Time Needed to Inform All Employees

Photo by Edvard Alexander Rølvaag on Unsplash

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:

Example 2:

Example 3:

Example 4:

Example 5:


  • 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.

We need to visit the children of each node of our tree structure to compute the total time from the root node to each leaf node. Eventually, we want to find the maximum value over all the root-to-leaf paths. We’ll keep summing the maximum of each level as we move down the tree. We can use the BFS algorithm here to traverse the tree after converting it into an Dictionary.

Let’s see how we might implement this functionality:

  1. Build the dictionary, where each key represent the parent and values are the immediate children.
  2. Create a Node containing the employee id and its cumulative time to reach that node, and insert it into a queue.
  3. Traverse over the nodes while taking the maximum value at each level. Keep inserting each node and update the max cost until the node we just inserted into the queue.
  4. Return the max cost.

Time complexity

The time complexity will be O(n) because all the nodes are traversed only once.

Space complexity

It will take O(n) space for children and O(n) for queue, so the total space complexity will be O(n).

Full Stack Programmer, love to solve problem’s during free time.