Is Maximum Profit in Job Scheduling easy or hard?

Maximum Profit in Job Scheduling is classified as a Hard problem because it combines interval scheduling with dynamic programming optimization. Recognizing the weighted interval scheduling pattern and integrating binary search for efficient transitions makes it challenging for many candidates.

Maximum Profit in Job Scheduling Python/Java solution

Implementations typically sort the jobs, store start times in a separate array, and use a DP array or memoized recursion. Python solutions often use bisect for binary search, while Java implementations use Arrays.binarySearch or manual binary search logic.

How to solve Maximum Profit in Job Scheduling in O(n log n)?

First combine the inputs into job tuples and sort them by start time. Use dynamic programming where dp[i] stores the maximum profit achievable starting from job i. For each job, perform a binary search to find the next job whose start time is at least the current job's end time, then compute max(skip, take).

What is the best approach for Maximum Profit in Job Scheduling?

Dynamic programming with binary search is the optimal approach. Sort jobs by start time, and for each job compute the best profit between skipping it or taking it and jumping to the next compatible job found via binary search. This reduces the transition lookup to O(log n), producing an overall time complexity of O(n log n).

Is Maximum Profit in Job Scheduling asked at Google/Amazon/Meta?

Maximum Profit in Job Scheduling is a classic weighted interval scheduling problem frequently asked by companies like Google, Amazon, and Meta. It tests dynamic programming, binary search optimization, and reasoning about overlapping intervals.

What data structure is used in Maximum Profit in Job Scheduling?

The solution primarily uses arrays or lists to store sorted job tuples and a dynamic programming array for memoization. Binary search on the sorted start times enables efficient lookup of the next compatible job, making the algorithm O(n log n).

What is the time complexity of Maximum Profit in Job Scheduling?

The optimal solution runs in O(n log n) time. Sorting the jobs takes O(n log n), and each dynamic programming state performs a binary search to locate the next non-overlapping job. Space complexity is O(n) for storing DP results.

Description Hints 3 Solution Video 14:45

1235. Maximum Profit in Job Scheduling

Hard54.8% acceptance22 companies

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.

Example 3:

Input: startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
Output: 6

Constraints:

1 <= startTime.length == endTime.length == profit.length <= 5 * 10⁴
1 <= startTime[i] < endTime[i] <= 10⁹
1 <= profit[i] <= 10⁴

Asked at top 20 of 22 companies

Airbnb Verkada Ponyai DoorDash DocuSign WeRide Snowflake PhonePe Palo Alto Networks Pinterest Meesho Swiggy

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Amazon Microsoft Bloomberg Uber Oracle Goldman Sachs

Infosys

Google

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Hints open one at a time

Hint 1

Think on DP.

Hint 2

Sort the elements by starting time, then define the dp[i] as the maximum profit taking elements from the suffix starting at i.

Hint 3

Use binarySearch (lower_bound/upper_bound on C++) to get the next index for the DP transition.

14:45

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