Bridges in Graph Tarjan's Algorithm in DSA Go - 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.

func dfs(u, parent int, graph [][]int, time *int, disc, low []int, bridges *[][2]int) {
    disc[u] = *time
    low[u] = *time
    (*time)++
    for _, v := range graph[u] {
        if disc[v] == 0 {
            dfs(v, u, graph, time, disc, low, bridges)
            if low[v] > disc[u] {
                *bridges = append(*bridges, [2]int{u, v})
            }
            if low[v] < low[u] {
                low[u] = low[v]
            }
        } else if v != parent {
            if disc[v] < low[u] {
                low[u] = disc[v]
            }
        }
    }
}

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The depth-first search (dfs) visits each node and explores its neighbors.
• How many times: Each node is visited once, and each edge is checked once during recursion.

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, 3000 edges	About 4000 operations

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

Final Time Complexity

Time Complexity: O(n + m)

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

Common Mistake

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

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

Interview Connect

Understanding this linear time complexity helps you explain how efficient graph algorithms work in practice, 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 identify edges that, if removed, increase the number of connected components

B. To find the shortest path between two nodes

C. To detect cycles in a graph

D. To count the total number of nodes in the graph

Solution

1. Step 1: Understand the definition of a bridge

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

2. Step 2: Purpose of Tarjan's algorithm for bridges

   Tarjan's algorithm efficiently finds all such edges (bridges) in a graph using DFS and low-link values.

3. Final Answer:

   To identify edges that, if removed, increase the number of connected components -> Option A

4. Quick Check:

   Bridges = edges increasing components when removed [OK]

Hint: Bridges split graph if removed, Tarjan finds them fast [OK]

Common Mistakes:

• Confusing bridges with cycles
• Thinking Tarjan finds shortest paths
• Mixing bridges with node counts

2. Which of the following Go code snippets correctly declares a slice to store adjacency lists for a graph with n nodes?

easy

A. adj := make([]int, n)

B. adj := make([][]int, n)

C. adj := [][]int{n}

D. adj := make([]int, 0, n)

Solution

1. Step 1: Understand adjacency list representation

   Adjacency list for n nodes is a slice of slices: each node has a slice of connected nodes.

2. Step 2: Correct Go syntax for adjacency list

   make([][]int, n) creates a slice of n slices of ints, perfect for adjacency lists.

3. Final Answer:

   adj := make([][]int, n) -> Option B

4. Quick Check:

   Adjacency list = slice of slices [OK]

Hint: Adjacency list needs slice of slices, not flat slice [OK]

Common Mistakes:

• Using make([]int, n) which is flat slice
• Incorrect slice literal syntax
• Confusing capacity with length

3. Consider the following Go code snippet implementing part of Tarjan's algorithm for bridges. What will be the output when run on the graph edges: (0-1), (1-2), (2-0), (1-3)?

package main
import "fmt"

var (
    time int
    adj [][]int
    disc []int
    low []int
    visited []bool
)

func dfs(u, parent int) {
    visited[u] = true
    disc[u] = time
    low[u] = time
    time++

    for _, v := range adj[u] {
        if v == parent {
            continue
        }
        if !visited[v] {
            dfs(v, u)
            if low[v] > disc[u] {
                fmt.Printf("Bridge found: %d - %d\n", u, v)
            }
            if low[v] < low[u] {
                low[u] = low[v]
            }
        } else {
            if disc[v] < low[u] {
                low[u] = disc[v]
            }
        }
    }
}

func main() {
    n := 4
    adj = make([][]int, n)
    adj[0] = []int{1, 2}
    adj[1] = []int{0, 2, 3}
    adj[2] = []int{0, 1}
    adj[3] = []int{1}

    disc = make([]int, n)
    low = make([]int, n)
    visited = make([]bool, n)

    dfs(0, -1)
}

medium

A. Bridge found: 2 - 0

B. Bridge found: 0 - 1
Bridge found: 1 - 3

C. No bridges found

D. Bridge found: 1 - 3

Solution

1. Step 1: Analyze the graph structure

   The graph has a triangle between nodes 0,1,2 and a single edge from 1 to 3. The triangle edges are not bridges because removing them doesn't disconnect the graph.

2. Step 2: Identify bridges using Tarjan's logic

   Edge 1-3 is a bridge because removing it disconnects node 3. The code prints "Bridge found: 1 - 3" only.

3. Final Answer:

   Bridge found: 1 - 3 -> Option D

4. Quick Check:

   Only edge 1-3 disconnects graph [OK]

Hint: Triangle edges are not bridges; check leaf edges [OK]

Common Mistakes:

• Marking triangle edges as bridges
• Ignoring parent check in DFS
• Confusing discovery and low values

4. The following Go code attempts to find bridges using Tarjan's algorithm but has a bug. What is the error?

func dfs(u, parent int) {
    visited[u] = true
    disc[u] = time
    low[u] = time
    time++

    for _, v := range adj[u] {
        if !visited[v] {
            dfs(v, u)
            if low[v] > disc[u] {
                fmt.Printf("Bridge: %d - %d\n", u, v)
            }
            if low[v] < low[u] {
                low[u] = low[v]
            }
        } else if v != parent {
            if disc[v] < low[u] {
                low[u] = disc[v]
            }
        }
    }
}

medium

A. The time variable is not incremented

B. The else condition should check v == parent instead of v != parent

C. The parent check is missing before recursive DFS call

D. The low[u] update is done incorrectly after DFS call

Solution

1. Step 1: Check parent skipping logic

   The code does not skip the parent node before calling DFS recursively, which can cause revisiting the parent and incorrect low values.

2. Step 2: Identify missing parent check before DFS

   Correct code should have "if v == parent { continue }" before DFS call to avoid revisiting parent edge.

3. Final Answer:

   The parent check is missing before recursive DFS call -> Option C

4. Quick Check:

   Skip parent before DFS to avoid cycles [OK]

Hint: Always skip parent node before DFS recursion [OK]

Common Mistakes:

• Confusing parent check placement
• Not incrementing time variable
• Misusing else if condition

5. Given a connected undirected graph with n nodes and m edges, how can Tarjan's algorithm be adapted to find all bridges efficiently? Choose the best approach.

hard

A. Use DFS once from any node, track discovery and low times, and identify edges where low[v] > disc[u]

B. Remove edges one by one and check connectivity each time

C. Use BFS to find shortest paths and mark edges not in shortest paths as bridges

D. Run DFS from each node separately without tracking discovery times

Solution

1. Step 1: Understand Tarjan's algorithm approach

   Tarjan's algorithm uses a single DFS traversal from any node, tracking discovery and low times to find bridges efficiently.

2. Step 2: Identify condition for bridges

   An edge u-v is a bridge if low[v] > disc[u], meaning v cannot reach u or its ancestors without that edge.

3. Final Answer:

   Use DFS once from any node, track discovery and low times, and identify edges where low[v] > disc[u] -> Option A

4. Quick Check:

   Single DFS + low/disc times = bridges [OK]

Hint: Single DFS with low/disc times finds bridges fast [OK]

Common Mistakes:

• Running DFS multiple times unnecessarily
• Using BFS which doesn't track low times
• Brute force edge removal checking connectivity