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.

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.

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.