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You are given a data structure of employee information, which includes the employee’s unique ID, their importance value, and their direct subordinates’ IDs. You need to implement a function to return the total importance value for a given employee and all their direct and indirect subordinates.

The employee structure is as follows:

class Employee:
    def __init__(self, id: int, importance: int, subordinates: List[int]):
        self.id = id
        self.importance = importance
        self.subordinates = subordinates

The function signature is:

def getImportance(employees: List['Employee'], id: int) -> int:

Note:

  1. One employee has at most one direct leader and may have several subordinates.
  2. The employees have unique IDs.

Clarifying Questions

  1. Can the list of employees be very large?
  2. Are the IDs for subordinates guaranteed to correspond to valid employees in the list?
  3. Is the employee list ordered or unordered with respect to IDs?

Strategy

To solve this problem, we can use either Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the employee hierarchy starting from the given employee ID.

Because we need to get the importance of an employee and all their subordinates, a map (dictionary) will be useful to quickly access each employee object by their ID.

Steps:

  1. Create a dictionary to map each employee’s ID to their corresponding Employee instance.
  2. Use DFS/BFS to traverse all the subordinates starting from the given employee ID.
  3. Accumulate the importance values of all the employees visited during the traversal.

We will use DFS in this solution for its simplicity in recursive implementation.

Code

Here’s the Python implementation of the strategy discussed:

from typing import List

class Employee:
    def __init__(self, id: int, importance: int, subordinates: List[int]):
        self.id = id
        self.importance = importance
        self.subordinates = subordinates

def getImportance(employees: List[Employee], id: int) -> int:
    # Create a dictionary to map employee id to Employee object
    employee_map = {employee.id: employee for employee in employees}
    
    def dfs(employee_id: int) -> int:
        employee = employee_map[employee_id]
        total_importance = employee.importance
        for sub_id in employee.subordinates:
            total_importance += dfs(sub_id)
        return total_importance
    
    return dfs(id)

Time Complexity

The time complexity of this solution is (O(N)), where (N) is the number of employees. This is because in the worst-case scenario, we might need to visit every employee once to sum up their importance. Each employee and their subordinates are visited exactly once.

The space complexity is also (O(N)) due to the recursion stack (DFS) which, in the worst-case scenario, can be as deep as the number of employees. The dictionary mapping will also use (O(N)) space to store employee mappings.

This implementation ensures that we efficiently compute the total importance by leveraging a map for quick lookups and DFS for hierarchy traversal.

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