Articulation Points in Graph in DSA Javascript - Time & Space Complexity

Time Complexity: Articulation Points in Graph

O(n + m)

Understanding Time Complexity

We want to understand how the time needed to find articulation points changes as the graph grows.

How does the algorithm's work increase when the graph has more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function dfs(u, parent) {
  visited[u] = true;
  disc[u] = low[u] = time++;
  let children = 0;
  for (const v of graph[u]) {
    if (!visited[v]) {
      children++;
      dfs(v, u);
      low[u] = Math.min(low[u], low[v]);
      if (parent === -1 && children > 1) articulation[u] = true;
      if (parent !== -1 && low[v] >= disc[u]) articulation[u] = true;
    } else if (v !== parent) {
      low[u] = Math.min(low[u], disc[v]);
    }
  }
}

This code finds articulation points using a depth-first search on the graph.

Identify Repeating Operations

Look for loops and recursive calls that repeat work.

• Primary operation: The depth-first search visits each node and checks all its neighbors.
• How many times: Each node and edge is visited once during the search.

How Execution Grows With Input

As the graph grows, the algorithm visits every node and edge once.

Input Size (n nodes, m edges)	Approx. Operations
10 nodes, 15 edges	About 25 visits (nodes + edges)
100 nodes, 200 edges	About 300 visits
1000 nodes, 5000 edges	About 6000 visits

Pattern observation: The work grows roughly in proportion to the number of nodes plus edges.

Final Time Complexity

Time Complexity: O(n + m)

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

Common Mistake

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

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

Interview Connect

Understanding this time complexity helps you explain how graph algorithms scale and why efficient traversal matters.

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 an articulation point in a graph?

easy

A. A vertex that, when removed, increases the number of connected components

B. A vertex with the highest degree in the graph

C. A vertex connected to all other vertices

D. A vertex that has no edges connected to it

Solution

1. Step 1: Understand the definition of articulation point

   An articulation point is a vertex that, if removed, increases the number of disconnected parts in the graph.

2. Step 2: Compare options with definition

   Only A vertex that, when removed, increases the number of connected components matches this definition exactly.

3. Final Answer:

   A vertex that, when removed, increases the number of connected components -> Option A

4. Quick Check:

   Articulation point = vertex increasing components when removed [OK]

Hint: Articulation point breaks graph connectivity when removed [OK]

Common Mistakes:

• Confusing articulation point with highest degree vertex
• Thinking articulation point is always connected to all vertices
• Assuming isolated vertices are articulation points

2. Which of the following is the correct way to initialize a graph adjacency list in JavaScript for articulation point detection?

easy

A. const graph = Array.from({ length: n }, () => []);

B. const graph = new Array(n).fill([]);

C. const graph = Array(n).fill(null);

D. const graph = new Map();

Solution

1. Step 1: Understand adjacency list initialization

   We need an array of separate empty arrays for each vertex to store neighbors.

2. Step 2: Analyze options

   const graph = new Array(n).fill([]); uses fill([]), which fills all positions with the same array reference (wrong). const graph = Array.from({ length: n }, () => []); uses Array.from with a function returning new arrays, so each vertex has its own array (correct). const graph = Array(n).fill(null); fills with null, not arrays. const graph = new Map(); uses Map, which is valid but not the typical adjacency list array.

3. Final Answer:

   const graph = Array.from({ length: n }, () => []); -> Option A

4. Quick Check:

   Separate arrays per vertex = const graph = Array.from({ length: n }, () => []); [OK]

Hint: Use Array.from with function to create separate arrays [OK]

Common Mistakes:

• Using fill([]) creates shared references causing bugs
• Initializing with null instead of arrays
• Confusing Map with adjacency list array

