Elixir

Task

Scheduling is how the processor decides which jobs (processes) get to use the processor and for how long. This can cause a lot of problems. For example, a long-running process might use the entire CPU and block all the other processes from executing. One solution is Shortest Job First (SJF), which you will implement in this challenge.

SJF works by letting the shortest jobs take the CPU first. If the jobs are the same size then break the tie with First In First Out (FIFO). The idea is that the shorter jobs will finish quicker, so theoretically jobs won't get frozen because of large jobs. (In practice they're frozen because of small jobs).

Specification

sjf(jobs, index)

Returns the clock-cycles (cc) of when the process will be executed for a given index

Parameters

jobs: Array - A non-empty array of positive integers representing the clock cycles needed to finish a job.

index: Integer - A positive integer that respresents the job we're interested in

Return Value

Integer - A number representing the clock cycles it takes to complete the job at index.

Examples

jobs	index	Return Value
[3,10,20,1,2]	0	6
[3,10,10,20,1,2]	2	26

Example

SJF([3, 10, 20, 1, 2], 0) 

Here, we have 5 jobs scheduled with times 3, 10, 20, 1 and 2. We want to determine when the job at the 0th index will complete, based on the second argument to the SJF function. The job we're interested in takes 3 time units to complete.

at 0cc [3, 10, 20, 1, 2] jobs[3] starts at 1cc [3, 10, 20, 0, 2] jobs[3] finishes, jobs[4] starts at 3cc [3, 10, 20, 0, 0] jobs[4] finishes, jobs[0] starts at 6cc [0, 10, 20, 0, 0] jobs[0] finishes 

so:

SJF([3,10,20,1,2], 0) == 6

TEST

defmodule TestSolution do use ExUnit.Case import Challenge, only: [sjf: 2] test "should handle the example" do assert sjf([3,10,20,1,2],0) == 6 assert sjf([3,10,10,20,1,2],2) == 26 assert sjf([10,10,10,10], 3) == 40 end end