elixirdev
How can I get immediate greater and lesser value of a number
value = 7
jobs = [1, 4, 9, 3, 8, 6, 2]
output should be 6,8
i have large and too many lists. please suggest me optimised solutions.
Marked As Solved
mudasobwa
jobs = [1, 4, 9, 3, 8, 6, 2]
num = 7
Enum.reduce(jobs, {nil, nil}, fn
e, {nil, max} when e < num -> {e, max}
e, {min, nil} when e > num -> {min, e}
e, {min, max} when e > min and e < num -> {e, max}
e, {min, max} when e < max and e > num -> {min, e}
_, min_max -> min_max
end)
The idea is you walk through the list narrowing down the difference. Maybe this looks shorter:
Enum.reduce(jobs, {nil, nil}, fn
e, {min, max} when (is_nil(min) or e > min) and e < num -> {e, max}
e, {min, max} when (is_nil(max) or e < max) and e > num -> {min, e}
_, min_max -> min_max
end)
Also Liked
NobbZ
So for any given n, do you always want to check if n-1 and n+1 are in the list, or do you search those x for that holds that there is no y greather than x and lesser than n (and the other way round for the other side)?
edit
Perhaps my last sentence could be better phrased as “…or do you want to search for the greatest number smaller than n as well as the smallest number greater than n?”
another edit
If it is the first, I’d roughly do
def search(n, list) do
Enum.filter(list, & &1 in [n-1, n+1])
endmudasobwa
Yeah, I know. I would even go with my first snippet because I value clarity way more than brevity.
Last Post!
NobbZ
So if you only want to see n-1 and n+1 my solution should suffice. You can microoptimise it though by using Enum.reduce or even List.fold. Also by precalculating n-1 and n+1 outside of the closure.
But I’m not sure if that will shave much…
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