What is the time complexity of Maximum Number of Weeks for Which You Can Work?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Maximum Number of Weeks for Which You Can Work?

This problem uses Array, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Number of Weeks for Which You Can Work?

Insight 1: The key is to realize that the longest project will dictate the maximum number of weeks. If the longest project's milestones are significantly greater than the sum of the milestones of other projects, some milestones from the longest project will remain unfinished. Insight 2: The condition `max_milestone > rest_sum + 1` determines if the longest project restricts the number of weeks. If true, we can't complete all milestones and the maximum weeks is determined by twice the sum of the other milestones plus one.

Which companies ask LeetCode 1953. Maximum Number of Weeks for Which You Can Work in interviews?

This problem is frequently asked at SAP, Wells Fargo.

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Maximum Number of Weeks for Which You Can Work - LeetCode 1953 Solution

Maximum Number of Weeks for Which You Can Work - Complete Solution Guide

Maximum Number of Weeks for Which You Can Work is LeetCode problem 1953, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 wo

Detailed Explanation

The problem asks us to find the maximum number of weeks we can work on 'n' projects, given an array 'milestones' representing the number of milestones for each project. We must work on exactly one milestone each week and cannot work on the same project in consecutive weeks. The goal is to maximize the total number of weeks worked under these constraints. We stop when all milestones are completed or if we can't proceed without violating the consecutive weeks rule.

Solution Approach

The solution calculates the total sum of all milestones and finds the maximum number of milestones in any single project. It then calculates the sum of all milestones except the largest one. If the largest milestone is greater than the sum of all other milestones plus one, it means we cannot complete all the milestones in the largest project due to the consecutive weeks constraint. In that case, the maximum number of weeks is 2 * (sum of other milestones) + 1. Otherwise, we can complete all milestones and the maximum number of weeks is the total sum of all milestones.

Step-by-Step Algorithm

Step 1: Calculate the total sum of all elements in the 'milestones' array.
Step 2: Find the maximum value in the 'milestones' array, representing the project with the most milestones.
Step 3: Calculate the sum of all milestones except for the maximum milestone (rest_sum).
Step 4: Check if the maximum milestone is greater than 'rest_sum + 1'.
Step 5: If it is, return '2 * rest_sum + 1'. Otherwise, return the 'total_sum'.

Key Insights

Insight 1: The key is to realize that the longest project will dictate the maximum number of weeks. If the longest project's milestones are significantly greater than the sum of the milestones of other projects, some milestones from the longest project will remain unfinished.
Insight 2: The condition `max_milestone > rest_sum + 1` determines if the longest project restricts the number of weeks. If true, we can't complete all milestones and the maximum weeks is determined by twice the sum of the other milestones plus one.
Insight 3: If the longest project is not significantly greater than the sum of other projects, we can complete all milestones from every project, and the maximum number of weeks is the sum of all milestones.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy.

Companies

Asked at: SAP, Wells Fargo.

Maximum Number of Weeks for Which You Can Work | LeetCode 1953

Medium

Array Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public long numberOfWeeks(int[] milestones) {
        long total_sum = 0;
        int max_milestone = 0;
        
        for (int milestone : milestones) {
            total_sum += milestone;
            max_milestone = Math.max(max_milestone, milestone);
        }
        
        long rest_sum = total_sum - max_milestone;
        
        if (max_milestone > rest_sum + 1) {
            return 2 * rest_sum + 1;
        } else {
            return total_sum;
        }
    }
}

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Problem Info

Description

Problem 1953: Maximum Number of Weeks for Which You Can Work - Detailed explanation and solution approaches.

Topics

Array Greedy

Companies

SAPWells Fargo

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1953

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Calculate the total sum of all elements in the 'milestones' array.
2
Step 2: Find the maximum value in the 'milestones' array, representing the project with the most milestones.
3
Step 3: Calculate the sum of all milestones except for the maximum milestone (rest_sum).
4
Step 4: Check if the maximum milestone is greater than 'rest_sum + 1'.
5
Step 5: If it is, return '2 * rest_sum + 1'. Otherwise, return the 'total_sum'.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The solution involves iterating through the array once to find the total sum and the maximum element. All other operations are constant time. Therefore, the time complexity is dominated by the array iteration, resulting in O(n). The space complexity is O(1) because we only use a few constant-size variables, regardless of the array size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorting-based approach

Sort the milestones array in descending order. Then iterate through the sorted array, always picking the largest remaining milestones and ensuring we don't pick the same project twice in a row. This becomes difficult to implement and has O(n log n) time complexity.

Time: O(n log n) due to sortingSpace: O(1) or O(n) depending on the sorting algorithm's space complexity.

Trade-offs: This approach is more complex to implement and less efficient than the current O(n) solution. It's generally not preferred.

✨

Key Insights

1Insight 1: The key is to realize that the longest project will dictate the maximum number of weeks. If the longest project's milestones are significantly greater than the sum of the milestones of other projects, some milestones from the longest project will remain unfinished.
2Insight 2: The condition `max_milestone > rest_sum + 1` determines if the longest project restricts the number of weeks. If true, we can't complete all milestones and the maximum weeks is determined by twice the sum of the other milestones plus one.
3Insight 3: If the longest project is not significantly greater than the sum of other projects, we can complete all milestones from every project, and the maximum number of weeks is the sum of all milestones.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: All milestones are 1: In this case, total_sum = n, and max_milestone = 1. Then rest_sum = n - 1. Since 1 <= n - 1 + 1 is always true, the code will return n which is correct.
Edge case 2: Milestones array with only one element: In this case, total_sum = milestones[0], and max_milestone = milestones[0]. Then rest_sum = 0. Since milestones[0] > 0 + 1 will be false for milestones[0] = 1. Hence the total_sum i.e., milestones[0] will be returned. For cases when milestones[0] > 1 it'll also return milestones[0].
Edge case 3: Milestones array with large values to cause integer overflow: Using long long or long data types (Java), this problem is avoided. In C, if integer overflow is a concern the data type used to accumulate the total sum and store individual milestones should be long long. However, it's handled automatically in the provided code.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the case where the largest project cannot be fully completed: Failing to account for the constraint that we cannot work on the same project in consecutive weeks leads to incorrect results when the longest project has significantly more milestones than the rest.

✗Common mistake 2: Using integer types that can overflow: If the sum of the milestones is large, using 'int' may lead to integer overflows and incorrect results. Using 'long' or 'long long' data types helps avoid this issue.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Task scheduling: Consider a project manager trying to schedule tasks among different teams. Each team has a certain number of tasks. The manager wants to maximize the number of weeks they can assign tasks, ensuring no team works on consecutive weeks.

🏦Real-world scenario 2: Resource allocation: Imagine allocating resources to different departments within an organization. Each department has a specific need for resources. You want to allocate resources over a period of time, ensuring no department receives resources in consecutive time periods to promote fairness and avoid bottlenecks.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 1, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Number of Weeks for Which You Can Work"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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