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Job Execution

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● HardAdobeFULLTIMEOA

Problem statement

There are jobs that can be executed in parallel on a processor, where the execution time of the j^th job is executionTime[j]. To speed up execution, the following strategy is used.

In one operation, a job is chosen, the major job, and is executed for x seconds. All other jobs are executed for y seconds where y < x.

A job is complete when it has been executed for at least executionTime[j] seconds, then it exits the pool. Find the minimum number of operations in which the processor can completely execute all the jobs if run optimally.

Function

AdobeGetMinimumOperations(executionTime: int[], x: int, y: int) → int

Complete the function getMinimumOperations in the editor.

getMinimumOperations has the following parameters:

  1. int executionTime[n]: the execution times of each job
  2. int x: the time for which the major job is executed
  3. int y: the time for which all other jobs are executed

Examples

Example 1

executionTime = [3, 4, 1, 7, 6]x = 4y = 2return = 3
The following strategy is optimal using 1-based indexing.
  • Choose job 4 as the major job and reduce the execution times of job 4 by x = 4 and of other jobs by y = 2. Now executionTime = [1, 2, -1, 3, 4]. Job 3 is complete, so it is removed.
  • Choose job 4, executionTime = [-1, 0, -, 1, 2]. So, jobs 1, 2, and 4 are now complete.
  • Choose job 5, executionTime = [-, -, -, -, 2]. Job 5 is complete.
    • It takes 3 operations to execute all the jobs so the answer is 3.

    Constraints

    Unknown for now

    More Adobe problems

    drafts saved locally
    public int AdobeGetMinimumOperations(int[] executionTime, int x, int y) {
  // write your code here
}
    executionTime[3, 4, 1, 7, 6]
    x4
    y2
    expected3
    checking account