How do you solve Count the Number of Complete Components on LeetCode?

Count the Number of Complete Components (LeetCode #2685) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + e) time complexity and O(n) space complexity. Key insight: Understanding the definition of connected components is crucial for solving graph problems.

What is the brute force solution for Count the Number of Complete Components?

The brute force approach for Count the Number of Complete Components is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each connected component of the graph and verifying if it is complete by comparing the number of edges to the expected number of edges in a complete graph. This method is straightforward but inefficient for larger graphs. The optimal Optimal Solution approach improves this to O(n + e).

What algorithm or data structure is used to solve Count the Number of Complete Components?

Count the Number of Complete Components uses the following concepts: Depth-First Search, Breadth-First Search, Union-Find, Graph Theory. The recommended patterns to study are: Graph Traversal (DFS/BFS), Union-Find (for connected components).

What are the interview tips for Count the Number of Complete Components?

Always clarify the definitions of terms like 'connected' and 'complete' before starting your solution. Think about edge cases, such as graphs with no edges or fully connected graphs. Practice both DFS and BFS approaches, as they can be useful in different scenarios.

What are common mistakes when solving Count the Number of Complete Components?

Confusing connected components with complete components. Not properly tracking visited nodes, leading to incorrect counts.

#2685

Count the Number of Complete Components

Medium

Depth-First SearchBreadth-First SearchUnion-FindGraph TheoryGraph Traversal (DFS/BFS)Union-Find (for connected components)

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + e)
Space	O(1)	O(n)

💡

Intuition

Time O(n + e)Space O(n)

The optimal approach uses a graph traversal technique (DFS or BFS) to efficiently explore connected components and check their completeness in a single pass. This reduces the need to repeatedly check edges and vertices, making it faster.

⚙️

Algorithm

5 steps

1Step 1: Create an adjacency list from the edges.
2Step 2: Use DFS or BFS to find all connected components.
3Step 3: For each connected component, count the number of vertices and edges.
4Step 4: Check if the number of edges equals (vertices * (vertices - 1)) / 2 to determine if it's complete.
5Step 5: Count and return the number of complete components.

solution.py34 lines

1# Full working Python code
2from collections import defaultdict
3
4def countCompleteComponents(n, edges):
5    graph = defaultdict(list)
6    for a, b in edges:
7        graph[a].append(b)
8        graph[b].append(a)
9
10    visited = set()
11    complete_count = 0
12
13    def dfs(node):
14        stack = [node]
15        component = []
16        while stack:
17            curr = stack.pop()
18            if curr not in visited:
19                visited.add(curr)
20                component.append(curr)
21                for neighbor in graph[curr]:
22                    if neighbor not in visited:
23                        stack.append(neighbor)
24        return component
25
26    for i in range(n):
27        if i not in visited:
28            component = dfs(i)
29            edges_count = sum(len(graph[v]) for v in component) // 2
30            if edges_count == (len(component) * (len(component) - 1)) // 2:
31                complete_count += 1
32
33    return complete_count
34

ℹ

Complexity note: This complexity is efficient because we traverse each vertex and edge only once, leading to a linear relationship with respect to the number of vertices and edges.

1Understanding the definition of connected components is crucial for solving graph problems.
2A complete component must have edges between every pair of vertices, which can be checked using the formula for complete graphs.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.