1376. Time Needed to Inform All Employees

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.

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

After reading the problem at the beginning, you can probably feel that DFS should be used to solve it. Use DFS to calculate the sum of each inform time from headID to a leaf, and finally take the maximum time.

However, because the manager it passes in is the parent of each node, such data is not conducive to DFS. One method is to use the manager to build a tree, and then use DFS to find all paths from top to bottom.

Another way is, for each node, use DFS to calculate the sum of the inform time from this node to headID, and finally take the maximum time. This is similar to the idea above, but the direction is bottom-up. In addition, a Map is needed to record the sum of the inform time from each node to headID, which can eliminate repeated calculations. For example, first summing the times from 4 to 2, it also sums the times from 0 to 2. Later, when calculating 0 to 2, you can get the value directly from the Map.

DFS

• Time: 
• Space: 

class Solution {
    public int numOfMinutes(int n, int headID, int[] manager, int[] informTime) {
        Map<Integer, Integer> memo = new HashMap<>();
        int ans = 0;
        for (int i = 0; i < n; i++) {
            ans = Math.max(ans, dfs(i, headID, manager, informTime, memo));
        }
        return ans;
    }

    private int dfs(int i, int headID, int[] manager, int[] informTime, Map<Integer, Integer> memo) {
        if (memo.containsKey(i)) {
            return memo.get(i);
        }
        if (i == headID) {
            return 0;
        }

        int parent = manager[i];
        int time = informTime[parent] + dfs(parent, headID, manager, informTime, memo);
        memo.put(i, time);
        return time;
    }
}

Reference

• 1376. Time Needed to Inform All Employees, LeetCode.
• 1376. Time Needed to Inform All Employees, LeetCode Solutions.