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I want to write a program that splits an array into two arrays, where any element in one array is smaller than any element in the other array.

The input that I have is:

a = [6, 45, 23, 65, 17, 48, 97, 32, 18, 9, 88]

And I'd like output like this:

[6, 23, 17, 18 , 9] < [45, 65, 48, 97, 32, 88]

I've tried:

i = 0
max = 0

while i < a.size
  if a[i] > max
    max = a[i]
  end
  i+=1
end

puts "this is the larger array: " + max.to_s

Which is completely off. As I am new to this, any help is appreciated.

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  • 1
    What is the criteria of where the two arrays should split? Should they be of equal length or any other criteria? Commented Nov 14, 2013 at 20:02
  • thanks hirolau, i should have mentioned that. Yes around the same size Commented Nov 14, 2013 at 20:06

5 Answers 5

Reset to default
4 
small, large = a.sort!.shift(a.size/2) ,a

p small, large 
#=> [6, 9, 17, 18, 23]
#=> [32, 45, 48, 65, 88, 97]
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3 Comments

Thanks hirolau, and is that ruby?
No Rails syntax here, there were some changes between 1.8 and 1.9 in how parallel assignment works, but I think it should work on 1.8+.
Thanks hirolau, that's perfect!
2

Try this:

newarray = a.sort.each_slice((a.size/2.0).round).to_a

It will give you an array containing your split array:

newarray = [[6,9,17,18,23,32],[45,48,65,88,97]]

In this case, if you have an odd number of elements in your array, the first array returned will always have the extra element. You can also save the arrays separately if you would like, but this way you can call each of the halves with newarray[0] and newarray[1]. If you want to split them simply add:

b = newarray[0]
c = newarray[1]
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2 Comments

I don't know how I overlooked this, thanks sabrams this is awesome!
No need to round. Just divide by 2 instead of 2.0.
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Don't use a while loop - sort the array and then split it in two

a.sort
a.in_groups_of( a.size/2)


a.sort.each_slice( a.size/2) probably does the trick without rails.
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8 Comments

This works, but only if you're using rails, ruby does not have in_groups_of
Your rails solution will work only if the array is even, check my solution below to see how to manage odd instances.
It's really not very helpful to offer a guess of what might work in a language one does not know.
it is helpful to suggest that the original algorithm (using a while loop) was the wrong idea and that the problem should be approached differently
In a comment, perhaps.
|
0 
a = [6, 45, 23, 65, 17, 48, 97, 32, 18, 9, 88]
a = a.sort
print a.shift(a.count/2), " < " , a 
#=> [6, 9, 17, 18, 23] < [32, 45, 48, 65, 88, 97]

Another variation

a = [6, 45, 23, 65, 17, 48, 97, 32, 18, 9, 88]
a = a.sort
print a.values_at(0..a.count/2), " < ", a.values_at((a.count/2)+1 .. -1)
#=> [6, 9, 17, 18, 23] < [32, 45, 48, 65, 88, 97]
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0

Assuming you want to preserve order, as in your example: 

def split_it(a,n)
  f = a.select {|e| e <= n}
  [f, a-f]
end

a = [6, 45, 23, 65, 17, 48, 97, 32, 18, 9, 88]

f, l = split_it(a,23)
puts "#{f} < #{l}" # => [6, 23, 17, 18, 9] < [45, 65, 48, 97, 32, 88]

If you want to preserve order and have the first subarray contain nbr elements, add this:

def split_nbr(a, nbr)
  n = 1
  loop do
    return [] if n > a.max
    b = split_it(a,n)
    return b if b.first.size == nbr
    n += 1
  end   
end

f, l = split_nbr(a,3)
puts "#{f} < #{l}" # => [6, 17, 9] < [45, 23, 65, 48, 97, 32, 18, 88]
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