Maximum Number Of Weeks For Which You Can Work - Leetcode Solution

Problem Statement:

You are given an integer n representing the total number of projects and an integer m representing the maximum number of weeks for which you can work on the projects. You are also given a list of integers tasks representing the number of weeks each project requires to complete.

You can only work on one project at a time and can switch between projects once a project is completed. However, you cannot start a new project if you do not have enough time to complete it within m weeks.

Return the maximum number of weeks you can work on the projects.

Solution:

In order to solve this problem, we can use a greedy approach. We can start by sorting the given list of tasks in descending order. This way, we can always choose the longest project first and finish it as soon as possible.

Next, we can iterate through the sorted list and keep track of the total number of weeks we have worked so far. If we have enough time to complete the current project, we subtract the required number of weeks from our remaining time and move on to the next project. If our remaining time is not sufficient to complete the current project, we stop and return the total number of weeks we have worked so far.

Let's look at the implementation:

def max_number_of_weeks(n: int, m: int, tasks: List[int]) -> int: """ :type n: int :type m: int :type tasks: List[int] :rtype: int """ # Sort the tasks in descending order tasks.sort(reverse=True)

# Keep track of the total number of weeks worked so far
total_weeks = 0

# Iterate through the tasks list
for task in tasks:
    # If we have enough time to complete the current project, subtract the required number of weeks
    # from our remaining time and move on to the next project.
    if total_weeks + task <= m:
        total_weeks += task
    # If our remaining time is not sufficient to complete the current project, stop and return the
    # total number of weeks we have worked so far.
    else:
        return total_weeks + (m - total_weeks)

# If we have completed all the projects within the given time, return the total number of weeks worked.
return total_weeks

Complexity Analysis:

The time complexity of this solution can be analyzed as follows:

1. Sorting the tasks array will take O(nlogn) time.

2. We are iterating through the sorted tasks array once, which will take O(n) time.

The overall time complexity of this solution is O(nlogn + n) = O(nlogn).

As we are using constant space to store only a few variables, the space complexity of this solution is O(1).

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