Employee Importance

Medium

Rippling

0 views

Topics:

ArraysGraphsRecursion

You have a data structure of employee information, including the employee's unique ID, importance value, and direct subordinates' IDs.

You are given an array of employees employees where:

employees[i].id is the ID of the i^th employee.
employees[i].importance is the importance value of the i^th employee.
employees[i].subordinates is a list of the IDs of the direct subordinates of the i^th employee.

Given an integer id that represents an employee's ID, return the total importance value of this employee and all their direct and indirect subordinates.

Example 1:

Input: employees = [[1,5,[2,3]],[2,3,[]],[3,3,[]]], id = 1
Output: 11
Explanation: Employee 1 has an importance value of 5 and has two direct subordinates: employee 2 and employee 3.
They both have an importance value of 3.
Thus, the total importance value of employee 1 is 5 + 3 + 3 = 11.

Example 2:

Input: employees = [[1,2,[5]],[5,-3,[]]], id = 5
Output: -3
Explanation: Employee 5 has an importance value of -3 and has no direct subordinates.
Thus, the total importance value of employee 5 is -3.

Constraints:

1 <= employees.length <= 2000
1 <= employees[i].id <= 2000
All employees[i].id are unique.
-100 <= employees[i].importance <= 100
One employee has at most one direct leader and may have several subordinates.
The IDs in employees[i].subordinates are valid IDs.

Employee Importance

Medium

Rippling

0 views

Topics:

ArraysGraphsRecursion

You have a data structure of employee information, including the employee's unique ID, importance value, and direct subordinates' IDs.

You are given an array of employees employees where:

employees[i].id is the ID of the i^th employee.
employees[i].importance is the importance value of the i^th employee.
employees[i].subordinates is a list of the IDs of the direct subordinates of the i^th employee.

Given an integer id that represents an employee's ID, return the total importance value of this employee and all their direct and indirect subordinates.

Example 1:

Input: employees = [[1,5,[2,3]],[2,3,[]],[3,3,[]]], id = 1
Output: 11
Explanation: Employee 1 has an importance value of 5 and has two direct subordinates: employee 2 and employee 3.
They both have an importance value of 3.
Thus, the total importance value of employee 1 is 5 + 3 + 3 = 11.

Example 2:

Input: employees = [[1,2,[5]],[5,-3,[]]], id = 5
Output: -3
Explanation: Employee 5 has an importance value of -3 and has no direct subordinates.
Thus, the total importance value of employee 5 is -3.

Constraints:

1 <= employees.length <= 2000
1 <= employees[i].id <= 2000
All employees[i].id are unique.
-100 <= employees[i].importance <= 100
One employee has at most one direct leader and may have several subordinates.
The IDs in employees[i].subordinates are valid IDs.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range for the employee IDs and importance values? Are they always positive integers?
Is it possible for the input list of employees to be empty or null?
Could there be cases where an employee's ID is not present in the list, or cases where an employee directly or indirectly reports to themselves, creating a cycle?
If an employee ID isn't found, should I return 0 or null, or throw an exception?
Can an employee appear multiple times in the input list, and if so, how should I handle it?

Brute Force Solution

Approach

The brute force approach calculates the total importance of an employee and all their direct and indirect subordinates. It works by checking every employee and recursively summing up the importance values of everyone below them in the organizational structure.

Here's how the algorithm would work step-by-step:

Start with the employee we want to find the total importance for.
Get that employee's importance value.
Check if that employee has any subordinates (employees who report directly or indirectly to them).
If they do have subordinates, repeat the process for each of those subordinates. Find their importance values and check *their* subordinates.
Keep doing this until you reach employees who have no subordinates.
Once you've gone through everyone, add up all the individual importance values you found along the way. This gives you the total importance.

Code Implementation

def employee_importance_brute_force(employees, employee_id):
    employee_map = {employee.id: employee for employee in employees}

    def get_total_importance(current_id):
        employee = employee_map[current_id]
        total_importance = employee.importance

        # Recursively sum importance of subordinates

        for subordinate_id in employee.subordinates:

            total_importance += get_total_importance(subordinate_id)

        return total_importance

    # Begin recursion to find total importance

    total_employee_importance = get_total_importance(employee_id)
    return total_employee_importance

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – In the worst-case scenario, the algorithm might need to traverse the entire employee hierarchy. Since we start with a specific employee and recursively explore their subordinates, the total number of employees visited could potentially be all 'n' employees. Even though it seems recursive, each employee's importance is only added once. Therefore, the time complexity is proportional to the number of employees, resulting in O(n).

