I need to return the smallest numbers (can be more than one) that occur in an array.
given [1,1,2,3,3,4]
expected [1,1]
given [8,8,2,3]
expected [2]
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1examples are clear for me, but what does `that occurs more than once` mean? – Ilya Apr 26 '17 at 11:11
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yes I wondered the same, isn't the second example wrong by that statement? – davegson Apr 26 '17 at 11:11
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2"To prevent confusion", you should simply say "smallest number". – Stefan Apr 26 '17 at 11:11
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@Stefan Hi. How would you word that exactly? I dont mind an edit. – Sylar Apr 26 '17 at 11:12
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Why `min(x)` is not possible.? – Roman Kiselenko Apr 26 '17 at 11:15
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@Sylar I'd replace largest / highest with smallest / lowest and it should be clear without further explanation. The fact that you treat `1` as the highest number in your project is irrelevant to the question. – Stefan Apr 26 '17 at 11:16
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@Зелёный I will not know the value of `n`. – Sylar Apr 26 '17 at 11:16
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If you want just the number of such elements: `arr.count { |i| i == arr.min }` – Sagar Pandya Apr 26 '17 at 11:46
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1@sagarpandya82: You could move `arr.min` out of the block to avoid calling it `n` times. – Eric Duminil Apr 26 '17 at 11:49
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3"To prevent confusion, the smallest number is the highest". Now, that prevented confusion! :) – Eric Duminil Apr 26 '17 at 11:49
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@EricDuminil good idea about `arr.min` and agreed about the confusion. – Sagar Pandya Apr 26 '17 at 11:56
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1on second thoughts no need for the block `arr.count(arr.min)`. – Sagar Pandya Apr 26 '17 at 12:06
4 Answers
6
What about this?
a = [1, 1, 2, 3, 3, 4]
min = a.min
a.select { |i| i == min }
=> [1, 1]
Ursus
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I think most would go with this; it's a shame it's not an answer in the question for which this question is a duplicate. – Sagar Pandya Apr 26 '17 at 12:17
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Another way:
a = [1,1,2,3,3,4]
a.group_by(&:itself).min.last
steenslag
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Slight variations `a.group_by(&:itself)[a.min]` and `a.group_by(&:itself).fetch(a.min)` – Sagar Pandya Apr 26 '17 at 11:53
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Here's one way to solve the problem:
- Find the smallest number
- Count how many times it appears
- Generate an array with the solution instead of selecting elements in the input
Snippet:
input = [1,1,2,3,3,4]
highest = input.min
[highest] * input.count(highest)
yoones
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Whilst this code snippet is welcome, and may provide some help, it would be [greatly improved if it included an explanation](//meta.stackexchange.com/q/114762) of *how* it addresses the question. Without that, your answer has much less educational value - remember that you are answering the question for readers in the future, not just the person asking now! Please [edit] your answer to add explanation, and give an indication of what limitations and assumptions apply. – Toby Speight Apr 26 '17 at 11:31
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Using this thinking we can do `Array.new(arr.count(arr.min), arr.min)` – Sagar Pandya Apr 26 '17 at 12:07
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You could, but then you would traverse the array 3 times when 2 suffice. – yoones Apr 26 '17 at 12:40
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Also, you can use take_while:
sorted = a.sort
sorted.take_while { |e| e == sorted[0] }
=> [1, 1]
Other possible way:
a = [1, 1, 2, 3, 3, 4]
Array.new(*a.each_with_object(Hash.new(0)) {|e, acc| acc[e] += 1}.min.reverse)
=> [1, 1]
Ilya
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