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

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

What data structures are used to solve Maximum Profit in Job Scheduling?

This problem uses Array, Binary Search, Dynamic Programming, Sorting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Profit in Job Scheduling?

Insight 1: Sort the jobs by their end times. This is crucial for efficiently determining which jobs can be scheduled together. Insight 2: Use Dynamic Programming (DP) to store the maximum profit achievable up to a certain end time. This allows us to build up the solution iteratively.

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Maximum Profit in Job Scheduling - LeetCode 1235 Solution

Q: Which companies ask LeetCode 1235. Maximum Profit in Job Scheduling in interviews?

This problem is frequently asked at Airbnb, Akuna Capital, DE Shaw, Databricks, DoorDash and 12 more companies.

Maximum Profit in Job Scheduling - Complete Solution Guide

Maximum Profit in Job Scheduling is LeetCode problem 1235, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 Explana

Detailed Explanation

The problem asks us to find the maximum profit achievable by selecting a subset of jobs, where each job has a start time, end time, and profit. The constraint is that no two selected jobs can have overlapping time ranges. If a job ends at time X, another job can start at time X. Essentially, we are trying to maximize the total profit while ensuring compatibility between selected jobs.

Solution Approach

The solution employs a dynamic programming approach combined with binary search. First, the jobs are sorted by their end times. Then, a DP array (represented as a list of pairs in the code, where each pair is `(end_time, max_profit)`) is used to store the maximum profit achievable up to a given end time. For each job, we use binary search to find the latest job that does not overlap with the current job. The profit of the current job is then added to the maximum profit achieved up to the end time of the non-overlapping job. If this combined profit is greater than the current maximum profit recorded for the end time of the current job, we update the DP array. The final result is the last element's profit in the DP array, representing the maximum profit achievable by scheduling jobs.

Step-by-Step Algorithm

Step 1: Represent each job as a tuple (start_time, end_time, profit).
Step 2: Sort the jobs by their end times in ascending order.
Step 3: Initialize a DP array (dp) with a base case: (0, 0), representing no job and no profit.
Step 4: Iterate through the sorted jobs:
Step 5: For each job (s, e, p), use binary search on the dp array to find the index 'i' of the latest job that does not overlap with the current job. Specifically, search for the rightmost job that ends before or at the start time 's' of the current job.
Step 6: Calculate the profit if the current job is included: current_profit = p + dp[i][1] (current job's profit plus the profit of the latest non-overlapping job).
Step 7: If the current_profit is greater than the last maximum profit recorded in dp, add a new entry (e, current_profit) to dp.
Step 8: After iterating through all jobs, the last element of the dp array contains the maximum profit achievable.

Key Insights

Insight 1: Sort the jobs by their end times. This is crucial for efficiently determining which jobs can be scheduled together.
Insight 2: Use Dynamic Programming (DP) to store the maximum profit achievable up to a certain end time. This allows us to build up the solution iteratively.
Insight 3: Use Binary Search to efficiently find the latest non-overlapping job to calculate the profit for the current job.

Complexity Analysis

Time Complexity: O(NlogN)

Space Complexity: O(N)

Topics

This problem involves: Array, Binary Search, Dynamic Programming, Sorting.

Companies

Asked at: Airbnb, Akuna Capital, DE Shaw, Databricks, DoorDash, Flipkart, PayPal, PhonePe, Pinterest, Pony.ai, Snowflake, Swiggy, Urban Company, Verkada, WeRide, Zeta, oyo.

Maximum Profit in Job Scheduling | LeetCode 1235

Hard

Array Binary Search Dynamic Programming Sorting

Time: O(NlogN)Space: O(N)

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    public int jobScheduling(int[] startTime, int[] endTime, int[] profit) {
        int n = startTime.length;
        int[][] jobs = new int[n][3];
        for (int i = 0; i < n; i++) {
            jobs[i][0] = startTime[i];
            jobs[i][1] = endTime[i];
            jobs[i][2] = profit[i];
        }