3. Consider the following JavaScript code snippet for finding articulation points in a graph with 4 vertices (0 to 3):

const graph = [[1,2],[0],[0,3],[2]];
const visited = [false, false, false, false];
const disc = [0, 0, 0, 0];
const low = [0, 0, 0, 0];
const parent = [-1, -1, -1, -1];
const ap = [false, false, false, false];
let time = 0;

function dfs(u) {
  visited[u] = true;
  disc[u] = low[u] = ++time;
  let children = 0;
  for (const v of graph[u]) {
    if (!visited[v]) {
      children++;
      parent[v] = u;
      dfs(v);
      low[u] = Math.min(low[u], low[v]);
      if (parent[u] === -1 && children > 1) ap[u] = true;
      if (parent[u] !== -1 && low[v] >= disc[u]) ap[u] = true;
    } else if (v !== parent[u]) {
      low[u] = Math.min(low[u], disc[v]);
    }
  }
}

dfs(0);
console.log(ap);

What will be the output printed?

medium

A. [false, false, false, false]

B. [true, false, true, false]

C. [true, true, false, false]

D. [false, true, true, false]

Solution

1. Step 1: Trace DFS and articulation point logic

   Start DFS at 0: children are 1 and 2. For root (0), if children > 1, ap[0] = true. So 0 is articulation point. For vertex 2, low[3] >= disc[2] triggers ap[2] = true.

2. Step 2: Determine articulation points array

   Vertices 0 and 2 are articulation points, others are not. So ap = [true, false, true, false].

3. Final Answer:

   [true, false, true, false] -> Option B

4. Quick Check:

   Root with >1 child and low[v]>=disc[u] mark articulation points [OK]

Hint: Root with multiple children and low[v]>=disc[u] mark articulation points [OK]

Common Mistakes:

• Ignoring root special case with multiple children
• Confusing low and disc arrays
• Not updating low[u] on back edges

4. The following code attempts to find articulation points but has a bug:

function dfs(u) {
  visited[u] = true;
  disc[u] = low[u] = ++time;
  let children = 0;
  for (const v of graph[u]) {
    if (!visited[v]) {
      children++;
      parent[v] = u;
      dfs(v);
      low[u] = Math.min(low[u], low[v]);
      if (low[v] >= disc[u]) ap[u] = true;
    } else if (v !== parent[u]) {
      low[u] = Math.min(low[u], disc[v]);
    }
  }
}

What is the bug in this code?

medium

A. It does not mark visited vertices

B. It uses '>' instead of '>=' in the articulation point check

C. It does not update parent array correctly

D. It misses the root vertex articulation point condition

Solution

1. Step 1: Analyze articulation point conditions

   The code checks if low[v] >= disc[u] to mark articulation points but misses the special case for root vertex which requires children > 1.

2. Step 2: Identify missing root condition

   Without checking if u is root and has multiple children, root articulation points won't be detected.

3. Final Answer:

   It misses the root vertex articulation point condition -> Option D

4. Quick Check:

   Root articulation point requires separate children > 1 check [OK]

Hint: Always check root vertex children count separately [OK]

Common Mistakes:

• Forgetting root vertex special case
• Using wrong comparison operator in articulation check
• Not updating parent array before DFS call

5. Given a connected undirected graph with 5 vertices and edges: 0-1, 1-2, 2-0, 1-3, 3-4, which vertices are articulation points?

hard

A. [1]

B. [0, 1, 3]

C. [1, 3]

D. [3, 4]

Solution

1. Step 1: Visualize the graph and connections

   Vertices 0,1,2 form a triangle. Vertex 1 connects to 3, which connects to 4.

2. Step 2: Identify articulation points

   Removing 1 disconnects 3 and 4 from the rest, so 1 is articulation point. Removing 3 disconnects 4, so 3 is articulation point. 0,2,4 are not articulation points.

3. Final Answer:

   [1, 3] -> Option C

4. Quick Check:

   Vertices connecting subgraphs are articulation points [OK]

Hint: Articulation points connect separate graph parts [OK]

Common Mistakes:

• Including vertices inside cycles as articulation points
• Missing leaf node articulation points
• Confusing articulation points with bridges