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

**Example 2:**

Input:4Output:"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
```