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.
Follow up:
Can you solve it in O(N) time and O(1) space?
class Solution:
    


    def backspaceCompare(self, S: str, T: str) -> bool:
        
        def helper(S):
            r = ''
            erase = 0
            for s in S[-1::-1]:
                if s == '#':
                    erase+=1
                elif s !='#' and erase == 0:
                    r=s+r
                elif s !='#' and erase >=1:
                    #r=s+r
                    erase-=1
                
                    
            return r
    
        return helper(S) == helper(T)
        
        

class Solution:
    


    def backspaceCompare(self, S: str, T: str) -> bool:
        
        def helper(S):
            r = ''
            erase = 0
            for s in S[-1::-1]:
                if s == '#':
                    erase+=1
                elif s !='#' and erase == 0:
                    r=s+r
                elif s !='#' and erase >=1:
                    #r=s+r
                    erase-=1
                
                    
            return r
    
        return helper(S) == helper(T)
        
        
