Bridges in Graph Tarjan's Algorithm in DSA C - 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 change when the number of nodes and edges increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void dfs(int u, int parent) {
    visited[u] = true;
    disc[u] = low[u] = ++time;
    for (int v : graph[u]) {
        if (!visited[v]) {
            dfs(v, u);
            low[u] = min(low[u], low[v]);
            if (low[v] > disc[u]) {
                // (u, v) is a bridge
            }
        } else if (v != parent) {
            low[u] = min(low[u], disc[v]);
        }
    }
}
    

This code runs a depth-first search to find bridges by tracking discovery and low times for each node.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The DFS visits each node once and checks all its edges.
• How many times: Each edge is checked at most twice (once from each end).

How Execution Grows With Input

As the number of nodes (n) and edges (m) grow, the algorithm visits each node once and explores all edges connected to it.

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, 5000 edges	About 6000 operations (1000 + 5000)

Pattern observation: The work grows roughly in a straight line with the number of nodes plus edges.

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 edge once or twice, not all pairs of nodes, so it does not do n squared work.

Interview Connect

Understanding this linear time complexity helps you explain why Tarjan's algorithm is efficient for large graphs and shows your grasp of graph traversal techniques.

Self-Check

"What if the graph was represented with an adjacency matrix instead of adjacency lists? 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 edges that, if removed, increase the number of connected components

C. To find the shortest path between two nodes

D. To count the total number of edges 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: Relate Tarjan's algorithm to bridges

   Tarjan's algorithm efficiently finds such edges by 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 parts when removed [OK]

Common Mistakes:

• Confusing bridges with shortest paths
• Thinking bridges detect cycles
• Assuming bridges count edges

2. Which of the following is the correct way to declare the array that stores discovery times in C for Tarjan's algorithm?

easy

A. int disc = new int[MAX];

B. int disc[];

C. int disc[MAX];

D. int[] disc = new int[MAX];

Solution

1. Step 1: Recall C array declaration syntax

   In C, arrays are declared with type and size like int arr[size];

2. Step 2: Match with discovery time array

   So, int disc[MAX]; is correct for fixed size MAX.

3. Final Answer:

   int disc[MAX]; -> Option C

4. Quick Check:

   C arrays use type name[size] [OK]

Hint: Use type name[size] for arrays in C [OK]

Common Mistakes:

• Using Java or C++ style array declarations in C
• Not specifying array size
• Using dynamic allocation syntax incorrectly

3. Consider the following code snippet for Tarjan's algorithm to find bridges. What will be printed?

void dfs(int u, int parent) {
  disc[u] = low[u] = ++time;
  for (int v : adj[u]) {
    if (disc[v] == -1) {
      dfs(v, u);
      low[u] = min(low[u], low[v]);
      if (low[v] > disc[u])
        printf("Bridge: %d - %d\n", u, v);
    } else if (v != parent) {
      low[u] = min(low[u], disc[v]);
    }
  }
}

// Graph edges:
// 0-1, 1-2, 2-0, 1-3
// Start dfs(0, -1);

medium

A. Bridge: 1 - 3

B. Bridge: 0 - 1

C. Bridge: 2 - 0

D. No bridges

Solution

1. Step 1: Analyze the graph structure

   The graph has a triangle 0-1-2 and an edge 1-3. Triangle edges are not bridges because removing them doesn't disconnect the graph.

2. Step 2: Apply Tarjan's condition for bridges

   Edge 1-3 is a bridge because removing it disconnects node 3.

3. Final Answer:

   Bridge: 1 - 3 -> Option A

4. Quick Check:

   Only edge 1-3 disconnects graph [OK]

Hint: Bridges disconnect nodes; triangle edges are safe [OK]

Common Mistakes:

• Thinking all edges in a cycle are bridges
• Ignoring parent check in DFS
• Misinterpreting low and disc values

4. The following code snippet is intended to find bridges using Tarjan's algorithm. What is the bug?

void dfs(int u, int parent) {
  disc[u] = low[u] = ++time;
  for (int v : adj[u]) {
    if (disc[v] == -1) {
      dfs(v, u);
      low[u] = min(low[u], low[v]);
      if (low[v] >= disc[u])
        printf("Bridge: %d - %d\n", u, v);
    } else if (v != parent) {
      low[u] = min(low[u], disc[v]);
    }
  }
}

medium

A. The condition should be low[v] > disc[u], not >=.

B. The disc array is not initialized.

C. The parent parameter is not used correctly.

D. The time variable is not incremented.

Solution

1. Step 1: Understand bridge condition in Tarjan's algorithm

   Bridge condition is low[v] > disc[u], meaning no back edge from v or its descendants to u or above.

2. Step 2: Identify the bug in condition

   Using >= incorrectly marks edges that are not bridges (like parent edges in cycles).

3. Final Answer:

   The condition should be low[v] > disc[u], not >=. -> Option A

4. Quick Check:

   Bridge if low[v] > disc[u] [OK]

Hint: Use strict greater than for bridge check [OK]

Common Mistakes:

• Using >= instead of > in bridge condition
• Not initializing discovery times
• Ignoring parent node in DFS

5. Given a connected undirected graph with 5 nodes and edges: 0-1, 1-2, 2-3, 3-4, 1-4, which edges are bridges according to Tarjan's algorithm?

hard

A. [0-1, 2-3]

B. [2-3, 3-4]

C. [1-4]

D. [0-1]

Solution

1. Step 1: Visualize the graph

   Edges: 0-1, 1-2, 2-3, 3-4, 1-4. The edge 1-4 creates a cycle with 1-2-3-4.

2. Step 2: Identify bridges

   Removing 0-1 disconnects node 0, so 0-1 is a bridge. Edges 2-3 and 3-4 are in cycle with 1-4, so not bridges.

3. Final Answer:

   [0-1] -> Option D

4. Quick Check:

   Only 0-1 disconnects graph [OK]

Hint: Edges in cycles are not bridges [OK]

Common Mistakes:

• Marking cycle edges as bridges
• Ignoring connectivity after edge removal
• Confusing bridge with articulation point