Space Complexity

O(N) – The brute force approach uses recursion. In the worst-case scenario, the organizational structure could resemble a linked list where each employee only reports to one manager, leading to a recursion depth of N, where N is the number of employees. Each recursive call adds a new frame to the call stack, storing local variables and the return address. Therefore, the maximum space used by the call stack would be proportional to N, giving a space complexity of O(N).

Optimal Solution

Approach

The goal is to find the total importance value of an employee and all their subordinates. The optimal approach involves efficiently searching through employee data to gather importance values, avoiding redundant checks by storing employee information in a way that facilitates quick lookups.

Here's how the algorithm would work step-by-step:

First, organize the employee information in a way that makes it quick to find an employee's information by their unique ID. Think of it like creating a well-organized filing cabinet.
Start with the employee whose total importance needs to be calculated. This is like picking a file from the cabinet to begin your search.
Add the employee's importance value to a running total. This is the starting value you are building upon.
Look at the employee's list of subordinates (the people who report to them).
For each subordinate, find their information in the organized employee data (the filing cabinet).
Add each subordinate's importance value to the running total. This continues building the value.
Repeat the process of looking up subordinates and adding their importance values for each of the current employee's subordinates.
Continue until you've gone through all employees and their subordinates linked to the initial employee. Then the running total contains the total importance.

Code Implementation

class Employee:
    def __init__(self, employee_id, importance, subordinates):
        self.employee_id = employee_id
        self.importance = importance
        self.subordinates = subordinates

def get_importance(employees, employee_id):
    employee_map = {employee.employee_id: employee for employee in employees}

    # Store employees in a map for quick ID-based lookup.
    employee_information = employee_map.get(employee_id)
    if not employee_information:
        return 0

    total_importance_value = 0

    def dfs(employee):
        nonlocal total_importance_value

        # Add the current employee's importance.
        total_importance_value += employee.importance

        # Traverse through subordinates to calculate total importance
        for subordinate_id in employee.subordinates:
            subordinate = employee_map.get(subordinate_id)
            if subordinate:
                dfs(subordinate)

    dfs(employee_information)

    return total_importance_value

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The solution involves iterating through each employee and their subordinates to calculate the total importance. Creating a dictionary (or hash map) allows us to look up employee information by ID in O(1) time. We visit each employee and their subordinates exactly once to calculate the total importance, effectively traversing the employee hierarchy. Therefore, the time complexity is directly proportional to the total number of employees, including subordinates, which is represented by n. Thus, the overall time complexity is O(n).

Space Complexity

O(N) – The algorithm utilizes a data structure (described as a 'filing cabinet') to store employee information, which allows for quick lookups by employee ID. This data structure, likely a hash map or dictionary, stores information for each employee, with the space used growing linearly with the number of employees. Therefore, the auxiliary space required to store employee data is proportional to N, where N is the number of employees. Recursive calls to process subordinates could, in the worst-case scenario where the employee hierarchy is very deep, lead to a recursion depth proportional to the number of employees, further contributing to O(N) space complexity.

Edge Cases

Case	How to Handle
Empty employee list	Return 0 since there are no employees to sum importance.
Employee list contains only one employee	Return the importance value of that single employee.
Employee ID not found in the list	Return 0 or throw an exception indicating invalid ID (depending on requirements).
Circular references in subordinates (cycles in the graph)	Use a 'visited' set during the traversal to prevent infinite loops and double-counting importance.
Maximum number of employees exceeding memory capacity	Consider iterative approach with bounded memory or using a database if data is too large.
Large importance values that may cause integer overflow	Use a larger integer type (long) or check for overflow before addition.
Employee with empty subordinates list	The algorithm should handle this case correctly since there's nothing to recurse/iterate on.
Duplicate employee IDs in the input list	The input data should be pre-validated to prevent duplicate IDs, or the first occurrence of an ID can be used and subsequent duplicates ignored.