Introduction

In this blog post, we will explore how to solve the LeetCode problem "1466" 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

There are n cities numbered from 0 to n - 1 and n - 1 roads such that there is only one way to travel between two different cities (this network form a tree). Last year, The ministry of transport decided to orient the roads in one direction because they are too narrow. Roads are represented by connections where connections[i] = [ai, bi] represents a road from city ai to city bi. This year, there will be a big event in the capital (city 0), and many people want to travel to this city. Your task consists of reorienting some roads such that each city can visit the city 0. Return the minimum number of edges changed. It's guaranteed that each city can reach city 0 after reorder. Example 1: Input: n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]] Output: 3 Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital). Example 2: Input: n = 5, connections = [[1,0],[1,2],[3,2],[3,4]] Output: 2 Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital). Example 3: Input: n = 3, connections = [[1,0],[2,0]] Output: 0 Constraints: 2 <= n <= 5 * 104 connections.length == n - 1 connections[i].length == 2 0 <= ai, bi <= n - 1 ai != bi

Explanation

Here's the breakdown of the solution:

• Represent the Graph: Represent the road network as an adjacency list. Crucially, store each edge as both a directed edge (a -> b) and its reverse (b -> a) to facilitate easy traversal and reorientation counting.
• Depth-First Search (DFS): Perform a DFS starting from the capital city (city 0). During the DFS, count the number of edges that need to be reversed. An edge needs to be reversed if, in the original connections list, it was oriented away from the capital (i.e., from the child node to the parent node).

• Optimized Counting: The presence of both forward and reverse edges in the adjacency list lets us efficiently determine if a road needs to be reoriented.

• Time & Space Complexity: O(N) time complexity, where N is the number of cities, due to DFS traversal. O(N) space complexity due to the adjacency list representation.

Code

    def minReorder(n: int, connections: list[list[int]]) -> int:
    """
    Calculates the minimum number of roads to reorient so that all cities can reach city 0.

    Args:
        n: The number of cities.
        connections: A list of road connections.

    Returns:
        The minimum number of edges changed.
    """

    adj = [[] for _ in range(n)]
    for a, b in connections:
        adj[a].append((b, 1))  # 1 indicates original direction
        adj[b].append((a, 0))  # 0 indicates reverse direction

    reorientations = 0

    def dfs(node, visited):
        nonlocal reorientations
        visited[node] = True
        for neighbor, direction in adj[node]:
            if not visited[neighbor]:
                reorientations += direction  # Add 1 if original direction
                dfs(neighbor, visited)

    visited = [False] * n
    dfs(0, visited)
    return reorientations