Bridges in Graph Tarjan's Algorithm in DSA Java - 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 with more nodes and edges?

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 (v == parent) continue;
        if (!visited[v]) {
            dfs(v, u);
            low[u] = Math.min(low[u], low[v]);
            if (low[v] > disc[u]) {
                // (u, v) is a bridge
            }
        } else {
            low[u] = Math.min(low[u], disc[v]);
        }
    }
}

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: DFS visits each node and explores all its edges once.
• How many times: Each node is visited once; each edge is checked once in the adjacency list.

How Execution Grows With Input

The algorithm visits every node and edge once, so if the graph doubles in size, the work roughly doubles too.

Input Size (n nodes, m edges)	Approx. Operations
10 nodes, 15 edges	About 25 operations
100 nodes, 150 edges	About 250 operations
1000 nodes, 1500 edges	About 2500 operations

Pattern observation: The work grows linearly with the number of nodes plus edges.

Final Time Complexity

Time Complexity: O(n + m)

This means the time grows in a straight line 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 is O(n²)."

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

Interview Connect

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

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 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 sort nodes in topological order

Solution

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

   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 is used to find such edges efficiently by using DFS and tracking discovery and low times.

3. Final Answer:

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

4. Quick Check:

   Bridges = edges increasing components [OK]

Hint: Bridges split graph into more parts when removed [OK]

Common Mistakes:

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

2. Which of the following Java code snippets correctly initializes the arrays used for discovery and low times in Tarjan's algorithm for a graph with n nodes?

easy

A. int[] disc = new int[n]; int[] low = new int[n]; Arrays.fill(disc, 0); Arrays.fill(low, 0);

B. int[] disc = new int[n]; int[] low = new int[n];

C. int disc[] = new int[n]; int low[] = new int[n]; disc = null; low = null;

D. int[] disc = new int[n]; int[] low = new int[n]; Arrays.fill(disc, -1); Arrays.fill(low, -1);

Solution

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

   Both arrays must be initialized with -1 to indicate unvisited nodes.

2. Step 2: Verify correct Java syntax and usage

   int[] disc = new int[n]; int[] low = new int[n]; Arrays.fill(disc, -1); Arrays.fill(low, -1); correctly creates arrays and fills them with -1 using Arrays.fill.

3. Final Answer:

   int[] disc = new int[n]; int[] low = new int[n]; Arrays.fill(disc, -1); Arrays.fill(low, -1); -> Option D

4. Quick Check:

   Arrays initialized with -1 = int[] disc = new int[n]; int[] low = new int[n]; Arrays.fill(disc, -1); Arrays.fill(low, -1); [OK]

Hint: Initialize discovery and low arrays with -1 for unvisited [OK]

Common Mistakes:

• Not filling arrays with -1, causing wrong visited checks
• Setting arrays to null after creation
• Filling arrays with 0 instead of -1

3. Consider the following Java code snippet implementing part of Tarjan's algorithm for bridges detection. What will be the output list of bridges for the given graph edges?

int n = 5;
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i < n; i++) graph.add(new ArrayList<>());
graph.get(0).add(1); graph.get(1).add(0);
graph.get(1).add(2); graph.get(2).add(1);
graph.get(2).add(0); graph.get(0).add(2);
graph.get(1).add(3); graph.get(3).add(1);
graph.get(3).add(4); graph.get(4).add(3);

// After running Tarjan's bridge detection, which edges are bridges?

medium

A. [[0, 1], [1, 2]]

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

C. [[1, 3], [3, 4]]

D. []

Solution

1. Step 1: Analyze the graph structure

   The graph has edges forming a triangle between nodes 0,1,2 and a chain 1-3-4.

2. Step 2: Identify bridges

   Edges in the triangle are not bridges because removing them doesn't disconnect the graph. Edges (1,3) and (3,4) are bridges because removing them increases components.

3. Final Answer:

   [[1, 3], [3, 4]] -> Option C

4. Quick Check:

   Bridges separate graph parts = [[1, 3], [3, 4]] [OK]

Hint: Edges in cycles are not bridges; chains are [OK]

Common Mistakes:

• Marking edges in cycles as bridges
• Ignoring edges that disconnect graph
• Confusing node order in edges

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

void dfs(int u, int parent) {
    disc[u] = time;
    low[u] = time;
    for (int v : graph.get(u)) {
        if (v == parent) continue;
        if (disc[v] == -1) {
            dfs(v, u);
            low[u] = Math.min(low[u], low[v]);
            if (low[v] > disc[u]) {
                bridges.add(new int[]{u, v});
            }
        } else {
            low[u] = Math.min(low[u], disc[v]);
        }
    }
}

medium

A. The time variable is not incremented correctly

B. The condition to add a bridge should be low[v] >= disc[u]

C. The else block should update low[u] with low[v], not disc[v]

D. The parent check should be if (v != parent) instead of if (v == parent)

Solution

1. Step 1: Check time increment usage

   The code assigns the current time to disc[u] and low[u] but fails to increment the time variable. This causes all nodes to receive the same discovery and low values, breaking the algorithm.

2. Step 2: Verify other potential issues

   The bridge condition (low[v] > disc[u]) is correct. The else block properly updates low[u] with disc[v] for back edges. The parent skip is correctly implemented.

3. Final Answer:

   The time variable is not incremented correctly -> Option A

4. Quick Check:

   Time increment missing [OK]

Hint: Check time increments carefully in DFS [OK]

Common Mistakes:

• Changing bridge condition incorrectly
• Updating low[u] with wrong values
• Misusing parent check condition

5. Given a connected undirected graph with n nodes and m edges, you want to find all bridges using Tarjan's algorithm. Which of the following modifications will correctly handle disconnected graphs as well?

hard

A. Only run DFS from node 0 since graph is connected

B. Run DFS from every node that is unvisited to cover all components

C. Reset discovery and low arrays to zero before each DFS call

D. Increase the time variable by 2 after each DFS call

Solution

1. Step 1: Understand the problem with disconnected graphs

   Tarjan's algorithm assumes DFS covers all nodes. For disconnected graphs, DFS from one node misses others.

2. Step 2: Correct approach for disconnected graphs

   Run DFS from every unvisited node to ensure all components are covered and bridges found.

3. Final Answer:

   Run DFS from every node that is unvisited to cover all components -> Option B

4. Quick Check:

   DFS from all unvisited nodes = Run DFS from every node that is unvisited to cover all components [OK]

Hint: Run DFS on all unvisited nodes for disconnected graphs [OK]

Common Mistakes:

• Assuming graph is always connected
• Resetting arrays incorrectly
• Changing time increments unnecessarily