## 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 *`n`

^{th}* term of the count-and-say sequence*.

**Example 1:**

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

**Example 2:**

Input:n = 4Output:"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).

## One Thought to “LeetCode 38. Count and Say”