# LeetCode 844. Backspace String Compare

## Description

https://leetcode.com/problems/backspace-string-compare/

Given two strings `S` and `T`, return 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 = "#a#c"
Output: true
Explanation: Both S and T become "c".
```

Example 4:

```Input: S = "a#c", T = "b"
Output: false
Explanation: S becomes "c" while T becomes "b".
```

Note:

• `1 <= S.length <= 200`
• `1 <= T.length <= 200`
• `S` and `T` only contain lowercase letters and `'#'` characters.

• Can you solve it in `O(N)` time and `O(1)` space?

## Explanation

use trie + backtracking

## Python Solution

``````class Solution:
def backspaceCompare(self, S: str, T: str) -> bool:

def helper(s):
result = []

for c in s:
if c != '#':
result.append(c)
elif result:
result.pop()

return "".join(result)

return helper(S) == helper(T)
``````
• Time Complexity: ~M + N
• Space Complexity: ~M + N

where M,N are the lengths of S and T respectively.