## Description

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

The 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".

## Explanation

A palindrome, and its reverse, are identical to each other.

## 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”