3612. Process String with Special Operations I

String
Simulation
Stack

TIME

O(n * L)

SPACE

O(L)

Intuition

There is no hidden structure, each character is a direct instruction on the running buffer. Just carry out the operations in order. Since s has at most 20 characters (and only the few # before the final state can grow it), the buffer stays small, so a plain simulation is more than fast enough.

Approach

Maintain result as a mutable list of characters and process each character of s:

Lowercase letter -> append it.
'*' -> pop the last character if the buffer is non-empty.
'#' -> double the buffer (result += result).
'%' -> reverse the buffer in place.

Join the list into a string at the end.

Complexity

Time: O(n * L), for each of the n characters, an op may touch up to the current length L (reverse/duplicate); bounded and tiny here
Space: O(L), the result buffer

Edge Cases

'*' on an empty buffer -> no-op
'#' on an empty buffer -> stays empty
'%' on an empty or single-character buffer -> unchanged
A string of only special characters -> result can end up empty

Code

class Solution:
    def processStr(self, s: str) -> str:
        """
        3612. Process String with Special Operations I
        Time: O(n * L)
        Space: O(L)
        """
        result = []

        for ch in s:
            if ch == "*":
                if result:
                    result.pop()
            elif ch == "#":
                result += result
            elif ch == "%":
                result.reverse()
            else:
                result.append(ch)

        return "".join(result)

/**
 * 3612. Process String with Special Operations I
 * Time: O(n * L)
 * Space: O(L)
 */
class Solution {
    public String processStr(String s) {
        StringBuilder result = new StringBuilder();

        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (ch == '*') {
                if (result.length() > 0) {
                    result.deleteCharAt(result.length() - 1);
                }
            } else if (ch == '#') {
                result.append(result);
            } else if (ch == '%') {
                result.reverse();
            } else {
                result.append(ch);
            }
        }

        return result.toString();
    }
}