844-backspace-string-compare

Try it on leetcode

Description

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 = "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.

 

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

Solution(Python)

class Solution:
    def backspaceCompare(self, s: str, t: str) -> bool:
        return self.twopointers(s, t)

    # Time Complexity: O(n+m)
    # Space Complexity: O(n+m)
    def stack(self, s, t):
        def build(string):
            ans = []
            for c in string:
                if c == "#":
                    ans.pop()
                else:
                    ans.append(c)
            return "".join(ans)

        return build(s) == build(t)

    # Time Complexity: O(n+m)
    # Space Complexity: O(1)
    def twopointers(self, s, t):
        def build(string):
            skip = 0
            for x in reversed(string):
                if x == "#":
                    skip += 1
                elif skip:
                    skip -= 1
                else:
                    yield x

        return all(x == y for x, y in zip_longest(build(s), build(t)))