# LeetCode 38. Count and Say

## Description

https://leetcode.com/problems/count-and-say/

The count-and-say sequence is a sequence of digit strings defined by the recursive formula:

• `countAndSay(1) = "1"`
• `countAndSay(n)` is the way you would “say” the digit string from `countAndSay(n-1)`, which is then converted into a different digit string.

To determine how you “say” a digit string, split it into the minimal number of groups so that each group is a contiguous section all of the same character. Then for each group, say the number of characters, then say the character. To convert the saying into a digit string, replace the counts with a number and concatenate every saying.

For example, the saying and conversion for digit string `"3322251"`:

Given a positive integer `n`, return the `nth` term of the count-and-say sequence.

Example 1:

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

Example 2:

```Input: n = 4
Output: "1211"
Explanation:
countAndSay(1) = "1"
countAndSay(2) = say "1" = one 1 = "11"
countAndSay(3) = say "11" = two 1's = "21"
countAndSay(4) = say "21" = one 2 + one 1 = "12" + "11" = "1211"
```

Constraints:

• `1 <= n <= 30`

## Explanation

The base case is when n = 1. For other case, we call countAndSay(n – 1) to get the previous string and build the count and say string by grouping and counting the consecutive numbers.

## Python Solution

``````class Solution:
def countAndSay(self, n: int) -> str:
if n == 1:
return "1"
else:
prev_sequence = self.countAndSay(n - 1)

result_sequence = ""
counter = {}
for i in range(0, len(prev_sequence)):
digit = prev_sequence[i]

if digit not in counter:
for key, value in counter.items():
result_sequence += str(value) + key
counter = {}
counter[digit] = 1

else:
counter[digit] += 1

for key, value in counter.items():
result_sequence += str(value) + key

return result_sequence``````
• Time complexity: O(N*M). M is the longest sequence length.
• Space complexity: O(N).