Introduction

In this blog post, we will explore how to solve the LeetCode problem "2030" 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, an integer k, a letter letter, and an integer repetition. Return the lexicographically smallest subsequence of s of length k that has the letter letter appear at least repetition times. The test cases are generated so that the letter appears in s at least repetition times. A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters. A string a is lexicographically smaller than a string b if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b. Example 1: Input: s = "leet", k = 3, letter = "e", repetition = 1 Output: "eet" Explanation: There are four subsequences of length 3 that have the letter 'e' appear at least 1 time: - "lee" (from "leet") - "let" (from "leet") - "let" (from "leet") - "eet" (from "leet") The lexicographically smallest subsequence among them is "eet". Example 2: Input: s = "leetcode", k = 4, letter = "e", repetition = 2 Output: "ecde" Explanation: "ecde" is the lexicographically smallest subsequence of length 4 that has the letter "e" appear at least 2 times. Example 3: Input: s = "bb", k = 2, letter = "b", repetition = 2 Output: "bb" Explanation: "bb" is the only subsequence of length 2 that has the letter "b" appear at least 2 times. Constraints: 1 <= repetition <= k <= s.length <= 5 * 104 s consists of lowercase English letters. letter is a lowercase English letter, and appears in s at least repetition times.

Explanation

Here's an efficient solution to find the lexicographically smallest subsequence with the specified constraints:

• Greedy Construction: Build the subsequence greedily, character by character. At each step, consider whether including the current character from s will lead to a lexicographically smaller subsequence while still satisfying the length k and repetition requirements for letter.

• Backtracking Prevention: If including a character would violate the constraints (length or repetition), skip it. This avoids generating all possible subsequences, which would be inefficient.

• Constraint Tracking: Keep track of the remaining length to build (k) and the remaining number of letter occurrences needed (repetition). Also, maintain information on how many letter characters are still available in the unprocessed part of s.

• Runtime & Storage Complexity: O(n), where n is the length of the input string s. Storage complexity is O(k).

Code

    def lexicographically_smallest_subsequence(s: str, k: int, letter: str, repetition: int) -> str:
    """
    Finds the lexicographically smallest subsequence of s of length k that has the letter appear at least repetition times.

    Args:
        s: The input string.
        k: The desired length of the subsequence.
        letter: The letter that must appear at least repetition times.
        repetition: The minimum number of times the letter must appear.

    Returns:
        The lexicographically smallest subsequence.
    """

    n = len(s)
    result = ""
    letter_count = 0
    for char in s:
        if char == letter:
            letter_count += 1

    for i in range(n):
        char = s[i]
        # Check the remaining length of s from the current index
        remaining_s_length = n - i - 1
        # Check if we can pop from the last character
        while result and result[-1] > char and k > len(result) and (
                result[-1] != letter or letter_count > repetition
        ) and len(result) + remaining_s_length >= k:
            popped_char = result[-1]
            result = result[:-1]
            if popped_char == letter:
                repetition += 1
            k -= 1

        # Add to the result when possible
        if len(result) < k:
            if char == letter:
                result += char
                repetition -= 1
            elif k - len(result) > repetition:
                result += char

        # Update letter_count by iterating
        if char == letter:
            letter_count -= 1

    return result