I've seen many versions of this problem in coding interviews, so I decided to solve it here, just for fun.

The problem goes like this:

Input: "aaaabbbcca"
Output: [("a", 4), ("b", 3), ("c", 2), ("a", 1)]

Write a function that converts the input to the output

One important thing to note here is that the input can have the same letter in different groups, making the solution a little bit tricky if, for example, you try to use dictionaries.

Here is my proposed solution:

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
            result[word[i-1]] = counter
            counter = 1

Pay attention to the line with Hash.new.compare_by_identity, that gives us the ability to use a hash with duplicated keys.

You can read the algorithm like this:

# 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

But calling that function, the output would look a little bit different than required.

puts count_in_a_row("aaaabbbcca")

Output:   {"a"=>4, "b"=>3, "c"=>2, "a"=>1}
Expected: [("a", 4), ("b", 3), ("c", 2), ("a", 1)]

I would ask the interviewer if she is ok with the output or if the format is a hard requirement, and she wants me to implement it. In that case, I would do something like this:

def format_output(raw_output)
    formatted_output = "["
    raw_output.each do |key, value|
        formatted_output += "(\"" + key + "\", " + value.to_s + "), "
    formatted_output.delete_suffix!(", ")
    formatted_output += "]"

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)]

And that's it, some final questions for you, what's the time complexity of this solution? Is there a better solution?

You can find the source code of this post here.

Image credits to @joshuaryanphoto on Unsplash.