Time Needed to Inform All Employees

We can solve this problem by performing a Depth-First Search (DFS) from the head of the company through the tree-like structure of employees. At each step, we need to keep track of the time taken to inform all the subordinates.

Here’s the code:

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from collections import defaultdict

class Solution:
    def numOfMinutes(self, n: int, headID: int, manager: List[int], informTime: List[int]) -> int:
        # Constructing a graph (tree) for the employees
        employee_graph = defaultdict(list)
        for i, m in enumerate(manager):
            if m != -1:
                employee_graph[m].append(i)

        # DFS function to calculate the time taken to inform all subordinates
        def dfs(employee_id):
            max_time = 0
            for subordinate in employee_graph[employee_id]:
                max_time = max(max_time, dfs(subordinate))
            return max_time + informTime[employee_id]

        return dfs(headID)

The function constructs a graph using a defaultdict, where each employee ID maps to a list of subordinate IDs. Then, a recursive DFS function is defined to traverse through the subordinates and calculate the total time taken to inform them all.

The time complexity of this solution is O(n), where n is the number of employees, and the space complexity is also O(n) to store the graph and the recursion stack in the worst case.

Identifying Problem Isomorphism

“Time Needed to Inform All Employees” can be approximated to “Network Delay Time”. Both problems are based on graph theory and require an understanding of shortest path algorithms.

In “Time Needed to Inform All Employees”, you are given a company structure as a tree where each node is an employee, and you need to calculate the time it will take to inform all employees starting from the head of the company. The time taken to inform an employee is given and it can take different amounts of time to inform different employees.

“Network Delay Time” has a similar structure. You have to find how long it will take for a signal to reach all nodes in the network, given the time it takes for the signal to travel along each edge.

Both problems require the calculation of the maximum time it takes for information to reach all nodes. However, the difference is in the structure of the graph. “Time Needed to Inform All Employees” is based on a tree (a special kind of graph) while “Network Delay Time” can involve a general graph, which can be more complex due to potential cycles and the presence of multiple paths between nodes. Thus, “Time Needed to Inform All Employees” is simpler compared to “Network Delay Time”. Solving “Time Needed to Inform All Employees” equips one with the skills needed for understanding the propagation of information across a network, a concept that is also necessary for “Network Delay Time”.

10 Prerequisite LeetCode Problems

“Time Needed to Inform All Employees” is a tree traversal problem where the tree is not necessarily binary, and the weight of the edges matters. Here are some related problems to prepare:

  1. 559. Maximum Depth of N-ary Tree: This problem involves determining the maximum depth of a tree, which is similar to the given problem in the sense that you need to find the maximum time taken to reach all employees.

  2. 690. Employee Importance: This problem also involves a hierarchy of employees, though the problem’s nature is slightly different. It would help you understand how to traverse such structures.

  3. 133. Clone Graph: This problem involves cloning a graph, which would help you understand how to traverse nodes in a non-binary tree or a graph.

  4. 207. Course Schedule: This problem is about determining if it is possible to finish all courses given some prerequisites. It is a typical application of the topological sorting in graph theory.

  5. 210. Course Schedule II: A follow-up to the Course Schedule problem, it requires you to find an order to finish all the courses.

  6. 102. Binary Tree Level Order Traversal: This problem could give a basic understanding of level order traversal, which is similar to the breadth-first search in a tree.

  7. 429. N-ary Tree Level Order Traversal: This is a variation of the previous problem but for N-ary trees.

  8. 743. Network Delay Time: This problem is very similar to the problem at hand, but it’s applied to a network (which can be seen as a weighted graph) instead of a hierarchy of employees.

  9. 684. Redundant Connection: This problem requires you to understand the structure of a tree and the concept of a cycle in a graph.

  10. 886. Possible Bipartition: This problem involves assigning people to two groups and could help you understand the concept of graph coloring.

These cover graph theory, tree traversal, and weighted edges, which will be useful in tackling the problem at hand.

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class Solution:
    def numOfMinutes(self, n: int, headID: int, manager: List[int], informTime: List[int]) -> int:
        def find(i):
            if manager[i] != -1:
                informTime[i] += find(manager[i])
                manager[i] = -1
            return informTime[i]
        return max(map(find, range(n)))

Problem Classification

  1. 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.

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.

Language Agnostic Coding Drills

  1. Variable and Function Definitions: Understanding how to define variables and functions.

  2. Recursion: The function find calls itself, which is a fundamental concept known as recursion. It’s used to solve problems that can be broken down into smaller, similar problems.

  3. Conditional Statements: Using if statements to control the flow of the program based on certain conditions.

  4. Lists and Indexing: Lists are a type of data structure in Python that can store multiple items in a single variable. Indexing is used to access the individual elements of the list.

  5. Looping Constructs: Understanding how to use loops (for loop in this case) to repeatedly execute a block of code.

  6. Built-in Functions: Using Python’s built-in functions, such as map, max, and range.

  7. Classes and Method Definitions: Defining classes and methods in Python. In this case, a class Solution is defined with a method numOfMinutes.

Targeted Drills in Python

  1. Variable and Function Definitions:
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def add_numbers(x, y):
    sum = x + y
    return sum

print(add_numbers(3, 5))  # Output: 8
  1. Recursion:
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def factorial(n):
    if n == 1:
        return 1
    else:
        return n * factorial(n-1)

print(factorial(5))  # Output: 120
  1. Conditional Statements:
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x = 10
if x > 0:
    print("Positive")  # Output: Positive
else:
    print("Non-positive")
  1. Lists and Indexing:
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numbers = [1, 2, 3, 4, 5]
print(numbers[2])  # Output: 3
  1. Looping Constructs:
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for i in range(5):
    print(i)  # Output: 0 1 2 3 4
  1. Built-in Functions:
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numbers = [1, 2, 3, 4, 5]
squares = list(map(lambda x: x ** 2, numbers))
print(max(squares))  # Output: 25
  1. Classes and Method Definitions:
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class Test:
    def greet(self):
        return "Hello, World!"

test = Test()
print(test.greet())  # Output: Hello, World!