# Solving Leetcode Interviews in Seconds with AI: Backspace String Compare


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "844" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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? 

	# Explanation
	Here's the breakdown of the approach, complexity, and the Python code:

*   **High-Level Approach:**
    *   Use two pointers, one for each string, starting from the end.
    *   Iterate backwards, skipping backspaced characters. A backspace count is maintained for each string.
    *   Compare the characters at the current pointers; if they don't match and both strings still have valid characters, the strings are not equal.

*   **Complexity:**
    *   Runtime Complexity: O(m + n), where m and n are the lengths of strings s and t, respectively.
    *   Storage Complexity: O(1)

	
	# Code
	```python
	def backspace_compare(s: str, t: str) -> bool:
    """
    Compares two strings after applying backspace operations.

    Args:
        s: The first string.
        t: The second string.

    Returns:
        True if the strings are equal after backspaces, False otherwise.
    """
    i = len(s) - 1
    j = len(t) - 1
    skip_s = 0
    skip_t = 0

    while i >= 0 or j >= 0:
        while i >= 0:
            if s[i] == '#':
                skip_s += 1
                i -= 1
            elif skip_s > 0:
                skip_s -= 1
                i -= 1
            else:
                break

        while j >= 0:
            if t[j] == '#':
                skip_t += 1
                j -= 1
            elif skip_t > 0:
                skip_t -= 1
                j -= 1
            else:
                break

        if i >= 0 and j >= 0:
            if s[i] != t[j]:
                return False
        else:
            if i >= 0 or j >= 0:
                return False

        i -= 1
        j -= 1

    return True
	```