Problem Description

You are given a string s that contains lowercase English letters and three special characters: *, #, and %.

Your task is to process the string from left to right and build a new string result by applying the following rules for each character:

• Lowercase letters: If the character is a lowercase English letter (a-z), append it to the result string.

• Asterisk (*): This acts as a backspace operation. It removes the last character from result if result is not empty. If result is empty, nothing happens.

• Hash (#): This duplicates the entire current result string and appends the copy to itself. For example, if result is currently "abc", after encountering #, it becomes "abcabc".

• Percent (%): This reverses the entire current result string. For example, if result is currently "abc", after encountering %, it becomes "cba".

After processing all characters in s according to these rules, return the final result string.

Example walkthrough:

If s = "ab*c#%", the processing would be:

1. 'a' → result = "a"
2. 'b' → result = "ab"
3. '*' → result = "a" (removes 'b')
4. 'c' → result = "ac"
5. '#' → result = "acac" (duplicates "ac")
6. '%' → result = "caca" (reverses "acac")

The final answer would be "caca".

Intuition

The problem asks us to simulate a series of operations on a string as we read characters one by one. This is a classic simulation problem where we need to maintain the current state and update it based on the input.

The key insight is that we need a data structure that supports:

• Adding elements to the end (for lowercase letters)
• Removing elements from the end (for *)
• Duplicating the entire content (for #)
• Reversing the order (for %)

A list (or dynamic array) is perfect for this because:

• append() adds to the end in O(1) time
• pop() removes from the end in O(1) time
• extend() can efficiently append a copy of the list to itself
• reverse() can reverse the list in-place in O(n) time

We can't use a regular string because strings are immutable in most programming languages, making operations like deletion and reversal inefficient. Each modification would create a new string, leading to unnecessary overhead.

The solution naturally follows a left-to-right scan of the input string s, where for each character we check its type and perform the corresponding operation on our result list. This straightforward approach works because each operation only depends on the current state of the result, not on future characters.

At the end, we simply join the list of characters into a final string. This approach is both intuitive and efficient, as we're performing each operation exactly once and in the order specified by the input.

Solution Approach

We implement the solution using a simulation approach with a list as our primary data structure. Here's how the implementation works:

Data Structure Choice: We use a list result = [] to store the characters of our result string. This choice allows us to efficiently perform all required operations.

Processing Loop: We iterate through each character c in the input string s and handle it based on its type:

1. Lowercase Letter Check: if c.isalpha()

   • When we encounter a lowercase letter, we simply append it to our result list using result.append(c)

2. Asterisk Handling: elif c == "*" and result

   • We check if c is an asterisk AND if the result list is not empty
   • If both conditions are true, we remove the last character using result.pop()
   • The check for non-empty result prevents errors when trying to pop from an empty list

3. Hash Handling: elif c == "#"

   • We duplicate the current result by using result.extend(result)
   • The extend() method takes an iterable (in this case, a copy of the current result) and adds all its elements to the end of the list
   • This effectively doubles the content: if result was ['a', 'b'], it becomes ['a', 'b', 'a', 'b']

4. Percent Handling: elif c == "%"

   • We reverse the result list in-place using result.reverse()
   • This operation modifies the list directly without creating a new one

Final Conversion: After processing all characters, we convert the list back to a string using "".join(result), which concatenates all characters in the list into a single string.

Time Complexity: O(n × m) where n is the length of the input string and m is the maximum length of the result during processing (due to the potential doubling operations with #).

Space Complexity: O(m) where m is the length of the final result string.

