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874. Backspace String Compare

Two Pointers String Stack Simulation

Problem - 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 = "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?

Solutions

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class Solution:
    def backspaceCompare(self, s: str, t: str) -> bool:
        i, j, skip1, skip2 = len(s) - 1, len(t) - 1, 0, 0
        while i >= 0 or j >= 0:
            while i >= 0:
                if s[i] == "#":
                    skip1 += 1
                    i -= 1
                elif skip1:
                    skip1 -= 1
                    i -= 1
                else:
                    break

            while j >= 0:
                if t[j] == "#":
                    skip2 += 1
                    j -= 1
                elif skip2:
                    skip2 -= 1
                    j -= 1
                else:
                    break

            if i >= 0 and j >= 0:
                if s[i] != t[j]:
                    return False
            elif i >= 0 or j >= 0:
                return False
            i -= 1
            j -= 1
        return True

Submission Stats:

  • Runtime: 0 ms (100.00%)
  • Memory: 18 MB (15.01%)