Introduction

In this blog post, we will explore how to solve the LeetCode problem "2876" 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 directed graph consisting of n nodes numbered from 0 to n - 1 and n directed edges. You are given a 0-indexed array edges where edges[i] indicates that there is an edge from node i to node edges[i]. Consider the following process on the graph: You start from a node x and keep visiting other nodes through edges until you reach a node that you have already visited before on this same process. Return an array answer where answer[i] is the number of different nodes that you will visit if you perform the process starting from node i. Example 1: Input: edges = [1,2,0,0] Output: [3,3,3,4] Explanation: We perform the process starting from each node in the following way: - Starting from node 0, we visit the nodes 0 -> 1 -> 2 -> 0. The number of different nodes we visit is 3. - Starting from node 1, we visit the nodes 1 -> 2 -> 0 -> 1. The number of different nodes we visit is 3. - Starting from node 2, we visit the nodes 2 -> 0 -> 1 -> 2. The number of different nodes we visit is 3. - Starting from node 3, we visit the nodes 3 -> 0 -> 1 -> 2 -> 0. The number of different nodes we visit is 4. Example 2: Input: edges = [1,2,3,4,0] Output: [5,5,5,5,5] Explanation: Starting from any node we can visit every node in the graph in the process. Constraints: n == edges.length 2 <= n <= 105 0 <= edges[i] <= n - 1 edges[i] != i

Explanation

Here's the approach to solve this problem efficiently:

• Cycle Detection: The core idea is to detect cycles in the graph. For each starting node, we traverse the graph until we either hit a cycle or reach a node that has already been visited from a different starting node.
• Memoization: To avoid redundant computations, we use memoization to store the length of the path starting from each node. This helps us to avoid recomputing paths that have already been calculated, significantly improving efficiency.

• Cycle Length Calculation: When a cycle is detected, we need to determine its length. We mark all nodes belonging to the cycle and store the cycle length as well. Any path leading into the cycle will have the length of the path to reach the cycle plus the length of the cycle itself.

• Time Complexity: O(N), where N is the number of nodes.

• Space Complexity: O(N)

Code

    def count_visited_nodes(edges):
    n = len(edges)
    answer = [0] * n
    visited = [0] * n
    path = [0] * n

    def dfs(node, start_node, path_len):
        if visited[node] == start_node:
            # Cycle detected
            cycle_len = path_len - path[node]
            # Update cycle nodes with the cycle length
            cycle_node = node
            while True:
                answer[cycle_node] = cycle_len
                cycle_node = edges[cycle_node]
                if cycle_node == node:
                    break
            return

        if visited[node] != 0:
            # Already visited from another starting node
            answer[start_node] = answer[node] + path_len - path[node]
            return

        visited[node] = start_node
        path[node] = path_len
        dfs(edges[node], start_node, path_len + 1)
        answer[start_node] = answer[node]
        return

    for i in range(n):
        if answer[i] == 0:  # Only process unvisited nodes
            dfs(i, i, 1)

    return answer