3614. Process String with Special Operations II

You are given a string s consisting of lowercase English letters and the special characters: '*', '#', and '%'.

You are also given an integer k.

Build a new string result by processing s according to the following rules from left to right:

• If the letter is a lowercase English letter append it to result.
• A '*' removes the last character from result, if it exists.
• A '#' duplicates the current result and appends it to itself.
• A '%' reverses the current result.

Return the kth character of the final string result. If k is out of the bounds of result, return '.'.

class Solution {
public:
    char processStr(string s, long long k) {
        long long len = 0, ma = 4e18;
        vector<long long> lens;
        for(auto& ch : s) {
            if(isalpha(ch)) len++;
            else if(ch == '*') len = max(len - 1, 0ll);
            else if(ch == '#') len = min(len * 2, ma);

            lens.push_back(len);
        }

        if(len <= k or len == 0) return '.';

        for(int i = s.length() - 1; i >= 0; i--) {
            if(isalpha(s[i])) {
                if(lens[i] - 1 == k) return s[i];
            } else if(s[i] == '#') {
                k = k % (lens[i] / 2);
            } else if(s[i] == '%') {
                k = lens[i] - k - 1;
            }
        }
        return '#';
    }
};