I have the following array
prime_arr = [2,3,5,7,11,13]
and I need to check whether the array has 3, 5 7 and return the value(s) differ based on presence
If it has 3 alone return "Yes" and 5 alone return "it is " and 7 alone return "done" and
If it has [3,5] return "Yes it is "
If it has [3,5,7] return "Yes it is done"
The return value should differ based on combination
Thanks in advance.
Since this looks a lot like homework (and since no one here is paid to write your code for you), I'm not going to give you the actual answer. I will however give you some pointers to pieces of the puzzle - hopefully you'll be able to piece them together.
You can find the intersection between two arrays using the #& method. So:
prime_arr = [2,3,5,7,11,13]
# => [2, 3, 5, 7, 11, 13]
prime_arr & [3]
# => [3]
prime_arr & [3,5]
# => [3, 5]
prime_arr & [3,5,7]
# => [3, 5, 7]
prime_arr & [4]
# => []
A case statement can toggle on Array values, like so:
a = [1,2]
case a
when [1]; "just one"
when [1,2]; "both"
end
# => "both"
Good luck!
Not sure about the other three cases you haven't made clear, so I am guessing.
[[3, 5, 7], ["Yes ", "it is ", "done"]]
.transpose.each.with_object(""){|(i, w), s| s << w if prime_arr.include?(i)}
words = ['Yes', 'it is', 'done']
prime_arr = [2, 3, 5, 7, 11, 13]
target = [3, 5, 7]
words.take((prime_arr & target).size).join(" ")
Edited:
prime_arr = [2, 3, 5, 7, 11, 13]
mapping = {
[3] => 'Yes',
[3, 5] => 'Yes it is',
[3, 5, 7] => 'Yes it is done'
}
mapping[prime_arr & [3, 5, 7]] # => 'Yes it is done'
mapping[prime_arr & [3, 5]] # => 'Yes it is'
mapping[prime_arr & [3]] # => 'Yes'
You get a nil when none of the keys in mapping matches.
