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Maximum Profit in Job Scheduling

Certification: Advanced Level Accuracy: 100% Submissions: 1 Points: 15

Write a Java program to find the maximum profit you can achieve by scheduling jobs. You have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i]. You need to find the maximum profit you can get if you can only work on one job at a time.

Example 1
    
  
  • Input: 
    startTime = [1, 2, 3, 3]
    endTime = [3, 4, 5, 6]
    profit = [50, 10, 40, 70]
    • 
  
  • Output: 120
    • 
  
  • Explanation: 
     The jobs are:

     - Job 1: Start at time 1, end at time 3, profit 50

     - Job 2: Start at time 2, end at time 4, profit 10

     - Job 3: Start at time 3, end at time 5, profit 40

     - Job 4: Start at time 3, end at time 6, profit 70

    The maximum profit can be achieved by scheduling Job 1 and Job 4.
    Job 1 starts at time 1 and ends at time 3, with profit 50.
    Job 4 starts at time 3 and ends at time 6, with profit 70.
    Total profit = 50 + 70 = 120.
    •
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 jobs are:

     - Job 1: Start at time 1, end at time 3, profit 20

     - Job 2: Start at time 2, end at time 5, profit 20

     - Job 3: Start at time 3, end at time 10, profit 100

     - Job 4: Start at time 4, end at time 6, profit 70

     - Job 5: Start at time 6, end at time 9, profit 60

    The maximum profit can be achieved by scheduling Job 1 and Job 4 and Job 5.
    Job 1 starts at time 1 and ends at time 3, with profit 20.
    Job 4 starts at time 4 and ends at time 6, with profit 70.
    Job 5 starts at time 6 and ends at time 9, with profit 60.
    Total profit = 20 + 70 + 60 = 150.
    •
Constraints
    
  
  • 1 ≤ startTime.length == endTime.length == profit.length ≤ 5 * 10^4
    • 
  
  • 1 ≤ startTime[i] < endTime[i] ≤ 10^9
    • 
  
  • 1 ≤ profit[i] ≤ 10^4
    • 
  
  • Time Complexity: O(n log n)
    • 
  
  • Space Complexity: O(n)
    •
ArraysDynamic Programming WiproDeloitte
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Hint
Solution

Solution Hints

    
  
  • Sort the jobs based on their end times
    • 
  
  • Use dynamic programming to find the maximum profit
    • 
  
  • For each job, you have two choices: either take the job or don't take it
    • 
  
  • If you take a job, you can't take jobs that overlap with it
    • 
  
  • Use binary search to find the latest job that finishes before the current job starts
    • 
  
  • Build a bottom-up DP array where dp[i] represents the maximum profit up to the ith job
    • 
  
  • For each job, update dp[i] = max(profit[i] + dp[j], dp[i-1]), where j is the latest non-overlapping job
    •

Steps to solve by this approach:

 Step 1: Create a jobs array where each job is represented as [startTime, endTime, profit].

 Step 2: Sort all jobs by their end times to process them in order of completion.

 Step 3: Initialize a dynamic programming array dp, where dp[i] represents the maximum profit up to the ith job.

 Step 4: For each job, we have two choices: either take the job or skip it.

 Step 5: If we take the job, we need to find the latest job that doesn't overlap with the current job and add its profit.

 Step 6: If we skip the job, the maximum profit is the same as the previous job's maximum profit.

 Step 7: For each job, dp[i] = max(profit[i] + dp[j], dp[i-1]), where j is the latest non-overlapping job.

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