Introduction

In this blog post, we will explore how to solve the LeetCode problem "1047" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a string s consisting of lowercase English letters. A duplicate removal consists of choosing two adjacent and equal letters and removing them. We repeatedly make duplicate removals on s until we no longer can. Return the final string after all such duplicate removals have been made. It can be proven that the answer is unique. Example 1: Input: s = "abbaca" Output: "ca" Explanation: For example, in "abbaca" we could remove "bb" since the letters are adjacent and equal, and this is the only possible move. The result of this move is that the string is "aaca", of which only "aa" is possible, so the final string is "ca". Example 2: Input: s = "azxxzy" Output: "ay" Constraints: 1 <= s.length <= 105 s consists of lowercase English letters.

Explanation

Here's the breakdown of the solution:

• Stack-based approach: Use a stack to keep track of characters. If the current character matches the top of the stack, it means we have a duplicate, so we pop from the stack. Otherwise, we push the current character onto the stack.
• Efficient Removal: This approach avoids repeated string manipulations, making it more efficient. It processes the string in a single pass.

• Result Generation: Finally, the characters remaining in the stack form the result string, which needs to be reversed as the stack stores characters in reverse order.

• Runtime Complexity: O(n), where n is the length of the string.

• Storage Complexity: O(n) in the worst case, where the stack might store all the characters if there are no duplicates.

Code

    def remove_duplicates(s: str) -> str:
    """
    Removes adjacent duplicate characters from a string.

    Args:
        s: The input string consisting of lowercase English letters.

    Returns:
        The final string after all duplicate removals have been made.
    """
    stack = []
    for char in s:
        if stack and stack[-1] == char:
            stack.pop()
        else:
            stack.append(char)
    return "".join(stack)