Solving the classic “aaaabbbcca” coding challenge

Photo by @joshuaryanphoto on Unsplash.
Input: “aaaabbbcca”
Output: [(“a”, 4), (“b”, 3), (“c”, 2), (“a”, 1)]
Write a function that converts the input to the output
def count_in_a_row(word)
counter = 1
result = Hash.new.compare_by_identity

for i in (1..word.length)
if word[i] == word[i-1]
counter += 1
else
result[word[i-1]] = counter
counter = 1
end
end

result
end
# Iterate the letters in the word, starting from the second one# # If the current letter is equal to the previous one, increase the counter# # If the current letter is different to the previous one,
# # save the key (the letter) with the value (the counter) in the hash,
# # and then reset the counter.
# Return the hash
puts count_in_a_row("aaaabbbcca")Output:   {"a"=>4, "b"=>3, "c"=>2, "a"=>1}
Expected: [("a", 4), ("b", 3), ("c", 2), ("a", 1)]
def format_output(raw_output)
formatted_output = "["
raw_output.each do |key, value|
formatted_output += "(\"" + key + "\", " + value.to_s + "), "
end
formatted_output.delete_suffix!(", ")
formatted_output += "]"
end
puts format_output(count_in_a_row("aaaabbbcca"))Output: [("a", 4), ("b", 3), ("c", 2), ("a", 1)]
Expected: [("a", 4), ("b", 3), ("c", 2), ("a", 1)]

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