0)]

D. [(1, 3), (3, 4)]

Solution

1. Step 1: Analyze graph components

   Nodes 0,1,2 form a cycle (triangle), so edges among them are not bridges. Nodes 3,4,5 form a cycle, so edges among them are not bridges.

2. Step 2: Identify bridge edges

   The edge (1,3) connects the two cycles. Removing (1,3) disconnects the graph, so it is a bridge.

3. Final Answer:

   [(1, 3)] -> Option B

4. Quick Check:

   Only edge connecting components is bridge [OK]

Hint: Bridges connect separate cycles or components [OK]

Common Mistakes:

• Marking cycle edges as bridges
• Ignoring connectivity between components
• Confusing bridges with articulation points