# 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.

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. `1 <= S.length <= 200`
2. `1 <= T.length <= 200`
3. `S` and `T` only contain lowercase letters and `'#'` characters.

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

## Explanation

generate two new strings and then compare

## Python Solution

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

string_s = []
for i in range(0, len(S)):
char = S[i]
print (char)
if char != '#':
string_s.append(char)
else:
if (len(string_s) > 0) :
string_s.pop()

string_t = []
for i in range(0, len(T)):
char = T[i]
if char != '#':
string_t.append(char)
else:
if (len(string_t) > 0) :
string_t.pop()

return string_s == string_t``````
• Time complexity: O(n).
• Space complexity: O(n).