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

Time Complexity: Bridges in Graph Tarjan's Algorithm

O(n + m)

Understanding Time Complexity

We want to know 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.

function findBridges(graph: number[][]): [number, number][] {
  const n = graph.length;
  const visited = new Array(n).fill(false);
  const disc = new Array(n).fill(-1);
  const low = new Array(n).fill(-1);
  const bridges: [number, number][] = [];
  let time = 0;

  function dfs(u: number, parent: number) {
    visited[u] = true;
    disc[u] = low[u] = time++;
    for (const v of 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]) bridges.push([u, v]);
      } else {
        low[u] = Math.min(low[u], disc[v]);
      }
    }
  }

  for (let i = 0; i < n; i++) {
    if (!visited[i]) dfs(i, -1);
  }

  return bridges;
}
    

This code finds all bridges (edges that if removed increase graph parts) using DFS and tracking discovery times.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

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

How Execution Grows With Input

As the number of nodes and edges grow, the algorithm checks each node and edge once.

Input Size (n nodes, m edges)	Approx. Operations
10 nodes, 15 edges	About 25 operations (visiting nodes + edges)
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.

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 path, so it must be exponential."

[OK] Correct: The DFS visits each node and edge only once, not all paths, so it runs efficiently in linear time.

Interview Connect

Understanding this linear time complexity helps you explain how graph algorithms handle large data efficiently.

Self-Check

"What if the graph was represented as 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 sort nodes in topological order

B. To find the shortest path between two nodes

C. To detect cycles in a graph

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

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 is used to find 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 D

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
• Mixing with topological sorting

2. Which of the following is the correct TypeScript syntax to declare the discovery time array for Tarjan's algorithm?

easy

A. let disc = Array(n).fill('undefined');

B. let disc: number = new Array(n);

C. let disc: number[] = new Array(n).fill(-1);

D. let disc: number[] = new Array(n);

Solution

1. Step 1: Understand array declaration with initial values in TypeScript

   To create an array of numbers with size n and initialize all values to -1, use new Array(n).fill(-1).

2. Step 2: Check type correctness

   TypeScript syntax requires disc to be number[] for an array of numbers.

3. Final Answer:

   let disc: number[] = new Array(n).fill(-1); -> Option C

4. Quick Check:

   Array<number> with fill(-1) = correct initialization [OK]

Hint: Use new Array(n).fill(-1) for initialized arrays [OK]

Common Mistakes:

• Using wrong type annotation
• Not initializing array values
• Filling with string instead of number

3. Consider the following TypeScript code snippet using Tarjan's algorithm to find bridges. What will be the output bridges for the given graph?

const graph = [[1, 2], [0, 2], [0, 1, 3], [2, 4], [3]];
// Edges: 0-1, 0-2, 1-2, 2-3, 3-4
// Find bridges
const bridges = findBridges(graph);
console.log(bridges);

medium

A. [[2, 3], [3, 4]]

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

C. [[0, 2], [2, 3]]

D. []

Solution

1. Step 1: Analyze the graph edges and connectivity

   The graph edges are 0-1, 0-2, 1-2 forming a triangle, and 2-3, 3-4 forming a chain. The triangle edges are not bridges because removing one does not disconnect the graph.

2. Step 2: Identify bridges

   Edges 2-3 and 3-4 are bridges because removing either disconnects node 4 or node 3 from the rest.

3. Final Answer:

   [[2, 3], [3, 4]] -> Option A

4. Quick Check:

   Bridges disconnect graph = [[2,3],[3,4]] [OK]

Hint: Bridges are edges not in any cycle [OK]

Common Mistakes:

• Including edges inside cycles as bridges
• Missing chain edges as bridges
• Confusing node indices

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

function dfs(u: number, parent: number) {
  visited[u] = true;
  disc[u] = low[u] = time++;
  for (const v of graph[u]) {
    if (!visited[v]) {
      dfs(v, u);
      low[u] = Math.min(low[u], low[v]);
      if (low[v] > disc[u]) {
        bridges.push([u, v]);
      }
    } else if (v !== parent) {
      low[u] = Math.min(low[u], disc[v]);
    }
  }
}

medium

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

B. The 'visited' array is not initialized before DFS

C. The 'time' variable is not incremented correctly

D. The parent node check 'v !== parent' is missing

Solution

1. Step 1: Check initialization of visited array

   The code uses 'visited[u]' but does not show initialization. Without initializing, visited may be undefined causing runtime errors.

2. Step 2: Verify other parts

   The condition 'low[v] > disc[u]' is correct for bridges. The parent check is present. The time increment is correct.

3. Final Answer:

   The 'visited' array is not initialized before DFS -> Option B

4. Quick Check:

   Uninitialized visited array causes errors [OK]

Hint: Always initialize visited before DFS [OK]

Common Mistakes:

• Changing correct bridge condition
• Ignoring visited initialization
• Removing parent check

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

hard

A. [[1, 3]]

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

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

D. []

Solution

1. Step 1: Analyze graph structure

   Edges 0-1, 1-2, 2-0 form a triangle (cycle). Edges 3-4, 4-5, 5-3 form another cycle. Edge 1-3 connects these two cycles.

2. Step 2: Identify bridges

   Only edge 1-3 is a bridge because removing it disconnects the two cycles. Other edges are part of cycles and not bridges.

3. Final Answer:

   [[1, 3]] -> Option A

4. Quick Check:

   Bridge connects two cycles = [[1, 3]] [OK]

Hint: Bridges connect separate cycles or components [OK]

Common Mistakes:

• Marking cycle edges as bridges
• Missing the connecting bridge edge
• Confusing node pairs