# Leetcode – Backspace String Compare Solution in Python

String manipulation is a cornerstone in the realm of algorithmic challenges, requiring a combination of logical reasoning, pattern recognition, and data structure knowledge. The “Backspace String Compare” problem from Leetcode embodies these elements, creating a captivating puzzle. This article endeavors to explore this problem, shedding light on its solution using Python.

1. Problem Statement
2. Deciphering the Problem
3. Potential Solution Approaches
4. Python Implementation
5. Analyzing the Solution
6. Possible Optimizations and Alternatives
7. Conclusion

### 1. Problem Statement

Given two strings S and T, return True if they are equal when both are typed into empty text editors. # means a backspace character. It is guaranteed that after backspacing characters, the strings on the text editors consist of lowercase English letters only.

### 2. Deciphering the Problem

For context, imagine typing into a text editor. Whenever you encounter the character #, it represents a backspace, which means the character before it gets deleted.

For example:

• S = "ab#c" translates to ac since the # deletes the b.
• T = "ad#c" translates to ac as well since the # deletes the d.

Given these translations, S and T are equivalent.

### 3. Potential Solution Approaches

There are multiple strategies one might consider:

1. Simulation: Directly simulate the input as if you are typing in a text editor. Use a stack or list to keep track of characters, and pop when you encounter a #.
2. Two Pointers: Traverse both strings from the end to the beginning (as the effect of backspace is from right to left). Skip characters which are removed by the backspace.

### 4. Python Implementation

Let’s explore the simulation approach using a stack:

def backspaceCompare(S, T):
def build(s):
stack = []
for char in s:
if char != '#':
stack.append(char)
elif stack:
stack.pop()
return ''.join(stack)

return build(S) == build(T)

### 5. Analyzing the Solution

• Function Build: This helper function is designed to process a string according to the problem’s rules. It uses a stack to manage the characters.
• If the character isn’t a #, it’s appended to the stack.
• If the character is a # and the stack isn’t empty, the last character is removed from the stack.
• The processed string is then constructed from the stack and returned.
• Finally, the processed versions of S and T are compared to check for equivalence.

### 6. Possible Optimizations and Alternatives

1. Space Efficiency with Two Pointers: Instead of using extra space to process the strings, we can use two pointers to traverse both strings from the end. While iterating, we can keep track of the number of backspaces (#) and skip that many characters. This method would be more space-efficient as it doesn’t require additional space proportional to the string lengths.
2. Extensions: The problem can be further complicated by introducing more editing operations, like forward delete, move cursor left, or move cursor right. Handling these would require a more intricate algorithm, perhaps combining both the simulation and two pointers approach.
3. Regular Expressions: Another approach might involve using regular expressions to identify and replace patterns. However, this might not necessarily be efficient, especially for large strings, but could serve as an interesting exercise in regex manipulations.

### 7. Conclusion

The “Backspace String Compare” problem beautifully blends string manipulation with stack operations, offering a realistic scenario (editing in a text editor). Python’s intuitive syntax and rich data structure support, like built-in lists acting as stacks, make solving such problems a smooth experience. This problem serves as a reminder that real-world tasks often involve layered operations, and algorithmic solutions can simplify seemingly complex tasks with elegance.