Скачать или смотреть Leetcode 844: Backspace String Compare - Java Solution with visualisation

Explanation: Backspace String Compare

Problem:
Given two strings `s` and `t`, both strings can be processed simultaneously. Each string may contain the character `'#'` which means a backspace character. When a backspace is encountered, the last character in the string is deleted. Return `true` if the strings are the same after backspace operations, otherwise return `false`.

Solution Approach:

1. Process Each String:
Iterate through each character in the string.
If the character is not a backspace (`'#'`), push it onto a stack.
If the character is a backspace:
Pop the top element from the stack (if the stack is not empty) to simulate backspacing.
2. Construct the Final Strings:
Pop characters from the stack and append them to a `StringBuilder` to form the processed string.
3. Compare the Processed Strings:
Compare the two processed strings. If they are equal, return `true`; otherwise, return `false`.

Time and Space Complexity:

Time Complexity: O(m + n), where m and n are the lengths of strings `s` and `t`, respectively.
Space Complexity: O(m + n) in the worst case, when all characters are pushed onto the stack.

Key Points:

The use of a stack effectively simulates the backspace operation.
By processing both strings and comparing the resulting strings, we determine if they are equal after backspace operations.

This approach provides a clear and efficient solution to the backspace string comparison problem.

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