Articulation Points in Graph in DSA C++ - 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 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;
  int children = 0;
  for (int v : graph[u]) {
    if (!visited[v]) {
      children++;
      dfs(v, u);
      low[u] = min(low[u], low[v]);
      if (parent != -1 && low[v] >= disc[u])
        articulation[u] = true;
    } else if (v != parent) {
      low[u] = min(low[u], disc[v]);
    }
  }
  if (parent == -1 && children > 1)
    articulation[u] = true;
}
    

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

Identify Repeating Operations

• Primary operation: The depth-first search (DFS) visits each node and explores all its edges.
• 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 and edges grows, the DFS explores more connections, increasing work roughly in proportion.

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

Pattern observation: The work grows roughly linearly with the sum of nodes and edges.

Final Time Complexity

Time Complexity: O(n + m)

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

Common Mistake

[X] Wrong: "The algorithm checks every possible path, so it must be exponential time."

[OK] Correct: The DFS visits each node and edge only once, avoiding repeated work, so it runs in linear time, not exponential.

Interview Connect

Understanding this linear time complexity helps you explain how graph algorithms efficiently find critical points, a valuable skill in many coding challenges.

Self-Check

"What if the graph is represented using 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 with the highest degree in the graph

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

C. A vertex connected to all other vertices

D. A vertex that forms a cycle with its neighbors

Solution

1. Step 1: Understand the role of articulation points

   An articulation point is a vertex that, if removed, breaks the graph into more pieces (connected components).

2. Step 2: Compare options with definition

   Only A vertex which, when removed, increases the number of connected components describes this property correctly; others describe different vertex properties.

3. Final Answer:

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

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 forms cycles
• Assuming articulation point connects to all vertices

2. Which of the following is the correct way to declare a vector of vectors for adjacency list in C++?

easy

A. vector adj;

B. vector adj<int>;

C. vector<int> adj[];

D. vector<vector> adj<int>;

Solution

1. Step 1: Recall correct syntax for 2D vector declaration

   In C++, a vector of vectors is declared as vector name; to represent adjacency lists.

2. Step 2: Check each option

   vector adj; matches correct syntax; A, B and D have syntax errors.

3. Final Answer:

   vector adj; -> Option A

4. Quick Check:

   2D vector syntax = vector [OK]

Hint: Use vector for adjacency lists in C++ [OK]

Common Mistakes:

• Using array syntax instead of vector
• Incorrect template syntax
• Missing angle brackets for nested vectors

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

int timer = 0;
vector<int> disc(4, -1), low(4, -1);
vector<bool> visited(4, false), isAP(4, false);
vector adj = {{1, 2}, {0, 2}, {0, 1, 3}, {2}};

void dfs(int u, int parent) {
    visited[u] = true;
    disc[u] = low[u] = timer++;
    int children = 0;
    for (int v : adj[u]) {
        if (v == parent) continue;
        if (!visited[v]) {
            children++;
            dfs(v, u);
            low[u] = min(low[u], low[v]);
            if (parent != -1 && low[v] >= disc[u])
                isAP[u] = true;
        } else {
            low[u] = min(low[u], disc[v]);
        }
    }
    if (parent == -1 && children > 1)
        isAP[u] = true;
}

dfs(0, -1);

vector<int> result;
for (int i = 0; i < 4; i++) {
    if (isAP[i]) result.push_back(i);
}
for (int x : result) cout << x << " ";

What will be the output?

medium

A. 2

B. 0 1 2

C. 1 2

D. 0 2

Solution

1. Step 1: Analyze the graph structure and DFS traversal

   The graph edges are: 0-1, 0-2, 1-2, 2-3 (triangle 0-1-2 + pendant 3 on 2). DFS from 0 visits 1 then 2 then 3; when back to 0, 2 already visited so root children=1 (not AP).

2. Step 2: Check articulation point conditions in code

   Root children=1 (<=1, not AP). For non-root u=2 (parent=1), low[3]=3 >= disc[2]=2 so isAP[2]=true. No others. Output is 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Root children=1 and vertex 2 disconnects 3 = 2 [OK]

Hint: Root with >1 child and low[v]>=disc[u] mark AP [OK]

Common Mistakes:

• Missing root articulation point condition
• Confusing low and disc arrays
• Ignoring parent check in DFS

4. The following code snippet is intended to find articulation points but has a bug:

void dfs(int u, int parent) {
    visited[u] = true;
    disc[u] = low[u] = timer++;
    int children = 0;
    for (int v : adj[u]) {
        if (!visited[v]) {
            children++;
            dfs(v, u);
            low[u] = min(low[u], low[v]);
            if (low[v] >= disc[u])
                isAP[u] = true;
        } else if (v != parent) {
            low[u] = min(low[u], disc[v]);
        }
    }
    if (parent == -1 && children > 1)
        isAP[u] = true;
}

What is the bug in this code?

medium

A. Not marking visited[u] as true at start

B. Incorrect update of low[u] inside else if block

C. Not incrementing timer before assigning disc[u]

D. Missing check for parent != -1 before marking isAP[u]

Solution

1. Step 1: Understand articulation point condition for non-root nodes

   For non-root nodes, isAP[u] should be set only if parent != -1 and low[v] >= disc[u]. The code misses the parent != -1 check before setting isAP[u].

2. Step 2: Verify other parts of code

   Timer is incremented correctly, visited[u] is set true, low[u] updated properly. So only missing parent check is the bug.

3. Final Answer:

   Missing check for parent != -1 before marking isAP[u] -> Option D

4. Quick Check:

   Parent check needed before marking articulation point [OK]

Hint: Add parent != -1 check before marking articulation point [OK]

Common Mistakes:

• Marking root as articulation point incorrectly
• Confusing order of visited and timer increment
• Ignoring parent in back edge condition

5. Given a connected 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: Analyze graph connectivity and removal effects

   Vertices 0,1,2 form a triangle (cycle), so none of them alone disconnects graph except vertex 1 which connects to 3 and 4. Removing 1 disconnects 3 and 4. Removing 3 disconnects 4. So 1 and 3 are articulation points.

2. Step 2: Confirm articulation points

   Vertex 1 connects two parts: cycle and chain (3-4). Vertex 3 connects 4 only. Removing 1 or 3 increases components. Vertex 0, 2, 4 are not articulation points.

3. Final Answer:

   1 3 -> Option C

4. Quick Check:

   Removing 1 or 3 disconnects graph = 1 3 [OK]

Hint: Check if removing vertex splits graph into parts [OK]

Common Mistakes:

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