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.

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
"""
Sr=[]
i=0
while i<len(S):
if S[i]=='#':
if Sr:
Sr.pop()
else:
Sr.append(S[i])
i+=1 Tr=[]
i=0
while i<len(T):
if T[i]=='#':
if Tr:
Tr.pop()
else:
Tr.append(T[i])
i+=1 return Tr==Sr

  

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