Construct String With Repeat Limit

lcpy gen -n 2182   # by problem number
lcpy gen -s construct_string_with_repeat_limit   # by problem name

Input: s = "cczazcc", repeatLimit = 3
Output: "zzcccac"
Explanation: We use all of the characters from s to construct the repeatLimitedString "zzcccac".
The letter 'a' appears at most 1 time in a row.
The letter 'c' appears at most 3 times in a row.
The letter 'z' appears at most 2 times in a row.
Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
The string is the lexicographically largest repeatLimitedString possible so we return "zzcccac".
Note that the string "zzcccca" is lexicographically larger but the letter 'c' appears more than 3 times in a row, so it is not a valid repeatLimitedString.

Input: s = "aababab", repeatLimit = 2
Output: "bbabaa"
Explanation: We use only some of the characters from s to construct the repeatLimitedString "bbabaa".
The letter 'a' appears at most 2 times in a row.
The letter 'b' appears at most 2 times in a row.
Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
The string is the lexicographically largest repeatLimitedString possible so we return "bbabaa".
Note that the string "bbabaaa" is lexicographically larger but the letter 'a' appears more than 2 times in a row, so it is not a valid repeatLimitedString.

class Solution:
    # Time: O(n + 26 log 26)
    # Space: O(26)
    def repeat_limited_string(self, s: str, repeat_limit: int) -> str:
        counts = [0] * 26
        for ch in s:
            counts[ord(ch) - ord("a")] += 1

        parts: list[str] = []
        big = 25
        while big >= 0:
            if counts[big] == 0:
                big -= 1
                continue
            use = min(counts[big], repeat_limit)
            parts.append(chr(ord("a") + big) * use)
            counts[big] -= use
            if counts[big] == 0:
                big -= 1
                continue
            small = big - 1
            while small >= 0 and counts[small] == 0:
                small -= 1
            if small < 0:
                break
            parts.append(chr(ord("a") + small))
            counts[small] -= 1
        return "".join(parts)