Solved! Leetcode 1376. Time Needed to Inform All Employees

source: https://leetcode.com/problems/time-needed-to-inform-all-employees/description/

Time Needed to Inform All Employees

Description

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

class Solution {

     public int numOfMinutes(int n, int headID, int[] manager, int[] informTime) {
        Map<Integer, List<Integer>> map = new HashMap<>();
        List<Integer> leafs = new ArrayList<>();
        for(int i=0;i<manager.length;i++){
            if(manager[i]!=-1){
                //adjacency list
                //key is the manager and the value is a list of subordinates
                List<Integer> l = map.getOrDefault(manager[i], new ArrayList<>());
                l.add(i);
                map.put(manager[i], l);
            }
            //if the informatime is zero 
            //an employee doesn't have subordinates
            //hence a leaf node
            if(informTime[i]==0){
                leafs.add(i);
            }
        }
        //timeMap to keep track of time to inform an employee
        Map<Integer, Integer> timeMap = new HashMap<>();
        timeMap.put(headID, 0);

        Queue<Integer> next = new LinkedList<>();
        //initialize the queue with the head of the employees
        next.add(headID);

        while(!next.isEmpty()){
            Queue<Integer> curr = next;
            next = new LinkedList<>();
            while(!curr.isEmpty()){
                int g = curr.poll();
                if(map.containsKey(g)){
                    for(int c: map.get(g)){
                        //time to inform an employee is equals 
                        //to the time taken by information to reach the manager of the employee
                        //plus time taken by the manager to inform his subordinates 
                        timeMap.put(c, timeMap.get(g) + informTime[g]);
                        next.offer(c);
                    }
                }
            }
        }

        int max = Integer.MIN_VALUE;
        //the maximum time taken by information to reach any of the leaf nodes
        for(int i: leafs){
            max = Math.max(max, timeMap.get(i));
        }
        return max;
    }    

}

class Solution {

public int numOfMinutes(int n, int headID, int[] manager, int[] informTime) {

Map<Integer, List<Integer>> map = new HashMap<>();

List<Integer> leafs = new ArrayList<>();

for(int i=0;i<manager.length;i++){

if(manager[i]!=-1){

//adjacency list

//key is the manager and the value is a list of subordinates

List<Integer> l = map.getOrDefault(manager[i], new ArrayList<>());

l.add(i);

map.put(manager[i], l);

}

//if the informatime is zero

//an employee doesn't have subordinates

//hence a leaf node

if(informTime[i]==0){

leafs.add(i);

}

//timeMap to keep track of time to inform an employee

Map<Integer, Integer> timeMap = new HashMap<>();

timeMap.put(headID, 0);

Queue<Integer> next = new LinkedList<>();

//initialize the queue with the head of the employees

next.add(headID);

while(!next.isEmpty()){

Queue<Integer> curr = next;

next = new LinkedList<>();

while(!curr.isEmpty()){

int g = curr.poll();

if(map.containsKey(g)){

for(int c: map.get(g)){

//time to inform an employee is equals

//to the time taken by information to reach the manager of the employee

//plus time taken by the manager to inform his subordinates

timeMap.put(c, timeMap.get(g) + informTime[g]);

next.offer(c);

}

int max = Integer.MIN_VALUE;

//the maximum time taken by information to reach any of the leaf nodes

for(int i: leafs){

max = Math.max(max, timeMap.get(i));

}

return max;

}

Rate this post

Coddicted