Leetcode 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. The head will inform his direct subordinates, and they will inform their subordinates, and so on until all employees know about the urgent news. Each employee needs informTime[i] minutes to inform their direct subordinates. Return the total time required to inform all employees.Problem
Approach
Steps
Complexity
Input: The input consists of an integer `n`, the number of employees in the company, the integer `headID`, which identifies the head of the company, an array `manager[]` of size `n` where `manager[i]` is the ID of the direct manager of employee `i`, and an array `informTime[]` of size `n` where `informTime[i]` is the time it takes for employee `i` to inform all of their direct subordinates.
Example: n = 6, headID = 2, manager = [2, 2, -1, 2, 2, 2], informTime = [0, 0, 1, 0, 0, 0]
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 employee i has no subordinates.
Output: The output is an integer representing the total time required for all employees to be informed about the urgent news.
Example: 1
Constraints:
• The output will be a single integer.
Goal: The goal is to determine the total time needed for all employees to receive the news, considering the time each employee takes to inform their subordinates.
Steps:
• Construct a graph representing the tree structure of employee-manager relationships.
• Use a depth-first search (DFS) approach to calculate the time required for each employee to inform their subordinates.
• For each employee, calculate the maximum time taken among their subordinates and add the time they take to inform them.
Goal: The problem needs to handle large inputs efficiently, ensuring the algorithm works within the constraints.
Steps:
• The tree structure is guaranteed to be valid.
• Each employee has at most one manager.
Assumptions:
• The manager-subordinate relationships form a tree structure, ensuring that all employees can be informed.
• Input: n = 6, headID = 2, manager = [2,2,-1,2,2,2], informTime = [0,0,1,0,0,0]
• Explanation: The head (employee 2) informs all the direct subordinates. The subordinates then take 1 minute to inform their own subordinates.
• Input: n = 5, headID = 0, manager = [-1, 0, 0, 1, 1], informTime = [0, 10, 0, 0, 0]
• Explanation: The head (employee 0) informs employees 1 and 2 in 10 minutes. Employee 1 further informs employees 3 and 4.
Approach: The problem is solved using depth-first search (DFS) to traverse the tree structure and calculate the time taken to inform all employees.
Observations:
• The problem essentially involves traversing a tree structure and calculating the total time recursively.
• Using DFS, we can compute the maximum time required to inform all subordinates for each employee.
Steps:
• Construct the tree using a graph representation.
• Use a recursive DFS function to traverse the tree and calculate the time taken to inform each employee's subordinates.
• Return the maximum time taken by the root (head) employee.
Empty Inputs:
• If the tree has only one employee (head), the result will be 0.
Large Inputs:
• Ensure the solution works efficiently for large values of `n` up to 100,000.
Special Values:
• If all employees take 0 minutes to inform others, the result will be 0.
Constraints:
• The problem constraints guarantee a valid tree structure and that all employees can be informed.
int numOfMinutes(int n, int headID, vector<int>& manager, vector<int>& informTime) {
map<int, vector<int>> gph;
for(int i = 0; i < manager.size(); i++)
gph[manager[i]].push_back(i);
return dfs(headID, gph, informTime);
}
int dfs(int cur, map<int, vector<int>> &gph, vector<int> &it) {
if(!gph.count(cur)) return 0;
int mx = 0;
for(auto x: gph[cur])
mx = max(mx, dfs(x, gph, it));
return mx + it[cur];
}
1 : Variable Initialization
int numOfMinutes(int n, int headID, vector<int>& manager, vector<int>& informTime) {
The function `numOfMinutes` is the main entry point. It takes the number of employees `n`, the head of the company `headID`, a list of managers for each employee, and a list of the time taken by each employee to inform others.
2 : Map Initialization
map<int, vector<int>> gph;
We declare a map `gph` where the key represents a manager's ID and the value is a vector of employees they manage.
3 : Looping
for(int i = 0; i < manager.size(); i++)
We loop through the list of managers and create the graph where each employee is linked to their manager.
4 : Graph Construction
gph[manager[i]].push_back(i);
For each employee `i`, we add them as a subordinate to their manager in the graph.
5 : Recursive Call
return dfs(headID, gph, informTime);
We initiate the depth-first search starting from the `headID`, using the constructed graph and the `informTime` list to calculate the total time.
6 : Recursive Function
int dfs(int cur, map<int, vector<int>> &gph, vector<int> &it) {
The `dfs` function performs a depth-first search to find the maximum time needed to inform all subordinates of a particular employee.
7 : Base Case
if(!gph.count(cur)) return 0;
If the current employee does not have any subordinates (i.e., they don't appear in the graph), return 0.
8 : Variable Declaration
int mx = 0;
We declare a variable `mx` to track the maximum time it takes to inform subordinates.
9 : Iterate Over Subordinates
for(auto x: gph[cur])
We iterate over all subordinates of the current employee.
10 : Recursive Call
mx = max(mx, dfs(x, gph, it));
For each subordinate, we recursively calculate the time needed to inform them and their subordinates, updating `mx` with the maximum time.
11 : Return Statement
return mx + it[cur];
After processing all subordinates, we return the maximum time plus the time it takes for the current employee to inform others.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) as each employee is visited once during DFS.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the recursion stack and the graph representation of the tree.
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