Time Needed to Inform All Employees

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TreesGraphsRecursion

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.

Time Needed to Inform All Employees

Medium

Google

3 views

Topics:

TreesGraphsRecursion

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

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 are the possible ranges for the employee IDs (headID and elements within managers)? Can I assume they are non-negative integers?
Can an employee be managed by multiple managers (i.e., can an employee ID appear multiple times in the `managers` array)?
Is it guaranteed that the `managers` array represents a valid employee hierarchy (i.e., no cycles, and every employee eventually reports to the `headID` or to an employee that reports to the `headID`)?
What happens if the `managers` array is empty or null? Should I return 0 in that case?
Are the `informTime` values always non-negative? Can they be zero?

Brute Force Solution

Approach

The brute force method for informing all employees involves simulating the information spreading process for every possible starting point and path. We explore all possible scenarios to find the longest time it takes for the information to reach everyone. This exhaustive search guarantees we find the absolute worst-case time, but might take a long time to finish.

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

Start with the first employee as the person who initially receives the information.
Trace how that employee informs their direct reports, and how those reports inform their reports, and so on, until everyone has the information. Keep track of how long this takes.
Now, repeat the process, but this time start with the second employee as the initially informed person. Again, trace the information spreading and track the time.
Do this for every employee; each one gets a turn to be the starting point for the information.
For each employee's scenario, consider all the possible ways the information could spread among the employees. For example, one employee may have multiple direct reports, and you need to try spreading the info to them in different orders.
After simulating the information spread from every employee and every possible path, compare the total time it took in each scenario.
The longest time it took for the information to reach everyone in any of the scenarios is the answer.

Code Implementation

def time_needed_to_inform_all_employees_brute_force(
    number_of_employees, manager_list, inform_time
):
    max_time = 0

    # Iterate through each employee as the starting point
    for starting_employee in range(number_of_employees):
        time_for_this_start = inform_time[starting_employee]
        informed = [False] * number_of_employees
        informed[starting_employee] = True
        current_time = 0
        total_informed = 1

        while total_informed < number_of_employees:
            newly_informed = []
            
            # Find employees who can be informed in this iteration
            for employee_index in range(number_of_employees):
                if informed[employee_index]:
                    for subordinate_index in range(number_of_employees):
                        if manager_list[subordinate_index] == employee_index:
                            if not informed[subordinate_index]:
                                newly_informed.append(subordinate_index)
            
            #If nobody was newly informed, it means we have a problem
            if not newly_informed: 
                current_time = float('inf')
                break

            max_inform_time = 0
            
            #Determine the max time it takes to inform newly informed people
            for employee_to_inform in newly_informed:
                max_inform_time = max(max_inform_time, inform_time[employee_to_inform])
            
            #mark new informs and increment totals
            for employee_to_inform in newly_informed:
                informed[employee_to_inform] = True
                total_informed += 1

            current_time += max_inform_time

        max_time = max(max_time, current_time)

    return max_time

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n!) – The algorithm simulates the information spreading process starting from each of the n employees. For each employee, it considers all possible ways the information could spread among their direct reports and their subsequent reports. This involves exploring different orders of informing employees, leading to a permutation of the subordinates. In the worst case, where each employee has many subordinates, the number of possible spreading paths grows factorially. Therefore, the overall time complexity is approximately O(n!), as for each of the n employees, we might need to explore up to n! different information spreading paths.

Space Complexity

O(N!) – The brute force method described explores all possible information spreading paths among employees. For each starting employee, it considers all permutations of their direct reports, and recursively continues this process down the hierarchy. In the worst-case, an employee might have many direct reports, leading to the algorithm needing to store information about the current path being explored. The number of such paths can grow factorially with the number of employees, N, leading to a recursion depth that may require storing O(N!) worth of information in the call stack as different orderings of message passing are tried. Therefore, the space complexity is O(N!).

Optimal Solution

Approach

The best way to solve this problem is to figure out how long it takes for the information to reach each employee from the head of the company. We need to efficiently explore the company structure to find the maximum time it takes for the information to spread.

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

Imagine the company structure as a series of connections, where each employee knows who they report to.
Start with the person who knows the information first (the head of the company).
Follow the connections to each employee down the chain of command.
As you move down, keep track of how much time it takes for the information to reach each person. Add the time it takes to inform each employee to the time their manager was informed.
If someone has multiple people reporting to them, the information spreads to each of them separately.
Keep going until you've reached every employee in the company.
The longest time it took for the information to reach any employee is the answer.

Code Implementation

def time_needed_to_inform_all_employees(number_of_employees, head_id, managers, inform_time):
    adjacency_list = [[] for _ in range(number_of_employees)]

    for employee, manager in enumerate(managers):
        if employee != head_id:
            adjacency_list[manager].append(employee)

    information_times = [0] * number_of_employees

    def depth_first_search(employee_id):
        # Use DFS to propagate information times down the hierarchy.
        for subordinate in adjacency_list[employee_id]:

            information_times[subordinate] = information_times[employee_id] + inform_time[employee_id]

            depth_first_search(subordinate)

    depth_first_search(head_id)

    # Find the maximum time it takes to inform all employees.
    max_time = 0
    for time in information_times:
        max_time = max(max_time, time)

    return max_time

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each employee to build the adjacency list representing the company structure, which takes O(n) time. Then, it performs a Depth-First Search (DFS) or Breadth-First Search (BFS) traversal starting from the head, visiting each employee once to calculate the information propagation time. Therefore, the time complexity is dominated by the graph traversal, visiting each of the n employees, making the overall time complexity O(n).

Space Complexity

O(N) – The algorithm implicitly builds a graph representation of the company hierarchy. In the worst case, the breadth-first or depth-first traversal, implied by steps 3-6, might require storing the entire company structure in memory to keep track of employees to visit and the time it takes to reach them. This means that in the worst-case scenario, we may need a queue or stack (depending on implementation details not specified in the plain English steps) to hold up to N employees (where N is the total number of employees). Therefore, the auxiliary space is proportional to N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
managers array is null or empty	Return 0 if managers array is null or empty, as there are no employees to inform.
informTime array is null or empty	Return 0 if informTime array is null or empty, as there's no time to inform anyone.
headID is invalid (out of range)	Return -1 if headID is out of the bounds of the managers array, indicating an error.
n is zero or negative	Return 0 if n (number of employees) is zero or negative, as there are no employees.
All employees report directly to the headID.	This creates a skewed tree; the solution should still correctly calculate the time based on the sum of informTime values of employees under headID.
informTime values are very large, potentially causing integer overflow.	Use long integers to prevent integer overflow during time calculation.
Circular dependencies in the manager-employee relationships.	Use a visited set to detect cycles and return -1 to indicate invalid input.
Employees with no subordinates.	The algorithm correctly handles employees with no subordinates because their informTime is added to the total path time of their managers.