Solving LeetCode 1235 Maximum Profit in Job Scheduling | Array | Binary Search | DP | Hard | MAANG

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LeetCode problem 1235. Maximum Profit in Job Scheduling [ 6 Jan 2024 ]

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📌 Video Overview:
Dive deep into the solution for the LeetCode challenge titled '1235. Maximum Profit in Job Scheduling'. Whether you're kickstarting your coding journey or aiming for roles at tech giants like Facebook, Amazon, Apple, Netflix, or Google, this tutorial is tailored for you. We'll break down the problem step-by-step, ensuring clarity and understanding.

🌟 Key Highlights:

A comprehensive walkthrough of the problem statement.
Intuitive explanation for both novice and intermediate coders.
Practical insights for acing interviews with top tech companies.
Code walkthrough to grasp the algorithmic approach.

📌 Chapters:
00:00 Problem Statement
04:50 Top Down Recursion
07:40 Overlapping subproblems
11:45 Improvement by Binary Search
14:01 Code

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Join me on this coding adventure! Let's learn, grow, and conquer challenges together.

LeetCode Problem Statement:

We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i].

You're given the startTime, endTime and profit arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range.

If you choose a job that ends at time X you will be able to start another job that starts at time X.

Example 1:

Input: startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
Output: 120
Explanation: The subset chosen is the first and fourth job.
Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.

Example 2:

Input: startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
Output: 150
Explanation: The subset chosen is the first, fourth and fifth job.
Profit obtained 150 = 20 + 70 + 60.

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Комментарии

Thank you for this. The top down recursion makes it so much easier to understand

rsKayiira

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