#1953

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Maximum Number of Weeks for Which You Can Work

Maximize the number of weeks you can work on projects with milestone constraints using a greedy approach and invariant validation.

array greedy

Problem Statement

You are given an integer array 'milestones', where each element represents the number of milestones for a particular project. The objective is to determine the maximum number of weeks you can work under these conditions: each week, you can work on one milestone from any project, but you cannot work on the same project's milestone consecutively, and you must work on the project with the largest number of remaining milestones. If no valid project remains, stop working.

To maximize the number of weeks, you must use a greedy strategy. You will alternate between projects with the most milestones, making sure no project violates the rule of not working on the same project consecutively. The challenge is balancing the milestone counts and ensuring you don’t exceed the available projects' limits, as the constraints may cause some milestones to remain unfinished.

Examples

Example 1

Input: milestones = [1,2,3]

Output: 6

One possible scenario is: - During the 1st week, you will work on a milestone of project 0.

During the 2nd week, you will work on a milestone of project 2.
During the 3rd week, you will work on a milestone of project 1.
During the 4th week, you will work on a milestone of project 2.
During the 5th week, you will work on a milestone of project 1.
During the 6th 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

One possible scenario is:

During the 1st week, you will work on a milestone of project 0.
During the 2nd week, you will work on a milestone of project 1.
During the 3rd week, you will work on a milestone of project 0.
During the 4th week, you will work on a milestone of project 1.
During the 5th week, you will work on a milestone of project 0.
During the 6th week, you will work on a milestone of project 2.
During the 7th 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 8th 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

Solution Approach

Greedy Approach

Start by selecting the project with the largest remaining milestones and work on its milestones until you cannot. Each time, choose the project with the next largest remaining milestone count, ensuring you don't break the constraint of consecutive project choices.

Invariant Validation

Throughout the solution, maintain an invariant that ensures at each step, the project with the largest remaining milestones is always chosen. Validate if working on the next milestone of a project violates any conditions, ensuring that you always have a valid project to work on.

Efficient Tracking

Use a max-heap or priority queue to efficiently track and update the project with the largest remaining milestones. This ensures that each week, the project with the maximum milestones is always selected quickly and efficiently.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the data structure used to track the largest remaining milestones (e.g., a heap or priority queue). Inserting, removing, and updating elements in the heap take O(log n) time. The space complexity is O(n) due to the storage required for the project milestones.

What Interviewers Usually Probe

The candidate should be comfortable with greedy algorithms and invariant management.
Look for efficiency in selecting and updating projects based on remaining milestones.
Candidates who suggest using a heap or priority queue demonstrate an understanding of efficient tracking.

Common Pitfalls or Variants

Common pitfalls

Failing to properly manage the condition of not working on the same project consecutively.
Not using efficient data structures, leading to slower performance as project numbers grow.
Overlooking edge cases where some projects may have fewer milestones than others.

Follow-up variants

What happens if the milestone values are extremely large (close to 10^9)?
What if projects have the same number of milestones?
How would the approach change if we were asked to minimize the number of weeks instead of maximizing them?

FAQ

What is the primary pattern for solving Maximum Number of Weeks for Which You Can Work?

The problem follows a greedy approach combined with invariant validation, where you select the project with the largest remaining milestones each week.

How do I track the largest project milestones efficiently?

Use a max-heap or priority queue to efficiently track and select the project with the largest remaining milestones.

Can I work on the same project consecutively?

No, you must alternate between projects with the largest remaining milestones and ensure no consecutive work on the same project.

What if two projects have the same number of milestones?

In case of a tie, you can choose either project, but ensure you maintain the alternating rule and always work on the project with the largest remaining milestones.

What happens if all projects have very few milestones?

If all projects have few milestones, you may be able to work on them for more weeks, but make sure no project violates the consecutive milestone rule.

terminal

Solution

Solution 1: Greedy

We consider under what circumstances we cannot complete all stage tasks. If there is a project $i$ whose number of stage tasks is greater than the sum of the number of stage tasks of all other projects plus $1$, then we cannot complete all stage tasks. Otherwise, we can definitely complete all stage tasks by interlacing between different projects.

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class Solution:
    def numberOfWeeks(self, milestones: List[int]) -> int:
        mx, s = max(milestones), sum(milestones)
        rest = s - mx
        return rest * 2 + 1 if mx > rest + 1 else s

Continue Practicing

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