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Problem Brief

Job Execution

INTERNOA
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There are n jobs that can be executed in parallel on a processor, where the execution time of the jth 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[i] 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 Description

Complete the function getMinimumOperations in the editor below.

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

Returns

int: the minimum number of operations in which the processor can complete the jobs

1Example 1

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

Constraints

Limits and guarantees your solution can rely on.

  • 1 ≤ n ≤ 105
  • 1 ≤ executionTime[i] ≤ 109
  • 1 ≤ y < x ≤ 109
public int getMinimumOperations(int[] executionTime, int x, int y) {
  // write your code here
}
Input

executionTime

[3, 4, 1, 7, 6]

x

6

y

2

Output

3

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