Introduction

In this blog post, we will explore how to solve the LeetCode problem "1209" 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 and an integer k, a k duplicate removal consists of choosing k adjacent and equal letters from s and removing them, causing the left and the right side of the deleted substring to concatenate together. We repeatedly make k duplicate removals on s until we no longer can. Return the final string after all such duplicate removals have been made. It is guaranteed that the answer is unique. Example 1: Input: s = "abcd", k = 2 Output: "abcd" Explanation: There's nothing to delete. Example 2: Input: s = "deeedbbcccbdaa", k = 3 Output: "aa" Explanation: First delete "eee" and "ccc", get "ddbbbdaa" Then delete "bbb", get "dddaa" Finally delete "ddd", get "aa" Example 3: Input: s = "pbbcggttciiippooaais", k = 2 Output: "ps" Constraints: 1 <= s.length <= 105 2 <= k <= 104 s only contains lowercase English letters.

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• High-Level Approach:

  • Use a stack to keep track of characters and their counts.
  • Iterate through the string, updating the stack. If the current character matches the top of the stack, increment its count. Otherwise, push the new character with a count of 1.
  • If any character's count reaches k, pop it from the stack.

• Complexity:

  • Runtime Complexity: O(n), where n is the length of the string.
  • Storage Complexity: O(n) in the worst case (e.g., string with no duplicates).

Code

    def remove_dupliates(s: str, k: int) -> str:    """
    Removes k duplicate characters from a string until no more removals can be made.

    Args:
        s: The input string.
        k: The number of adjacent duplicate characters to remove.

    Returns:
        The final string after all duplicate removals.
    """
    stack = []  # Stack of (char, count) tuples

    for char in s:
        if stack and stack[-1][0] == char:
            stack[-1] = (char, stack[-1][1] + 1)
            if stack[-1][1] == k:
                stack.pop()
        else:
            stack.append((char, 1))

    result = ""
    for char, count in stack:
        result += char * count

    return result