Introduction

In this blog post, we will explore how to solve the LeetCode problem "2685" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer n. There is an undirected graph with n vertices, numbered from 0 to n - 1. You are given a 2D integer array edges where edges[i] = [ai, bi] denotes that there exists an undirected edge connecting vertices ai and bi. Return the number of complete connected components of the graph. A connected component is a subgraph of a graph in which there exists a path between any two vertices, and no vertex of the subgraph shares an edge with a vertex outside of the subgraph. A connected component is said to be complete if there exists an edge between every pair of its vertices. Example 1: Input: n = 6, edges = [[0,1],[0,2],[1,2],[3,4]] Output: 3 Explanation: From the picture above, one can see that all of the components of this graph are complete. Example 2: Input: n = 6, edges = [[0,1],[0,2],[1,2],[3,4],[3,5]] Output: 1 Explanation: The component containing vertices 0, 1, and 2 is complete since there is an edge between every pair of two vertices. On the other hand, the component containing vertices 3, 4, and 5 is not complete since there is no edge between vertices 4 and 5. Thus, the number of complete components in this graph is 1. Constraints: 1 <= n <= 50 0 <= edges.length <= n * (n - 1) / 2 edges[i].length == 2 0 <= ai, bi <= n - 1 ai != bi There are no repeated edges.

Explanation

Here's a breakdown of the solution:

• Identify Connected Components: Use Depth First Search (DFS) to find all connected components in the graph.
• Check for Completeness: For each connected component, verify if it's a complete graph. A graph with k nodes is complete if it has k * (k - 1) / 2 edges.

• Count Complete Components: Increment the count for each connected component that satisfies the completeness condition.

• Runtime Complexity: O(n + m), where n is the number of nodes and m is the number of edges.

• Storage Complexity: O(n), primarily due to the adjacency list and visited set in DFS.

Code

    def solve():
    def is_complete(component, adj):
        num_nodes = len(component)
        num_edges = 0
        for u in component:
            for v in component:
                if u != v and v in adj[u]:
                    num_edges += 1
        num_edges //= 2  # Each edge is counted twice
        return num_edges == num_nodes * (num_nodes - 1) // 2

    def dfs(node, adj, visited, component):
        visited.add(node)
        component.add(node)
        for neighbor in adj[node]:
            if neighbor not in visited:
                dfs(neighbor, adj, visited, component)

    n, edges = params

    adj = [set() for _ in range(n)]
    for u, v in edges:
        adj[u].add(v)
        adj[v].add(u)

    visited = set()
    complete_components = 0

    for i in range(n):
        if i not in visited:
            component = set()
            dfs(i, adj, visited, component)
            if is_complete(component, adj):
                complete_components += 1

    return complete_components


def complete_components(n: int, edges: list[list[int]]) -> int:
    global params
    params = (n, edges)
    return solve()