Introduction

In this blog post, we will explore how to solve the LeetCode problem "2316" 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 nodes, 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 nodes ai and bi. Return the number of pairs of different nodes that are unreachable from each other. Example 1: Input: n = 3, edges = [[0,1],[0,2],[1,2]] Output: 0 Explanation: There are no pairs of nodes that are unreachable from each other. Therefore, we return 0. Example 2: Input: n = 7, edges = [[0,2],[0,5],[2,4],[1,6],[5,4]] Output: 14 Explanation: There are 14 pairs of nodes that are unreachable from each other: [[0,1],[0,3],[0,6],[1,2],[1,3],[1,4],[1,5],[2,3],[2,6],[3,4],[3,5],[3,6],[4,6],[5,6]]. Therefore, we return 14. Constraints: 1 <= n <= 105 0 <= edges.length <= 2 * 105 edges[i].length == 2 0 <= ai, bi < n ai != bi There are no repeated edges.

Explanation

Here's the breakdown of the solution:

• Connected Components: The core idea is to identify the connected components within the graph. Nodes within the same connected component are reachable from each other.
• Counting Unreachable Pairs: Once we have the sizes of each connected component, we can calculate the number of unreachable pairs. For each component, we multiply its size by the sum of the sizes of all subsequent components.

• Depth-First Search (DFS): DFS is used to efficiently explore and identify each connected component.

• 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 representation of the graph and the visited set in DFS.

Code

    def count_unreachable_pairs(n: int, edges: list[list[int]]) -> int:
    """
    Calculates the number of pairs of nodes that are unreachable from each other in a graph.

    Args:
        n: The number of nodes in the graph.
        edges: A list of edges, where each edge is a list of two integers representing the connected nodes.

    Returns:
        The number of unreachable pairs of nodes.
    """

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

    visited = [False] * n
    component_sizes = []

    def dfs(node):
        visited[node] = True
        size = 1
        for neighbor in adj[node]:
            if not visited[neighbor]:
                size += dfs(neighbor)
        return size

    for i in range(n):
        if not visited[i]:
            component_sizes.append(dfs(i))

    unreachable_pairs = 0
    total_nodes = 0
    for size in component_sizes:
        unreachable_pairs += size * (n - total_nodes - size)
        total_nodes += size

    return unreachable_pairs