Here is an approach :
def sum_to_n?(a,n)
!!a.find{|e| a.include?(n-e)}
end
a = [1,2,3,4,5]
sum_to_n?(a,9) # => true
sum_to_n?(a,11) # => false
If you want to get those 2 elements:
def sum_to_n?(a,n)
num=a.find{|e| a.include?(n-e)}
unless num
puts "not exist"
else
p [n-num,num]
end
end
a = [1,2,3,4,5]
sum_to_n?(a,9)
# >> [5, 4]
sum_to_n?(a,11)
# >> not exist
Logic
Enumerable#find
method passing one array element per iteration.Now for any iteration,say I have an element e
,and I subtracted it from n. Now I was just testing that (n-e)
is present in the source array.If I found a match #find
will stop finding,and immediately will return e
.If not found,then it will go for next iteration. If #find
completes its iteration,but didn't find (n-e)
,as per the documentation it will return nil
.