Maximum Number of Weeks for Which You Can Work

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Greedy AlgorithmsArrays

There are n projects numbered from 0 to n - 1. You are given an integer array milestones where each milestones[i] denotes the number of milestones the i^th project has.

You can work on the projects following these two rules:

Every week, you will finish exactly one milestone of one project. You must work every week.
You cannot work on two milestones from the same project for two consecutive weeks.

Once all the milestones of all the projects are finished, or if the only milestones that you can work on will cause you to violate the above rules, you will stop working. Note that you may not be able to finish every project's milestones due to these constraints.

Return the maximum number of weeks you would be able to work on the projects without violating the rules mentioned above.

Example 1:

Input: milestones = [1,2,3]
Output: 6
Explanation: One possible scenario is:
- During the 1^st week, you will work on a milestone of project 0.
- During the 2^nd week, you will work on a milestone of project 2.
- During the 3^rd week, you will work on a milestone of project 1.
- During the 4^th week, you will work on a milestone of project 2.
- During the 5^th week, you will work on a milestone of project 1.
- During the 6^th week, you will work on a milestone of project 2.
The total number of weeks is 6.

Example 2:

Input: milestones = [5,2,1]
Output: 7
Explanation: One possible scenario is:
- During the 1^st week, you will work on a milestone of project 0.
- During the 2^nd week, you will work on a milestone of project 1.
- During the 3^rd week, you will work on a milestone of project 0.
- During the 4^th week, you will work on a milestone of project 1.
- During the 5^th week, you will work on a milestone of project 0.
- During the 6^th week, you will work on a milestone of project 2.
- During the 7^th week, you will work on a milestone of project 0.
The total number of weeks is 7.
Note that you cannot work on the last milestone of project 0 on 8^th week because it would violate the rules.
Thus, one milestone in project 0 will remain unfinished.

Constraints:

n == milestones.length
1 <= n <= 10⁵
1 <= milestones[i] <= 10⁹

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the `milestones` array, and what is the maximum value of any element in the array?
Can the elements in the `milestones` array be zero or negative?
If the input array is empty, what value should I return?
Could you provide a small example of an input and the expected output?
Is there any specific tie-breaking criteria if multiple assignments lead to the same maximum number of weeks?

Brute Force Solution

Approach

To find the maximum number of weeks you can work, the brute force approach is like trying every single possible work schedule. We consider every combination of tasks we could do each week, without worrying about efficiency.

Here's how the algorithm would work step-by-step:

Start by considering all possible ways to work on a single task for one week.
Then, consider all possible ways to work on the tasks for two weeks.
Keep repeating this, each time increasing the number of weeks we consider.
For each number of weeks, try every possible combination of assigning tasks to each week, making sure we don't work on any task more times than is available.
As we explore these combinations, we keep track of the best possible work schedule we have found so far (the one that allows us to work the most weeks).
Finally, after considering all possibilities up to a certain point (like using all available tasks), we choose the best work schedule found.

Code Implementation

def max_weeks_brute_force(tasks):
    number_of_tasks = len(tasks)
    max_weeks_possible = 0

    for weeks in range(1, sum(tasks) + 1):
        is_possible = False
        import itertools
        for assignment in itertools.product(range(number_of_tasks), repeat=weeks):
            task_counts = [0] * number_of_tasks
            for task_index in assignment:
                task_counts[task_index] += 1

            valid_assignment = True
            for task_index in range(number_of_tasks):
                if task_counts[task_index] > tasks[task_index]:
                    valid_assignment = False
                    break

            if valid_assignment:
                is_possible = True
                break

        if is_possible:
            max_weeks_possible = weeks
        else:
            # If no combination works, we have exceeded maximum possiblen           break

    # We need to figure out the maximum number of weeksn   return max_weeks_possible

Big(O) Analysis

Time Complexity

O(exponential) – The brute force approach explores every possible work schedule for each possible number of weeks. Let n be the number of available tasks. In the worst case, the algorithm explores combinations akin to assigning tasks to weeks where the number of combinations grows exponentially with both n (number of tasks) and the number of weeks considered. Because of the intensive exploration of all combinations of task assignments, the overall time complexity is O(exponential), reflecting the combinatorial explosion.

