# Solving Leetcode Interviews in Seconds with AI: Smallest K-Length Subsequence With Occurrences of a Letter


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```