# Time Needed to Inform All Employees

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.

**Example 3:**

**Input:** n = 7, headID = 6, manager = [1,2,3,4,5,6,-1], informTime = [0,6,5,4,3,2,1]

**Output:** 21

**Explanation:** The head has id = 6. He will inform employee with id = 5 in 1 minute.

The employee with id = 5 will inform the employee with id = 4 in 2 minutes.

The employee with id = 4 will inform the employee with id = 3 in 3 minutes.

The employee with id = 3 will inform the employee with id = 2 in 4 minutes.

The employee with id = 2 will inform the employee with id = 1 in 5 minutes.

The employee with id = 1 will inform the employee with id = 0 in 6 minutes.

Needed time = 1 + 2 + 3 + 4 + 5 + 6 = 21.

**Example 4:**

**Input:** n = 15, headID = 0, manager = [-1,0,0,1,1,2,2,3,3,4,4,5,5,6,6], informTime = [1,1,1,1,1,1,1,0,0,0,0,0,0,0,0]

**Output:** 3

**Explanation:** The first minute the head will inform employees 1 and 2.

The second minute they will inform employees 3, 4, 5 and 6.

The third minute they will inform the rest of employees.

**Example 5:**

**Input:** n = 4, headID = 2, manager = [3,3,-1,2], informTime = [0,0,162,914]

**Output:** 1076

**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.

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:

- Build the dictionary, where each key represent the parent and values are the immediate children.
- Create a Node containing the employee
`id`

and its cumulative time to reach that node, and insert it into a*queue*. - 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.
- 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*).