Backspace String Compare - Leetcode Solution

The Backspace String Compare problem on LeetCode is a fairly simple problem that requires string manipulation. The problem statement is as follows:

Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.

For example, S = "ab#c", and T = "ad#c" would return true because both become "ac". Similarly, S = "a#c" and T = "b" would also return true because both become "" (empty strings).

One way to solve this problem is to use two stacks to represent the two strings S and T. We iterate through each string character by character, and for each character in the string, we push it onto the stack unless it is a backspace character, in which case we pop the top element from the stack. We repeat this process for both strings, and at the end, we compare the elements in the two stacks.

Here is the detailed algorithm:

1. Create two stacks, one for S and one for T.
2. For each character in S, perform the following steps: a. If the character is a backspace ("#"), pop the top element from the S stack. b. Otherwise, push the character onto the S stack.
3. Repeat step 2 for string T.
4. Compare the remaining elements in the S stack and T stack. a. If the stacks are equal, return true. b. Otherwise, return false.

Here is the Python code implementation of this solution:

class Solution:
    def backspaceCompare(self, S: str, T: str) -> bool:
        stack_s = []
        stack_t = []

        # process string S
        for char in S:
            if char == "#":
                if stack_s:
                    stack_s.pop()
            else:
                stack_s.append(char)

        # process string T
        for char in T:
            if char == "#":
                if stack_t:
                    stack_t.pop()
            else:
                stack_t.append(char)

        return stack_s == stack_t

In summary, the Backspace String Compare problem on LeetCode can be solved by using stacks to represent the two strings and performing character-by-character comparison. The time complexity of this solution is O(n) where n is the length of the input strings, and the space complexity is also O(n) due to the use of stacks.

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