Introduction

In this blog post, we will explore how to solve the LeetCode problem "2858" 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 is a simple directed graph with n nodes labeled from 0 to n - 1. The graph would form a tree if its edges were bi-directional. You are given an integer n and a 2D integer array edges, where edges[i] = [ui, vi] represents a directed edge going from node ui to node vi. An edge reversal changes the direction of an edge, i.e., a directed edge going from node ui to node vi becomes a directed edge going from node vi to node ui. For every node i in the range [0, n - 1], your task is to independently calculate the minimum number of edge reversals required so it is possible to reach any other node starting from node i through a sequence of directed edges. Return an integer array answer, where answer[i] is the minimum number of edge reversals required so it is possible to reach any other node starting from node i through a sequence of directed edges. Example 1: Input: n = 4, edges = [[2,0],[2,1],[1,3]] Output: [1,1,0,2] Explanation: The image above shows the graph formed by the edges. For node 0: after reversing the edge [2,0], it is possible to reach any other node starting from node 0. So, answer[0] = 1. For node 1: after reversing the edge [2,1], it is possible to reach any other node starting from node 1. So, answer[1] = 1. For node 2: it is already possible to reach any other node starting from node 2. So, answer[2] = 0. For node 3: after reversing the edges [1,3] and [2,1], it is possible to reach any other node starting from node 3. So, answer[3] = 2. Example 2: Input: n = 3, edges = [[1,2],[2,0]] Output: [2,0,1] Explanation: The image above shows the graph formed by the edges. For node 0: after reversing the edges [2,0] and [1,2], it is possible to reach any other node starting from node 0. So, answer[0] = 2. For node 1: it is already possible to reach any other node starting from node 1. So, answer[1] = 0. For node 2: after reversing the edge [1, 2], it is possible to reach any other node starting from node 2. So, answer[2] = 1. Constraints: 2 <= n <= 105 edges.length == n - 1 edges[i].length == 2 0 <= ui == edges[i][0] < n 0 <= vi == edges[i][1] < n ui != vi The input is generated such that if the edges were bi-directional, the graph would be a tree.

Explanation

Here's a breakdown of the solution:

• Build Adjacency List: Create an adjacency list to represent the graph. For each edge (u, v), store both (v, 0) and (u, 1) in the adjacency lists of u and v respectively. The 0 denotes a forward edge (no reversal needed), while 1 denotes a backward edge (reversal needed).
• Depth-First Search (DFS): Perform DFS from each node i to calculate the minimum number of reversals needed to reach all other nodes. Keep track of visited nodes to avoid cycles.

• Efficient Reversal Counting: During the DFS, increment the reversal count only when traversing a backward edge.

• Runtime Complexity: O(n^2), where n is the number of nodes.

• Storage Complexity: O(n), for the adjacency list and visited set.

Code

    def min_edge_reversals(n: int, edges: list[list[int]]) -> list[int]:
    """
    Calculates the minimum number of edge reversals required to reach any other node
    starting from each node in a directed graph.

    Args:
        n: The number of nodes in the graph.
        edges: A list of directed edges, where edges[i] = [ui, vi].

    Returns:
        A list of integers, where answer[i] is the minimum number of edge reversals
        required to reach any other node starting from node i.
    """

    adj = [[] for _ in range(n)]
    for u, v in edges:
        adj[u].append((v, 0))  # Forward edge (no reversal)
        adj[v].append((u, 1))  # Backward edge (reversal needed)

    def dfs(start_node: int) -> int:
        """
        Performs DFS to calculate the minimum reversals from a starting node.

        Args:
            start_node: The starting node for the DFS.

        Returns:
            The minimum number of edge reversals.
        """

        reversals = 0
        visited = {start_node}
        stack = [(start_node, 0)]  # (node, cost)

        while stack:
            node, cost = stack.pop()
            reversals += cost

            for neighbor, edge_type in adj[node]:
                if neighbor not in visited:
                    visited.add(neighbor)
                    stack.append((neighbor, edge_type))
        return reversals

    answer = []
    for i in range(n):
        reversals = 0
        visited = {i}
        stack = [(i, 0)]

        q = [(i, 0, {i})]  # (node, cost, visited)

        total_reversals = 0
        all_visited = set()
        all_visited.add(i)

        def dfs2(node, reversals_so_far, visited_so_far):
            nonlocal total_reversals
            if len(visited_so_far) == n:
                total_reversals = reversals_so_far
                return

            min_reversals = float('inf')


            for neighbor, edge_type in adj[node]:
                if neighbor not in visited_so_far:
                    new_visited = visited_so_far.copy()
                    new_visited.add(neighbor)


                    dfs2(neighbor, reversals_so_far + edge_type, new_visited)



        def bfs(start_node):
            reversals = 0
            visited = {start_node}
            q = [(start_node, 0, visited)]
            min_reversals = float('inf')

            while q:
                curr_node, curr_reversals, curr_visited = q.pop(0)

                if len(curr_visited) == n:
                    min_reversals = curr_reversals
                    return min_reversals

                for neighbor, edge_type in adj[curr_node]:
                    if neighbor not in curr_visited:
                        new_visited = curr_visited.copy()
                        new_visited.add(neighbor)
                        q.append((neighbor, curr_reversals + edge_type, new_visited))
            return min_reversals

        answer.append(bfs(i))

    return answer