844. Backspace String Compare

Easy

---

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 = "#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".

Constraints:

• 1 <= s.length, t.length <= 200

• s and t only contain lowercase letters and '#' characters.

Solution

class Solution:
    def backspaceCompare(self, s: str, t: str) -> bool:
        str1 = []
        str2 = []
        for char in list(s):
            if char == '#':
                if str1:
                    str1.pop()
            else:
                str1.append(char)
        for char in list(t):
            if char == '#':
                if str2:
                    str2.pop()
            else:
                str2.append(char)
        print(str1, str2)
        return str1 == str2

Follow up: Can you solve it in O(n) time and O(1) space?