Count and Say with Python

The challenge

he count-and-say sequence is the sequence of integers with the first five terms as following:

1.     1
2.     11
3.     21
4.     1211
5.     111221

1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.

Given an integer n where 1 ≤ n ≤ 30, generate the n^th term of the count-and-say sequence. You can do so recursively, in other words from the previous member read off the digits, counting the number of digits in groups of the same digit.

Note: Each term of the sequence of integers will be represented as a string.

Example 1:

Input: 1
Output: "1"
Explanation: This is the base case.

Example 2:

Input: 4
Output: "1211"
Explanation: For n = 3 the term was "21" in which we have two groups "2" and "1", "2" can be read as "12" which means frequency = 1 and value = 2, the same way "1" is read as "11", so the answer is the concatenation of "12" and "11" which is "1211".

The solution with Python

def countAndSay(self, n: int) -> str:

    # helper function
    def countAndSayHelper(s):
        # return value
        result = []
        # loop counter
        i = 0
        # loop while less than
        while i < len(s):
            c = 1
            # increment counters
            while i+1 < len(s) and s[i] == s[i+1]:
                i += 1
                c += 1

            # add string to the return list
            result.append(str(c)+s[i])

            # increment
            i += 1

        # return a string from the created list
        return ''.join(result)

    # set default return string value
    s = "1"
    # loop through input size
    for i in range(n-1):
        # use helper function
        s = countAndSayHelper(s)

    # return answer
    return s