Introduction

In this blog post, we will explore how to solve the LeetCode problem "2785" 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

Given a 0-indexed string s, permute s to get a new string t such that: All consonants remain in their original places. More formally, if there is an index i with 0 <= i < s.length such that s[i] is a consonant, then t[i] = s[i]. The vowels must be sorted in the nondecreasing order of their ASCII values. More formally, for pairs of indices i, j with 0 <= i < j < s.length such that s[i] and s[j] are vowels, then t[i] must not have a higher ASCII value than t[j]. Return the resulting string. The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in lowercase or uppercase. Consonants comprise all letters that are not vowels. Example 1: Input: s = "lEetcOde" Output: "lEOtcede" Explanation: 'E', 'O', and 'e' are the vowels in s; 'l', 't', 'c', and 'd' are all consonants. The vowels are sorted according to their ASCII values, and the consonants remain in the same places. Example 2: Input: s = "lYmpH" Output: "lYmpH" Explanation: There are no vowels in s (all characters in s are consonants), so we return "lYmpH". Constraints: 1 <= s.length <= 105 s consists only of letters of the English alphabet in uppercase and lowercase.

Explanation

Here's a breakdown of the solution:

• Identify and Separate: Iterate through the input string, separating vowels and consonants. Store the vowels in a separate list.
• Sort Vowels: Sort the extracted vowels list in ascending order based on their ASCII values.

• Reconstruct String: Rebuild the string by iterating through the original string again. If a character is a consonant, place it directly in the new string. If it's a vowel, take the next sorted vowel and place it in the new string.

• Runtime Complexity: O(n log n) due to the sorting of vowels, where n is the length of the string.

• Storage Complexity: O(n) because, in the worst case, we could have all vowels in the string which we store.

Code

    def sort_vowels(s: str) -> str:
    """
    Given a 0-indexed string s, permute s to get a new string t such that:

    All consonants remain in their original places. More formally, if there is an index i with 0 <= i < s.length such that s[i] is a consonant, then t[i] = s[i].
    The vowels must be sorted in the nondecreasing order of their ASCII values. More formally, for pairs of indices i, j with 0 <= i < j < s.length such that s[i] and s[j] are vowels, then t[i] must not have a higher ASCII value than t[j].

    Return the resulting string.

    The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in lowercase or uppercase.
    Consonants comprise all letters that are not vowels.

    Example 1:
    Input: s = "lEetcOde"
    Output: "lEOtcede"
    Explanation: 'E', 'O', and 'e' are the vowels in s; 'l', 't', 'c', and 'd' are all consonants. The vowels are sorted according to their ASCII values, and the consonants remain in the same places.

    Example 2:
    Input: s = "lYmpH"
    Output: "lYmpH"
    Explanation: There are no vowels in s (all characters in s are consonants), so we return "lYmpH".

    Constraints:
    1 <= s.length <= 105
    s consists only of letters of the English alphabet in uppercase and lowercase.
    """

    vowels = []
    for char in s:
        if char in "aeiouAEIOU":
            vowels.append(char)

    vowels.sort()

    result = ""
    vowel_index = 0
    for char in s:
        if char in "aeiouAEIOU":
            result += vowels[vowel_index]
            vowel_index += 1
        else:
            result += char

    return result