# 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 nth 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)