# Backspace String Compare

Clarify the problem:

- The problem requires comparing two strings after applying backspace operations.
- We need to implement a function that takes two strings as input and returns True if the strings are equal after applying backspace operations, and False otherwise.

Analyze the problem:

- Input: Two strings
`s`

and`t`

. - Output: True if
`s`

and`t`

are equal after applying backspace operations, False otherwise. - Constraints:
- The strings can contain lowercase English letters and '#' characters.
- The '#' character represents a backspace.

- Input: Two strings
Design an algorithm:

- We can solve this problem using a two-pointer approach.
- Initialize two pointers,
`p1`

and`p2`

, to the end of strings`s`

and`t`

, respectively. - While
`p1`

or`p2`

is greater than or equal to 0:- Skip any backspace characters by finding the next valid character or reaching the beginning of the string.
- If
`p1`

or`p2`

is out of bounds, return False. - If the characters at positions
`p1`

and`p2`

are not equal, return False. - Move
`p1`

and`p2`

one position to the left.

- If both
`p1`

and`p2`

are out of bounds, return True.

Explain your approach:

- We will implement a function called
`backspaceCompare`

that takes two strings,`s`

and`t`

, as input. - We will use two pointers,
`p1`

and`p2`

, to track the positions in`s`

and`t`

, respectively. - We will iterate from the end of the strings towards the beginning, following the two-pointer approach.
- While iterating, we will skip any backspace characters and compare the characters at the current positions.
- If at any point the characters are not equal, we will return False.
- Finally, if both pointers are out of bounds, we will return True.

- We will implement a function called
Write clean and readable code:

python`def backspaceCompare(s, t): p1 = len(s) - 1 p2 = len(t) - 1 while p1 >= 0 or p2 >= 0: count_backspace_s = 0 while p1 >= 0 and (s[p1] == '#' or count_backspace_s > 0): if s[p1] == '#': count_backspace_s += 1 else: count_backspace_s -= 1 p1 -= 1 count_backspace_t = 0 while p2 >= 0 and (t[p2] == '#' or count_backspace_t > 0): if t[p2] == '#': count_backspace_t += 1 else: count_backspace_t -= 1 p2 -= 1 if p1 >= 0 and p2 >= 0 and s[p1] != t[p2]: return False p1 -= 1 p2 -= 1 return p1 < 0 and p2 < 0`

Test your code:

python`# Test case 1 # s = "ab#c", t = "ad#c" # After applying backspace, both strings become "ac". # Return True assert backspaceCompare("ab#c", "ad#c") == True # Test case 2 # s = "ab##", t = "c#d#" # After applying backspace, both strings become "". # Return True assert backspaceCompare("ab##", "c#d#") == True # Test case 3 # s = "a##c", t = "#a#c" # After applying backspace, both strings become "c". # Return True assert backspaceCompare("a##c", "#a#c") == True # Test case 4 # s = "a#c", t = "b" # After applying backspace, s becomes "c" and t becomes "b". # Return False assert backspaceCompare("a#c", "b") == False`

Optimize if necessary:

- The current solution has a time complexity of O(m + n), where m and n are the lengths of strings s and t, respectively.
- We are iterating through both strings once, skipping the backspace characters.
- This solution is already optimal in terms of time complexity.

Handle error cases:

- The code handles the case when the strings s and t are empty or contain only backspace characters.
- In such cases, both pointers will be out of bounds, and the function will return True.

Discuss complexity analysis:

- The time complexity of the solution is O(m + n), where m and n are the lengths of strings s and t, respectively.
- The space complexity is O(1) since we are not using any additional data structures.
- The best-case scenario occurs when the lengths of both strings are zero or the strings are equal after applying backspace. In this case, the function returns True in constant time.
- The worst-case scenario occurs when the lengths of both strings are large and we need to iterate through the entire strings. In this case, the function takes O(m + n) time.
- The average-case time complexity is also O(m + n), assuming random inputs.