I need to count the number of values that both arrays have.
def process_2arrays(arr1, arr2)
length1 = arr1.count
length2 = arr2.count
arr3 = []
i = 0
while length1 >= i do
ci = arr1[i]
if arr2.include?(ci)
arr3 << ci
damn = arr3.count
i = i + 1
end
return [(damn), (2), (3), (4)]
end
end
When I pass the values to the function it returns [nil, 2, 3, 4] Whats the problem here?
To find elements that exist in both arrays, use the set intersection method &.
http://ruby-doc.org/core-2.2.0/Array.html#method-i-26
def count_same_elements(a1,a2)
(array1 & array2).length
end
Example
count_same_elements([1,2,3,4],[2,3,4,5])
=> 3
damn is initialized within a do .. end block, specifically the while block. Therefore, its value will live within that scope, and when you call the variable outside the block its value is nil.
If you want to preserve the value, you must initialize the variable to nil outside the block.
i = 0
damn = nil
...
As a side note, your code is lacking the most basic Ruby standards. In Ruby you generally use an iterator, not the while. Moreover, you don't use the return at the end of a method.
This is how you would write your method in Ruby using the iterators and taking advantage of some methods from the core library.
def process_2arrays(arr1, arr2)
arr3 = arr1.select { |e| arr2.include?(e) }
[arr3.size, 2, 3, 4]
end
Changing completely approach, you can use
def process_2arrays(arr1, arr2)
(arr1 & arr2).size
end
