Bridges in Graph Tarjan's Algorithm in DSA Python - 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 does the algorithm's work increase when we add more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def find_bridges(graph):
    n = len(graph)
    time = 0
    disc = [-1] * n
    low = [-1] * n
    bridges = []

    def dfs(u, parent):
        nonlocal time
        disc[u] = low[u] = time
        time += 1
        for v in graph[u]:
            if disc[v] == -1:
                dfs(v, u)
                low[u] = min(low[u], low[v])
                if low[v] > disc[u]:
                    bridges.append((u, v))
            elif v != parent:
                low[u] = min(low[u], disc[v])

    for i in range(n):
        if disc[i] == -1:
            dfs(i, -1)
    return bridges

This code finds all bridges in an undirected graph using Tarjan's algorithm with DFS traversal.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Depth-first search (DFS) visits each node and edge once.
• How many times: Each node is visited once, and each edge is checked once during DFS.

How Execution Grows With Input

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

Input Size (n, m)	Approx. Operations
10 nodes, 15 edges	About 25 operations (10 nodes + 15 edges)
100 nodes, 200 edges	About 300 operations (100 + 200)
1000 nodes, 3000 edges	About 4000 operations (1000 + 3000)

Pattern observation: The work grows roughly in proportion to the sum of nodes and edges.

Final Time Complexity

Time Complexity: O(n + m)

This means the algorithm runs in time proportional to the number of nodes plus the number of 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 edge once during DFS, not every pair of nodes, so it is faster than O(n²).

Interview Connect

Understanding this time complexity helps you explain how graph traversal algorithms efficiently handle large networks.

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 find the shortest path between two nodes

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

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

4. Quick Check:

   Bridges = edges increasing components when removed [OK]

Hint: Bridges disconnect graph if removed [OK]

Common Mistakes:

• Confusing bridges with cycles
• Thinking it finds shortest paths
• Assuming it counts nodes

2. Which of the following is the correct way to initialize the discovery time and low arrays in Tarjan's algorithm in Python?

easy

A. discovery = [-1] * n; low = [-1] * n

B. discovery = [0] * n; low = [0] * n

C. discovery = list(range(n)); low = list(range(n))

D. discovery = None; low = None

Solution

1. Step 1: Understand initialization purpose

   Discovery and low arrays store times; initializing with -1 means unvisited nodes.

2. Step 2: Choose correct initialization

   Using [-1] * n marks all nodes as unvisited initially, which is standard practice.

3. Final Answer:

   discovery = [-1] * n; low = [-1] * n -> Option A

4. Quick Check:

   Unvisited nodes marked with -1 [OK]

Hint: Use -1 to mark unvisited nodes [OK]

Common Mistakes:

• Initializing with zeros (means visited)
• Using range values (incorrect meaning)
• Setting to None (causes errors)

3. What will be the output bridges for the following undirected graph using Tarjan's algorithm?

graph = {0: [1, 2], 1: [0, 2], 2: [0, 1, 3], 3: [2, 4], 4: [3]}

medium

A. []

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

C. [(2, 3), (3, 4)]

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

Solution

1. Step 1: Analyze graph connections

   Nodes 0,1,2 form a triangle (cycle), so no bridges there. Edges (2,3) and (3,4) connect to nodes outside the cycle.

2. Step 2: Identify bridges

   Removing edge (2,3) or (3,4) disconnects the graph, so these are bridges.

3. Final Answer:

   [(2, 3), (3, 4)] -> Option C

4. Quick Check:

   Edges breaking connectivity = bridges [OK]

Hint: Edges outside cycles are likely bridges [OK]

Common Mistakes:

• Marking cycle edges as bridges
• Missing edges connecting components
• Confusing node pairs order

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

def dfs(u, parent):
    visited[u] = True
    disc[u] = time
    low[u] = time
    time += 1
    for v in graph[u]:
        if not visited[v]:
            dfs(v, u)
            low[u] = min(low[u], low[v])
            if low[v] > disc[u]:
                bridges.append((u, v))
        elif v != parent:
            low[u] = min(low[u], disc[v])

medium

A. visited array is not updated correctly

B. The condition low[v] > disc[u] should be low[v] >= disc[u]

C. The parent check should be if v == parent, not v != parent

D. time variable is not declared global or passed, causing UnboundLocalError

Solution

1. Step 1: Check variable usage

   The variable 'time' is incremented inside dfs but not declared global or passed, causing an error.

2. Step 2: Understand Python variable scope

   Without 'global time' or passing time as argument, Python treats 'time' as local, causing UnboundLocalError.

3. Final Answer:

   time variable is not declared global or passed, causing UnboundLocalError -> Option D

4. Quick Check:

   Modify time scope to fix error [OK]

Hint: Declare time global or pass as argument [OK]

Common Mistakes:

• Ignoring variable scope errors
• Changing condition operators incorrectly
• Misunderstanding parent node check

5. Given a graph with nodes 0 to 5 and edges:

edges = [(0,1), (1,2), (2,0), (1,3), (3,4), (4,5), (5,3)]

Using Tarjan's algorithm, which edges are bridges?

hard

A. [(1, 3)]

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

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

D. []

Solution

1. Step 1: Analyze graph structure

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

2. Step 2: Identify connecting edges

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

3. Final Answer:

   [(1, 3)] -> Option A

4. Quick Check:

   Bridge connects two cycles [OK]

Hint: Bridges connect separate cycles or components [OK]

Common Mistakes:

• Marking cycle edges as bridges
• Including all edges in bridge list
• Ignoring connectivity impact