Backspace String Compare

Backspace String Compare

What is this problem about?

The Backspace String Compare interview question gives you two strings containing lowercase letters and '#' characters, where '#' represents a backspace. You need to determine if the two strings are equal after the backspaces are processed. For example, "ab#c" becomes "ac". This Backspace String Compare coding problem is about string processing and stack-like behavior.

Why is this asked in interviews?

This is a very popular question at companies like Google, Meta, and Bloomberg. It tests whether you can solve the problem in O(N) time and O(1) space. While using a stack is easy (O(N) space), solving it with two pointers from the end of the strings is the "Gold Standard" for senior candidates.

Algorithmic pattern used

This follows the Two Pointers, String, Simulation, Stack interview pattern.

• Stack approach: Iterate forward, push chars to stack, pop on '#'.
• Two Pointers approach: Iterate backward. Maintain a skip counter for each string. If you see '#', increment skip. If you see a char and skip > 0, skip the char. Otherwise, compare the char with the other string's current char.

Example explanation

S = "ab#c", T = "ad#c"

1. S backward: 'c' (compare), then '#' (skip next), then 'a' (compare). Final: "ac".
2. T backward: 'c' (compare), then '#' (skip next), then 'a' (compare). Final: "ac". Result: True.

Common mistakes candidates make

• Forward Traversal with Pointers: Trying to use two pointers from the beginning. This is very difficult because you don't know if a character will be deleted until you see the '#' later.
• Handling Multiple Backspaces: Failing to handle cases like "a###b", where multiple backspaces might clear a string or do nothing if the string is already empty.
• Space Complexity: Settling for a O(N) space solution when the interviewer asks for O(1).

Interview preparation tip

Always try to process "deletions" or "backspaces" from the end of the string. This allows you to know immediately if a character should be ignored, avoiding the need for a temporary storage like a stack.