Count and Say

#Brute Force approach
"""
Starting from "1", we build each term by reading the previous string and grouping consecutive identical characters.
While scanning from left to right, we compare each character with its previous one; if they match we increase the count, and when they differ we append the count and the previous character to form the next sequence.
"""
class Solution:
    def countAndSay(self, n: int) -> str:
        res = "1"

        for _ in range(n - 1):
            temp = ""
            count = 1

            for i in range(1, len(res)):
                if res[i] == res[i - 1]:
                    count += 1
                else:
                    temp += str(count) + res[i - 1]
                    count = 1

            # Add last group
            temp += str(count) + res[-1]
            res = temp

        return res