## 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.

**Example 1:**

Input:S = "ab#c", T = "ad#c"Output:trueExplanation: Both S and T become "ac".

**Example 2:**

Input:S = "ab##", T = "c#d#"Output:trueExplanation: Both S and T become "".

**Example 3:**

Input:S = "a##c", T = "#a#c"Output:trueExplanation: Both S and T become "c".

**Example 4:**

Input:S = "a#c", T = "b"Output:falseExplanation: 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.

**Follow up:**

- Can you solve it in
`O(N)`

time and`O(1)`

space?

## Explanation

Create a helper function to compare if the backspace results from two strings are matched.

## 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: O(n).
- Space complexity: O(n).

## One Thought to “LeetCode 844. Backspace String Compare”