# Simple Iterative Ruby solution w/ explanation (accepted)

• I broke up the problem into 2 functions. This is merely an application of the Cartesian Product generalized to n sets where n >= 2. Here is a great resource http://ndp.jct.ac.il/tutorials/discrete/node28.html

Using cartesian product for the input of "23" we get ["a", "b", "c"] x ["d", "e", "f"] = ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]

Say we want "234" then we would have the following

["a", "b", "c"] x ( ["d", "e", "f"] x ["g", "h", "i"] ).

Therefore we only need to write a function that returns the cartesian product of two arrays. We then can generalize to n > 2 arrays by using the reduce function. The code follows

``````def letter_combinations(digits)
dict = {
"1" => [],
"2" => %w(a b c),
"3" => %w(d e f),
"4" => %w(g h i),
"5" => %w(j k l),
"6" => %w(m n o),
"7" => %w(p q r s),
"8" => %w(t u v),
"9" => %w(w x y z)
}
digits.split("").reverse.map do |digit|
dict[digit]
end.reduce([]) do |result, arr|
cartesian_product(arr, result)
end
end

def cartesian_product(arrA, arrB)
return arrB if arrA.length == 0
return arrA if arrB.length == 0
result = []
for i in (0..arrA.length - 1)
for j in (0..arrB.length - 1)
result += [arrA[i] + arrB[j]]
end
end
result
end
``````

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