Example Walkthrough

Let's trace through the example s = "ab*c#%" step by step to see how our solution works:

Initial Setup:

• Input string: s = "ab*c#%"
• Result list: result = []

Step-by-step Processing:

1. Character 'a':

   • Check: c.isalpha() → True (it's a lowercase letter)
   • Action: result.append('a')
   • Result: ['a']

2. Character 'b':

   • Check: c.isalpha() → True (it's a lowercase letter)
   • Action: result.append('b')
   • Result: ['a', 'b']

3. Character '*':

   • Check: c == '*' and result → True (asterisk and result is not empty)
   • Action: result.pop() (removes last element 'b')
   • Result: ['a']

4. Character 'c':

   • Check: c.isalpha() → True (it's a lowercase letter)
   • Action: result.append('c')
   • Result: ['a', 'c']

5. Character '#':

   • Check: c == '#' → True
   • Action: result.extend(result) (extends with a copy of itself)
   • Before extend: ['a', 'c']
   • After extend: ['a', 'c', 'a', 'c']

6. Character '%':

   • Check: c == '%' → True
   • Action: result.reverse() (reverses the list in-place)
   • Before reverse: ['a', 'c', 'a', 'c']
   • After reverse: ['c', 'a', 'c', 'a']

Final Conversion:

• Join the list: "".join(['c', 'a', 'c', 'a']) → "caca"
• Return: "caca"

This walkthrough demonstrates how each special character modifies the result list: the asterisk acts as backspace, the hash duplicates the content, and the percent reverses everything. Using a list allows us to perform these operations efficiently compared to working with immutable strings.

Solution Implementation

Time and Space Complexity

Time Complexity: O(2^n), where n is the length of string s.

The analysis breaks down as follows:

• Iterating through the string takes O(n) time
• For each character c in s:
  • If c is alphabetic: append() operation is O(1)
  • If c is *: pop() operation is O(1)
  • If c is #: extend(result) operation is O(len(result)), which doubles the current list
  • If c is %: reverse() operation is O(len(result))

In the worst-case scenario, if the string contains multiple # characters, each # operation doubles the length of result. If we have k occurrences of #, the size of result can grow to 2^k in the worst case. Since k can be proportional to n, the time complexity becomes O(2^n) when considering the cumulative cost of all operations, particularly the extend() and reverse() operations on increasingly large lists.

Space Complexity: O(2^n), where n is the length of string s.

The space complexity is determined by the maximum size of the result list. In the worst case with multiple # operations, the result list can grow exponentially to size 2^n. The reference answer states O(1) when ignoring the space consumption of the answer itself, meaning if we don't count the output list result as part of the space complexity (only considering auxiliary space), then the space complexity would be O(1) as no additional data structures are used beyond the output.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Duplication with # Operation

The Pitfall: A critical bug exists in the hash operation implementation. Using result.extend(result) or result.extend(result[:]) creates an infinite loop issue in Python. When extending a list with itself, Python continues to read from the list as it grows, causing the operation to never terminate properly.

Example of the Problem:

result = ['a', 'b']
result.extend(result)  # This will cause issues!
# Expected: ['a', 'b', 'a', 'b']
# Actual: Unpredictable behavior or infinite extension

The Solution: Create a complete copy of the list before extending:

elif char == "#":
    # Create a copy first, then extend
    temp = result.copy()  # or result[:]
    result.extend(temp)

2. Not Handling Empty Result for Asterisk Operation

The Pitfall: Forgetting to check if the result is empty before attempting to pop can cause an IndexError:

elif char == "*":
    result.pop()  # IndexError if result is empty!

The Solution: Always check if the list has elements before popping:

elif char == "*" and result:
    result.pop()

3. Performance Issues with Large Duplications

The Pitfall: Multiple # operations can cause exponential growth of the result string. For example, if the input is "a####", the result grows from 1 → 2 → 4 → 8 → 16 characters. This can quickly consume memory and time.

The Solution: Consider adding bounds checking or limits:

elif char == "#":
    if len(result) > MAX_LENGTH // 2:  # Prevent overflow
        raise ValueError("Result string would exceed maximum length")
    temp = result.copy()
    result.extend(temp)

Corrected Complete Solution:

class Solution:
    def processStr(self, s: str) -> str:
        result = []
      
        for char in s:
            if char.isalpha():
                result.append(char)
            elif char == "*" and result:
                result.pop()
            elif char == "#":
                # Fix: Create a copy before extending
                temp = result.copy()
                result.extend(temp)
            elif char == "%":
                result.reverse()
      
        return "".join(result)