# Mastering the Backspace String Compare on LeetCode: A Comprehensive Guide

## Dive deep into solving the intriguing LeetCode problem "Backspace String Compare" with expert strategies and Python solutions.

In the world of software engineering interviews, the ability to dissect and solve problems is invaluable. Today, I'm excited to explore the "Backspace String Compare" (LeetCode 844) problem from LeetCode.

This challenge not only tests your understanding of string manipulation but also your ability to think critically about operations that affect data structure.

## Introduction to the Problem

Imagine you're typing on a text editor that supports a peculiar feature: a backspace character represented by '#'. You type two strings, but because of your frequent use of the backspace, the final text might look different from what you initially intended.

The question is, after all the backspacing, do the two strings end up being the same?

Given two strings

`s`

and`t`

, return`true`

if they are equal when both are typed into empty text editors.`'#'`

means a backspace character.Note that after backspacing an empty text, the text will continue empty.

Example 1:`Input: s = "ab#c", t = "ad#c" Output: true Explanation: Both s and t become "ac".`

Example 2:`Input: s = "ab##", t = "c#d#" Output: true Explanation: Both s and t become "".`

Example 3:`Input: s = "a#c", t = "b" Output: false Explanation: s becomes "c" while t becomes "b".`

### Considering Different Approaches

**The Stack Approach**: Stacks naturally follow the Last In, First Out (LIFO) principle, which aligns perfectly with the backspace functionality. We can iterate over each character in the strings, using a stack to build the final string post-backspacing.**The Two-Pointer Approach**: This approach involves iterating from the end of both strings towards the beginning, simulating the backspace operation in reverse. This method is more space-efficient as it requires no extra data structure.**Direct Comparison with In-Place Modification**: Here, we modify the strings in place, effectively "backspacing" by overwriting characters and comparing lengths and contents directly afterward.

### Solution Explanation and Complexity Analysis

**The Stack Approach**offers an intuitive solution. We iterate through each character of the strings, pushing characters onto a stack unless the character is a '#', in which case we pop the last character off the stack. The time complexity is O(N+M) and the space complexity is also O(N+M), where N and M are the lengths of the strings s and t, respectively.**The Two-Pointer Approach**requires iterating through each string backwards, decrementing a pointer whenever a '#' is encountered and skipping the next character as needed. This method has a time complexity of O(N+M) but reduces the space complexity to O(1), as it doesn't require additional data structures.**The****Direct Comparison with In-Place Modification**requires modifying the strings in space and overwriting characters. Since it doesn't use ay additional data structures the space complexity is O(1). The time complexity is O(N+M) because we must examine every character in both strings.

## The Stack Solution in Python

```
def backspaceCompare(s: str, t: str) -> bool:
def buildString(input_str):
stack = []
for char in input_str:
if char != '#':
stack.append(char)
elif stack:
stack.pop()
return "".join(stack)
# Compare the processed strings
return buildString(s) == buildString(t)
```

## The **Direct Comparison with In-Place Modification** in Python

This solution demonstrates the in-place modification strategy. We process the backspaces directly within the input strings, adjusting their effective length, and then compare the results.

```
def backspaceCompare(s: str, t: str) -> bool:
def backspace_process(input_str):
k = 0
for char in input_str:
if char != '#':
input_str[k] = char
k += 1
else:
k = max(k-1, 0)
return k
s, t = list(s), list(t)
s_length = backspace_process(s)
t_length = backspace_process(t)
if s_length != t_length:
return False
for i in range(s_length):
if s[i] != t[i]:
return False
return True
```

## Conclusion:

Solving the "Backspace String Compare" problem efficiently requires understanding the underlying principles of stacks and two-pointer techniques. Both methods have their merits, with the stack approach being more intuitive and the two-pointer method being more space-efficient.

Regardless of the approach, the essence of solving such problems lies in recognizing the applicable patterns and data structures. This problem is a great example of how understanding basic concepts can be applied to seemingly complex challenges, making it an excellent practice for engineers preparing for interviews.

I hope this comprehensive guide has illuminated the path to mastering this intriguing LeetCode problem. Happy coding, and may you approach your next software engineering interview with confidence!