Time Needed to Inform All Employees

Tested Python solution for LeetCode 1376 with 23 pytest cases. Generate a practice environment with lcpy.

LeetCode 1376, Medium. Topics: Tree, Depth-First Search, Breadth-First Search. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 23 parametrized pytest cases, and a playground notebook:

lcpy gen -n 1376   # by problem number
lcpy gen -s time_needed_to_inform_all_employees   # by problem name

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Problem

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.

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Examples

Input: n = 1, headID = 0, manager = [-1], informTime = [0]
Output: 0

Explanation: The head of the company is the only employee in the company.

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.

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Constraints

• 1 <= n <= 10^5
• 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.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(n)
    # Space: O(n)
    def num_of_minutes(
        self, n: int, head_id: int, manager: list[int], inform_time: list[int]
    ) -> int:
        children: list[list[int]] = [[] for _ in range(n)]
        for employee, boss in enumerate(manager):
            if boss != -1:
                children[boss].append(employee)

        total = 0
        stack = [(head_id, inform_time[head_id])]
        while stack:
            employee, elapsed = stack.pop()
            total = max(total, elapsed)
            for child in children[employee]:
                stack.append((child, elapsed + inform_time[child]))
        return total

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Complexity

Time	Space
O(n)	O(n)

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Tags

NeetCode All.