Backspace String Compare

Problem Statement

The problem involves two strings, s and t, both of which simulate textual input into a text editor that supports the backspace character. Our task is to determine if, after fully processing both strings with all backspace characters taking effect, the resulting output from both strings is the same.

A backspace character, represented here as '#', results in the deletion of the character immediately preceding it, if there is one. If there is no character to delete (i.e., the string is empty or there have been multiple backspaces in succession), the backspace has no effect.

The challenge is to process these strings according to the rules of typing and backspacing in a text editor and to then compare the final forms of both strings to see if they are identical.

Examples

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.

Approach and Intuition

Based on the problem and examples provided, constructing the resulting string from the given input string while respecting the deletion caused by '#' characters is crucial. The core intuition revolves around simulating this typing and backspacing process effectively. Here’s a basic approach to handle this:

1. Initialize Stack for Each String: Use a stack data structure to simulate the effect of typing and backspacing on both strings s and t. Stacks are ideal for this as they easily allow us to add and remove characters at the end of the string.

2. Process Each Character of Both Strings:

   • For every character in the string:
     • If the character is not '#', push it onto its respective stack.
     • If the character is '#' and the stack is not empty, pop the top character of the stack. This simulates the backspace operation.

3. Convert Stack to String: Once all characters have been processed through the stack, convert the stack back to a string for both s and t.

4. Compare the Two Resulting Strings: If the strings obtained from both stacks are identical, return true as they represent the same text after all backspaces are taken into account. Otherwise, return false.

By leveraging a stack to process each string, we ensure that we manage the backspace operations efficiently, effectively modeling the behavior of a text editor. This approach is straightforward and ensures that our solution respects the constraints provided, specifically the bounds on the length of s and t and their character content. And this simulative approach fundamentally ensures that each character and backspace is processed as it would be in a real-time typing scenario.

Solutions

class Solution {
    public boolean compareStrings(String str1, String str2) {
        int index1 = str1.length() - 1, index2 = str2.length() - 1;
        int backCount1 = 0, backCount2 = 0;
    
        while (index1 >= 0 || index2 >= 0) {
            while (index1 >= 0) {
                if (str1.charAt(index1) == '#') {backCount1++; index1--;}
                else if (backCount1 > 0) {backCount1--; index1--;}
                else break;
            }
            while (index2 >= 0) {
                if (str2.charAt(index2) == '#') {backCount2++; index2--;}
                else if (backCount2 > 0) {backCount2--; index2--;}
                else break;
            }
            if (index1 >= 0 && index2 >= 0 && str1.charAt(index1) != str2.charAt(index2))
                return false;
            if ((index1 >= 0) != (index2 >= 0))
                return false;
            index1--; index2--;
        }
        return true;
    }
}