# 1376. 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.

&#x20;**Example 1:**

<pre><code><strong>Input: n = 1, headID = 0, manager = [-1], informTime = [0]
</strong><strong>Output: 0
</strong><strong>Explanation: The head of the company is the only employee in the company.
</strong></code></pre>

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/02/27/graph.png)

<pre><code><strong>Input: n = 6, headID = 2, manager = [2,2,-1,2,2,2], informTime = [0,0,1,0,0,0]
</strong><strong>Output: 1
</strong><strong>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.
</strong>The tree structure of the employees in the company is shown.
</code></pre>

&#x20;**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.

***My Solutions:***

方法1：bottom up dfs

```java
class Solution {
    public int numOfMinutes(int n, int headID, int[] manager, int[] informTime) {
        int res = 0;
        for (int i = 0; i < n; i++) {
            res = Math.max(res, bottomUpHelper(i, manager, informTime));
        }
        return res;
    }

    private int bottomUpHelper(int i, int[] manager, int[] time) {
        // 如果当前人物有manager，继续往上找他的manager的manager
        if (manager[i] != -1) {
            time[i] += bottomUpHelper(manager[i], manager, time);
            manager[i] = -1;
        }
        return time[i];
    }
}
```

方法2：top down dfs。先建立一个manger-小兵-小小兵的map，然后从head开始往下dfs所有的小兵。

```java
class Solution {
   public int numOfMinutes(int n, int headID, int[] manager, int[] informTime) {
        HashMap<Integer, List<Integer>> map = new HashMap<>();
        int count = 0;
        for (int i = 0; i < manager.length; i++)  {
            int m = manager[i];
            if (!map.containsKey(m)) map.put(m, new ArrayList<>());
            map.get(m).add(i);
        }
        return topDownHelper(map, informTime, headID);
    }

    private int topDownHelper(HashMap<Integer, List<Integer>> map, int[] time, int curr) {
        int max = 0;
        // 如果当前人物不是manager，直接返回max
        if (!map.containsKey(curr)) return max;
        // 如果当前人物是manager，遍历ta所有管理的人
        for (int i = 0; i < map.get(curr).size(); i++) {
            max = Math.max(max, topDownHelper(map, time, map.get(curr).get(i)));
        }
        return max + time[curr];
    }
}
```