Space Complexity

O(N!) – The brute force approach necessitates exploring every possible work schedule. To achieve this, implicit data structures are required to store the current combination of task assignments for each week under consideration. Because the algorithm evaluates all possible combinations and permutations of assigning tasks to weeks, the number of possible work schedules grows factorially with the number of tasks (N, where N is the number of tasks). This leads to auxiliary space usage proportional to the number of combinations, which can be described by O(N!).

Optimal Solution

Approach

The goal is to find the most weeks you can work given a set of tasks, where you can only work on one task per week. The clever approach is recognizing that the limiting factor is usually the task with the largest number of units, and comparing that to the total units of all other tasks.

Here's how the algorithm would work step-by-step:

First, find the largest number of units among all the tasks.
Then, calculate the total number of units from all the tasks excluding the one with the largest number of units.
If the largest number of units is less than or equal to the sum of all the other units, you can complete all the tasks by alternating between the largest task and the others. The answer is the sum of all units.
If the largest number of units is greater than the sum of all the other units, then the number of weeks you can work is equal to twice the sum of the other tasks plus one. This is because you alternate between the largest task and other tasks until you've exhausted the other tasks, leaving only the largest task. Then, the remaining units from largest task cannot be used.

Code Implementation

def maximum_weeks(milestones):
    largest_milestone = 0
    other_milestones_sum = 0

    for milestone in milestones:
        if milestone > largest_milestone:
            other_milestones_sum += largest_milestone
            largest_milestone = milestone
        else:
            other_milestones_sum += milestone

    # If the largest is smaller, all tasks can be completed.
    if largest_milestone <= other_milestones_sum:
        return largest_milestone + other_milestones_sum

    # If the largest is bigger, we use the other tasks as much as possible.
    return 2 * other_milestones_sum + 1

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n to find the largest number of units. Then, it iterates through the array again (also of size n) to calculate the sum of all the other units, excluding the largest one. These are two independent linear operations with respect to the input size n. Therefore, the overall time complexity is O(n).

Space Complexity

O(1) – The algorithm described only uses a few variables to store intermediate values like the largest number of units and the sum of the other units. These variables require a constant amount of extra space regardless of the number of tasks (N). Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty or null milestones array	Return 0, as there are no projects to work on.
Milestones array with a single element	Return the single element's value, as you can work on that project for that many weeks.
All milestones have a value of 1	Return the length of the array, as each project can be worked on for one week.
One milestone is significantly larger than the sum of all others.	Calculate the sum of all milestones excluding the largest, and return min(sum, total - sum) * 2 + 1.
Two milestones are very large, but roughly equal, while others are small.	The greedy algorithm should still function correctly as long as we prioritize smaller tasks against the large ones, and each week utilize the most abundant category.
Integer overflow when calculating the sum of milestones	Use a 64-bit integer type to store the sum to prevent overflow, or return maximum integer value if overflow is unavoidable.
Milestones array contains zero values	The zero values will not affect the overall calculation as adding them to other values and comparison will yield correct results, so no extra handling needed.
Large array sizes that could impact performance	The core logic involves sorting and summation, which should scale reasonably well (O(n log n) or O(n) if max heap is used), and the algorithm does not rely on deeply nested loops which makes it efficient.

Empty or null milestones array

How to Handle:

Return 0, as there are no projects to work on.

Milestones array with a single element

How to Handle:

Return the single element's value, as you can work on that project for that many weeks.

All milestones have a value of 1

How to Handle:

Return the length of the array, as each project can be worked on for one week.

One milestone is significantly larger than the sum of all others.

How to Handle:

Calculate the sum of all milestones excluding the largest, and return min(sum, total - sum) * 2 + 1.

Two milestones are very large, but roughly equal, while others are small.

How to Handle:

The greedy algorithm should still function correctly as long as we prioritize smaller tasks against the large ones, and each week utilize the most abundant category.

Integer overflow when calculating the sum of milestones

How to Handle:

Use a 64-bit integer type to store the sum to prevent overflow, or return maximum integer value if overflow is unavoidable.

Milestones array contains zero values

How to Handle:

The zero values will not affect the overall calculation as adding them to other values and comparison will yield correct results, so no extra handling needed.

Large array sizes that could impact performance

How to Handle:

The core logic involves sorting and summation, which should scale reasonably well (O(n log n) or O(n) if max heap is used), and the algorithm does not rely on deeply nested loops which makes it efficient.