Backspace String Compare

The problem is about comparing two strings s and t, considering that the character ‘#’ acts as a backspace. We want to determine if the two strings are equal after applying the backspaces.

Here’s a step-by-step guide to solving this problem:

1. Create a Function to Apply Backspaces: To handle backspaces, we’ll create a function that takes a string as input and returns a new string after applying the backspaces. This function will iterate through the string and use a stack to handle the characters and backspaces.

2. Apply Backspaces to Both Strings: We’ll use the function created in step 1 to get the final strings for both s and t.

3. Compare the Final Strings: Finally, we’ll compare the two strings obtained after applying the backspaces and return True if they are equal, otherwise False.

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class Solution:
    def backspaceCompare(self, s: str, t: str) -> bool:
        # Function to apply backspaces to a string
        def process_string(s: str) -> str:
            stack = []
            for char in s:
                if char != '#':
                    stack.append(char)
                elif stack:
                    stack.pop()
            return ''.join(stack)

        # Apply backspaces to both strings
        final_s = process_string(s)
        final_t = process_string(t)

        # Compare the final strings
        return final_s == final_t

Example:

• Input: s = "ab#c", t = "ad#c"
• Output: true
• Explanation: After applying backspaces, both s and t become "ac", so the function returns true.

Insights:

• Time Complexity: The code runs in (O(m + n)) time, where (m) and (n) are the lengths of the strings s and t, respectively.
• Space Complexity: The space complexity is also (O(m + n)) due to the use of stacks.
• Use of Stack: Utilizing a stack simplifies the handling of backspaces, as we can easily add characters to the stack and remove them when a backspace is encountered.
• Separation of Concerns: Separating the logic to handle backspaces into a separate function improves the code’s readability and maintainability.

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# @param {String} s
# @param {String} t
# @return {Boolean}

def final_string(s)
  output = []
  s.chars.each do |c|
    if c == '#'
      output.pop
    else
      output << c
    end
  end
  output.join
end

def backspace_compare(s, t)
  first = final_string(s)
  second = final_string(t)
  
  first == second    
end