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Time Needed to Inform All Employees

A company has n employees with a unique ID for each employee from 0 to n - 1. The head of the company is the one with headID.

Each employee has one direct manager given in the manager array where manager[i] is the direct manager of the i-th employee, manager[headID] = -1. Also, it is guaranteed that the subordination relationships have a tree structure.

The head of the company wants to inform all the company employees of an urgent piece of news. He will inform his direct subordinates, and they will inform their subordinates, and so on until all employees know about the urgent news.

The i-th employee needs informTime[i] minutes to inform all of his direct subordinates (i.e., After informTime[i] minutes, all his direct subordinates can start spreading the news).

Return the number of minutes needed to inform all the employees about the urgent news.

 

Example 1:

Input: n = 1, headID = 0, manager = [-1], informTime = [0]
Output: 0
Explanation: The head of the company is the only employee in the company.

Example 2:

Input: n = 6, headID = 2, manager = [2,2,-1,2,2,2], informTime = [0,0,1,0,0,0]
Output: 1
Explanation: The head of the company with id = 2 is the direct manager of all the employees in the company and needs 1 minute to inform them all.
The tree structure of the employees in the company is shown.

 

Constraints:

  • 1 <= n <= 105
  • 0 <= headID < n
  • manager.length == n
  • 0 <= manager[i] < n
  • manager[headID] == -1
  • informTime.length == n
  • 0 <= informTime[i] <= 1000
  • informTime[i] == 0 if employee i has no subordinates.
  • It is guaranteed that all the employees can be informed.

Solution Explanation: Time Needed to Inform All Employees

This problem describes a scenario where information needs to be disseminated through a hierarchical company structure. The goal is to find the total time required to inform all employees. The most efficient approach is using Depth-First Search (DFS).

Approach: DFS

  1. Graph Representation: The input manager array implicitly defines a directed acyclic graph (DAG). Each employee is a node, and an edge from employee i to employee j exists if manager[j] == i (i.e., i is j's manager). We'll represent this graph using an adjacency list for efficient neighbor lookups.

  2. DFS Function (dfs): This recursive function calculates the time needed to inform all subordinates of a given employee.

    • Base Case: If an employee has no subordinates (their list in the adjacency list is empty), the time is 0.

    • Recursive Step: For each subordinate j of employee i, recursively call dfs(j) to get the time needed to inform all of that subordinate's team. Add informTime[i] (the time employee i takes to inform their direct reports) to the maximum time it takes among all subordinates. This is because the total time is determined by the longest path in the tree.

  3. Main Function: Start the DFS from the headID (root of the tree). The result of dfs(headID) will be the total time needed to inform all employees.

Time and Space Complexity

  • Time Complexity: O(N), where N is the number of employees. Each employee is visited exactly once during the DFS traversal.

  • Space Complexity: O(N) in the worst case, due to the recursive call stack during DFS and the space used to store the adjacency list. The adjacency list's space usage is proportional to the number of edges in the graph, which is at most N-1 in a tree structure.

Code Implementation (Python)

from collections import defaultdict
 
class Solution:
    def numOfMinutes(self, n: int, headID: int, manager: List[int], informTime: List[int]) -> int:
        graph = defaultdict(list)  # Adjacency list to represent the company hierarchy
        for i, m in enumerate(manager):
            graph[m].append(i)
 
        def dfs(employee_id):
            max_time = 0
            for subordinate in graph[employee_id]:
                max_time = max(max_time, dfs(subordinate))
            return max_time + informTime[employee_id]
 
        return dfs(headID)

The other languages (Java, C++, Go, TypeScript, C#) follow a very similar structure, differing only in syntax and data structure implementations. The core logic of the DFS algorithm remains consistent.