        Arrays.sort(jobs, Comparator.comparingInt(a -> a[1]));

        java.util.List<Pair> dp = new java.util.ArrayList<>();
        dp.add(new Pair(0, 0));

        for (int[] job : jobs) {
            int s = job[0];
            int e = job[1];
            int p = job[2];

            int i = binarySearchRight(dp, s) - 1;

            int prev_profit = dp.get(i).profit;
            int current_profit = p + prev_profit;

            int last_max_profit = dp.get(dp.size() - 1).profit;

            if (current_profit > last_max_profit) {
                dp.add(new Pair(e, current_profit));
            }
        }

        return dp.get(dp.size() - 1).profit;
    }

    private int binarySearchRight(java.util.List<Pair> dp, int target) {
        int low = 0;
        int high = dp.size();

        while (low < high) {
            int mid = low + (high - low) / 2;
            if (dp.get(mid).endTime <= target) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }

        return low;
    }

    static class Pair {
        int endTime;
        int profit;

        public Pair(int endTime, int profit) {
            this.endTime = endTime;
            this.profit = profit;
        }
    }
}

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Problem Info

Description

Problem 1235: Maximum Profit in Job Scheduling - Detailed explanation and solution approaches.

Topics

Array Binary Search Dynamic Programming Sorting

Companies

AirbnbAkuna CapitalDE ShawDatabricksDoorDashFlipkartPayPalPhonePe

Stats

Difficulty:Hard

Languages:4

Companies:17

Problem #:1235

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

8 steps

1
Step 1: Represent each job as a tuple (start_time, end_time, profit).
2
Step 2: Sort the jobs by their end times in ascending order.
3
Step 3: Initialize a DP array (dp) with a base case: (0, 0), representing no job and no profit.
4
Step 4: Iterate through the sorted jobs:
5
Step 5: For each job (s, e, p), use binary search on the dp array to find the index 'i' of the latest job that does not overlap with the current job. Specifically, search for the rightmost job that ends before or at the start time 's' of the current job.
6
Step 6: Calculate the profit if the current job is included: current_profit = p + dp[i][1] (current job's profit plus the profit of the latest non-overlapping job).
7
Step 7: If the current_profit is greater than the last maximum profit recorded in dp, add a new entry (e, current_profit) to dp.
8
Step 8: After iterating through all jobs, the last element of the dp array contains the maximum profit achievable.

💡

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(NlogN)

💾

Space Complexity

O(N)

📝

Why?

The time complexity is dominated by sorting and the binary search performed within the loop. The space complexity arises from storing the DP array, which grows proportionally with the number of jobs.

📖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

Recursion with Memoization

A recursive approach can be used to solve this problem. After sorting the jobs by end time, a recursive function can explore two choices for each job: either include it in the schedule or exclude it. Memoization can be used to store the results of subproblems and avoid redundant computations.

Time: O(NlogN)Space: O(N)

Trade-offs: Recursion with memoization can be more intuitive for some, but it might lead to stack overflow issues for large input sizes. The iterative DP solution is generally preferred due to its better space management and reduced risk of stack overflow.

✨

Key Insights

1Insight 1: Sort the jobs by their end times. This is crucial for efficiently determining which jobs can be scheduled together.
2Insight 2: Use Dynamic Programming (DP) to store the maximum profit achievable up to a certain end time. This allows us to build up the solution iteratively.
3Insight 3: Use Binary Search to efficiently find the latest non-overlapping job to calculate the profit for the current job.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: No jobs are given (empty input arrays). The solution implicitly handles this as the initialization of dp with (0,0) will ensure a return value of 0, which is the correct profit.
Edge case 2: Jobs with the same end time. The solution correctly handles this as it compares end times for sorting and properly schedules the tasks in order to choose the optimal solution.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the jobs by end time. Sorting by start time or profit will not lead to an optimal solution because it doesn't allow for efficient checking of non-overlapping intervals.

✗Common mistake 2: Not using binary search for finding the latest non-overlapping job. A linear search would increase the time complexity to O(N^2), which is not efficient for large input sizes.

✗Common mistake 3: Incorrectly implementing binary search, leading to wrong indexes and consequently wrong profit calculation

✅

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: Resource allocation. Scheduling tasks on a computer system to maximize resource utilization while avoiding conflicts (e.g., two processes trying to access the same resource at the same time).

🏦Real-world scenario 2: Project scheduling. Deciding which projects to undertake to maximize profit, considering deadlines and resource constraints. A construction company deciding which buildings to build to maximize profit before certain dates and avoid resource contention across jobs.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 3, 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

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

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

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.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 Profit in Job Scheduling"?

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

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