How do you solve Maximum Number of Weeks for Which You Can Work on LeetCode?

Maximum Number of Weeks for Which You Can Work (LeetCode #1953) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(1) space complexity. Key insight: The brute force approach involves simulating each week and selecting milestones from projects while adhering to the constraints. This method is straightforward but inefficient as it checks all possible combinations.

What algorithm or data structure is used to solve Maximum Number of Weeks for Which You Can Work?

Maximum Number of Weeks for Which You Can Work uses the following concepts: Array, Greedy. The recommended patterns to study are: .

Maximum Number of Weeks for Which You Can Work — LeetCode #1953 (Medium)

Tags: Array, Greedy

Related patterns:

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves simulating each week and selecting milestones from projects while adhering to the constraints. This method is straightforward but inefficient as it checks all possible combinations.

The time complexity is O(n²) because we sort the milestones array in each iteration, which takes O(n log n), and we may do this for each milestone, resulting in a quadratic relationship.

Step 1: Initialize a week counter to zero.
Step 2: While there are milestones left in any project, select the project with the highest remaining milestones that can be worked on without violating the consecutive week rule.
Step 3: Increment the week counter and update the milestones for the selected project. Repeat until no valid selections are possible.

For milestones = [1, 2, 3]: 1. Week 1: Select project 2 (3 left) -> [1, 2, 2] 2. Week 2: Select project 1 (2 left) -> [1, 1, 2] 3. Week 3: Select project 0 (1 left) -> [0, 1, 2] 4. Week 4: Select project 2 (2 left) -> [0, 1, 1] 5. Week 5: Select project 1 (1 left) -> [0, 0, 1] 6. Week 6: Select project 2 (1 left) -> [0, 0, 0] Total weeks = 6.

#1953

Maximum Number of Weeks for Which You Can Work

Medium

Array↗Greedy↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

⚙️

Algorithm

3 steps

1Step 1: Initialize a week counter to zero.
2Step 2: While there are milestones left in any project, select the project with the highest remaining milestones that can be worked on without violating the consecutive week rule.
3Step 3: Increment the week counter and update the milestones for the selected project. Repeat until no valid selections are possible.

solution.py18 lines

1# Full working Python code
2from collections import Counter
3
4def maxWeeks(milestones):
5    weeks = 0
6    while True:
7        milestones.sort(reverse=True)
8        if milestones[0] == 0:
9            break
10        for i in range(len(milestones)):
11            if milestones[i] > 0:
12                weeks += 1
13                milestones[i] -= 1
14                if i + 1 < len(milestones) and milestones[i + 1] > 0:
15                    milestones[i + 1] -= 1
16                break
17    return weeks
18

ℹ

Complexity note: The time complexity is O(n²) because we sort the milestones array in each iteration, which takes O(n log n), and we may do this for each milestone, resulting in a quadratic relationship.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.