algoadvance

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 to be 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".

Clarifying Questions

1. Characters Consideration: Are we assuming that the strings only consist of lowercase letters and the '#' character?
   • Yes, the strings will only consist of lowercase letters and the '#' character.
2. String Length: Is there any constraint on the length of the strings s and t?
   • The problem does not specify specific constraints, but typical LeetCode constraints apply where strings can be up to a reasonable limit (e.g., 200 characters).

Strategy

To solve this problem, we need to simulate the typing of the given strings considering the backspace (i.e., '#' character).

Here’s the strategy we’ll use:

1. We will process each string to effectively handle the backspaces, resulting in a final string that is equivalent to the text editor’s state.
2. We can create a helper function build which constructs the final string after processing the backspaces.
3. Compare the results for both strings.

Code

def backspaceCompare(s: str, t: str) -> bool:
    def build(string):
        result_stack = []
        for char in string:
            if char != '#':
                result_stack.append(char)
            elif result_stack:
                result_stack.pop()
        return ''.join(result_stack)
    
    return build(s) == build(t)

# Test Cases
print(backspaceCompare("ab#c", "ad#c"))  # Should print True
print(backspaceCompare("ab##", "c#d#"))  # Should print True
print(backspaceCompare("a#c", "b"))      # Should print False

Time Complexity

• The build function iterates through the given string once, processing each character.
• Thus, each call to build is O(n), where n is the length of the string.
• Since we call build twice (once for each string), the overall time complexity is O(n + m), where n is the length of s and m is the length of t.

Explanation

1. Helper Function build(string):
   • Uses a stack to build the resultant string.
   • For each character in the string:
     • Append the character to the stack if it is not a backspace ('#').
     • If it is a backspace and the stack is not empty, pop the last character from the stack.
   • Finally, join the stack to form the resultant processed string.
2. Comparison:
   • Use the helper function to process both strings s and t.
   • Compare the results of both processed strings directly. If they are equal, the function returns true; otherwise, it returns false.

This solution effectively captures the key requirement of handling backspace characters and efficiently compares the resultant strings.

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