LeetCode-by-Swift-Python-C-SQL/Python/String/0038_count_and_say.py at master · lee00286/LeetCode-by-Swift-Python-C-SQL · GitHub

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class Solution:
    def countAndSay(self, n: int) -> str:
        # Base Case: 1
        # 2 = "11", 3 = "21", 4 = "1211", 5 = "111221"

        def countCon(num: str) -> str:
            i = 0
            cons = ""
            latest = ""
            count = 0

            # Count number of consecutives
            for char in num:
                # First consecutive
                if (latest == ""):
                    latest = char
                    count += 1
                else:
                    # If consecutive
                    if (latest == char):
                        count += 1
                    # If not consecutive
                    else:
                        cons += str(count) + latest
                        latest = char
                        count = 1

            cons += str(count) + latest

            return cons

        # None Case
        if (not n or n <= 0):
            return ''
        # Base Case
        if (n == 1):
            return '1'
        # Recursive Case
        else:
            # Count consecutives
            return countCon(self.countAndSay(n - 1))