Backspace String Compare.py

'''
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 <= 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?



'''

class Solution(object):
    def backspaceCompare(self, S, T):
        """
        :type S: str
        :type T: str
        :rtype: bool
        """
        s = []
        for c in S:
            if c == '#':
                if s:
                    s.pop()
            else:
                s.append(c)
        
        t = []
        for c in T:
            if c == '#':
                if t:
                    t.pop()
            else:
                t.append(c)
                
        return s == t