How do you solve Process String with Special Operations I on LeetCode?

Process String with Special Operations I (LeetCode #3612) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Stack usage simplifies operations

What is the time complexity of Process String with Special Operations I?

The optimal solution for Process String with Special Operations I runs in O(n) time and uses O(n) space. The stack allows for efficient operations, maintaining a linear time complexity as each operation is handled in constant time.

What is the brute force solution for Process String with Special Operations I?

The brute force approach for Process String with Special Operations I is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach processes each character in the string and applies operations directly to a result string. It can lead to inefficient string manipulations, especially with duplication and reversal. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Process String with Special Operations I?

Process String with Special Operations I uses the following concepts: String, Simulation. The recommended patterns to study are: Stack, String Manipulation.

What are the interview tips for Process String with Special Operations I?

Explain your thought process clearly Discuss edge cases Practice similar string manipulation problems

What are common mistakes when solving Process String with Special Operations I?

Not handling empty stack for '*' operation Confusing '#' and '%' operations

#3612

Process String with Special Operations I

Medium

StringSimulationStackString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

This approach uses a stack to efficiently manage the result string, allowing for quick operations without excessive string manipulation.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty stack.
2Step 2: Iterate through each character in the input string.
3Step 3: For letters, push to stack; for '*', pop; for '#', duplicate the stack; for '%', reverse the stack.

solution.py12 lines

1def process_string(s):
2    stack = []
3    for char in s:
4        if char.isalpha():
5            stack.append(char)
6        elif char == '*':
7            if stack: stack.pop()
8        elif char == '#':
9            stack += stack
10        elif char == '%':
11            stack.reverse()
12    return ''.join(stack)

ℹ

Complexity note: The stack allows for efficient operations, maintaining a linear time complexity as each operation is handled in constant time.

1Stack usage simplifies operations
2In-place operations reduce time complexity

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.