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Formatted question description: https://leetcode.ca/all/1376.html
1376. Time Needed to Inform All Employees (Medium)
A company has n
employees with a unique ID for each employee from 0
to n  1
. The head of the company has 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 ith
employee, manager[headID] = 1
. Also it's guaranteed that the subordination relationships have a tree structure.
The head of the company wants to inform all the employees of the company 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 ith
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 <= 10^5
0 <= headID < n
manager.length == n
0 <= manager[i] < n
manager[headID] == 1
informTime.length == n
0 <= informTime[i] <= 1000
informTime[i] == 0
if employeei
has no subordinates. It is guaranteed that all the employees can be informed.
Related Topics:
Depthfirst Search
Similar Questions:
Solution 1. BFS
Build the tree then get the maximum aggregated informTime (i.e. the sum of informTime starting from headID to the current node) of each (leaf) node.
// OJ: https://leetcode.com/problems/timeneededtoinformallemployees/
// Time: O(N)
// Space: O(N)
class Solution {
public:
int numOfMinutes(int n, int headID, vector<int>& manager, vector<int>& informTime) {
unordered_map<int, vector<int>> m; // manager => subordinates
for (int i = 0; i < n; ++i) m[manager[i]].push_back(i);
queue<pair<int, int>> q;
int ans = 0;
q.emplace(headID, informTime[headID]);
while (q.size()) {
int cnt = q.size();
while (cnt) {
auto p = q.front();
q.pop();
ans = max(ans, p.second);
for (int sub : m[p.first]) q.push({ sub, p.second + informTime[sub] });
}
}
return ans;
}
};
Solution 2. DFS
Same idea as Solution 1 but using DFS.
BTW, using vector<vector<int>>
can boost the speed a lot comparing with unordered_map<int, vector<int>>
.
// OJ: https://leetcode.com/problems/timeneededtoinformallemployees/
// Time: O(N)
// Space: O(N)
class Solution {
vector<vector<int>> m; // manager => subordinates
int dfs(int id, vector<int>& time) {
int ans = 0;
for (int n : m[id]) ans = max(ans, dfs(n, time));
return time[id] + ans;
}
public:
int numOfMinutes(int n, int headID, vector<int>& manager, vector<int>& informTime) {
m.resize(n);
for (int i = 0; i < n; ++i) {
if (manager[i] == 1) continue;
m[manager[i]].push_back(i);
}
return dfs(headID, informTime);
}
};

class Solution { public int numOfMinutes(int n, int headID, int[] manager, int[] informTime) { Map<Integer, Set<Integer>> map = new HashMap<Integer, Set<Integer>>(); for (int i = 0; i < n; i++) { int curManager = manager[i]; if (curManager >= 0) { Set<Integer> set = map.getOrDefault(curManager, new HashSet<Integer>()); set.add(i); map.put(curManager, set); } } int minutes = 0; Queue<Integer> employeeQueue = new LinkedList<Integer>(); Queue<Integer> timeQueue = new LinkedList<Integer>(); employeeQueue.offer(headID); timeQueue.offer(0); while (!employeeQueue.isEmpty()) { int employee = employeeQueue.poll(); int time = timeQueue.poll(); minutes = Math.max(minutes, time); int totalTime = time + informTime[employee]; Set<Integer> subordinates = map.getOrDefault(employee, new HashSet<Integer>()); for (int subordinate : subordinates) { employeeQueue.offer(subordinate); timeQueue.offer(totalTime); } } return minutes; } }

// OJ: https://leetcode.com/problems/timeneededtoinformallemployees/ // Time: O(N) // Space: O(N) class Solution { public: int numOfMinutes(int n, int headID, vector<int>& manager, vector<int>& informTime) { unordered_map<int, vector<int>> m; // manager => subordinates for (int i = 0; i < n; ++i) m[manager[i]].push_back(i); queue<pair<int, int>> q; int ans = 0; q.emplace(headID, informTime[headID]); while (q.size()) { int cnt = q.size(); while (cnt) { auto p = q.front(); q.pop(); ans = max(ans, p.second); for (int sub : m[p.first]) q.push({ sub, p.second + informTime[sub] }); } } return ans; } };

class Solution: def numOfMinutes( self, n: int, headID: int, manager: List[int], informTime: List[int] ) > int: def dfs(i): ans = 0 for j in g[i]: ans = max(ans, informTime[i] + dfs(j)) return ans g = defaultdict(list) for i, m in enumerate(manager): g[m].append(i) return dfs(headID)