Is Sort Vowels by Frequency easy or hard?

The problem is typically rated Medium because it combines multiple concepts: identifying vowels, counting frequencies, sorting or bucket grouping, and reconstructing the string. The individual operations are simple, but combining them efficiently requires careful implementation.

Sort Vowels by Frequency Python/Java solution

Implement the solution by scanning the string, storing vowel counts in a dictionary (Python) or HashMap (Java), sorting vowels by frequency, and rebuilding the result. Replace vowels during a second pass through the string to maintain consonant positions.

How to solve Sort Vowels by Frequency in O(n)?

Use a bucket sort strategy. First count vowel frequencies with a hash map, then place vowels into buckets indexed by their frequency. Traverse buckets from highest frequency to lowest to build the sorted vowel list, then replace vowels during a second pass of the string. This avoids comparison sorting and runs in linear time.

What is the best approach for Sort Vowels by Frequency?

The most practical approach uses a hash map to count vowel frequencies and then sorts the vowels based on their counts. After sorting, rebuild the vowel sequence and place it back into the original string positions. This solution runs in O(n log k) time where k is the number of distinct vowels and uses O(k) extra space.

Is Sort Vowels by Frequency asked at Google/Amazon/Meta?

Frequency counting and character reordering problems appear frequently in interviews at companies like Amazon, Google, and Meta. While the exact problem name may vary, interviewers often test string manipulation combined with hash maps and sorting patterns.

What data structure is used in Sort Vowels by Frequency?

The main data structure is a hash map for counting vowel frequencies. Depending on the approach, the solution may also use a sorting array, a bucket array for frequency grouping, or a priority queue (max heap) to retrieve vowels in descending frequency order.

What is the time complexity of Sort Vowels by Frequency?

The common solution runs in O(n log k) time due to sorting the distinct vowels after counting frequencies. Counting vowels requires O(n) time and sorting at most k vowel types adds the log factor. A bucket sort variant can reduce this to O(n) time when frequencies are bounded by the string length.

3913. Sort Vowels by Frequency - Practice Online

You are given a string s consisting of lowercase English characters.

Rearrange only the vowels in the string so that they appear in non-increasing order of their frequency.

If multiple vowels have the same frequency, order them by the position of their first occurrence in s.

Return the modified string.

Vowels are 'a', 'e', 'i', 'o', and 'u'.

The frequency of a letter is the number of times it occurs in the string.

Example 1:

Input: s = "leetcode"

Output: "leetcedo"

Explanation:

Vowels in the string are ['e', 'e', 'o', 'e'] with frequencies: e = 3, o = 1.
Sorting in non-increasing order of frequency and placing them back into the vowel positions results in "leetcedo".

Example 2:

Input: s = "aeiaaioooa"

Output: "aaaaoooiie"

Explanation:

Vowels in the string are ['a', 'e', 'i', 'a', 'a', 'i', 'o', 'o', 'o', 'a'] with frequencies: a = 4, o = 3, i = 2, e = 1.
Sorting them in non-increasing order of frequency and placing them back into the vowel positions results in "aaaaoooiie".

Example 3:

Input: s = "baeiou"

Output: "baeiou"

Explanation:

Each vowel appears exactly once, so all have the same frequency.
Thus, they retain their relative order based on first occurrence, and the string remains unchanged.

Constraints:

1 <= s.length <= 10⁵
s consists of lowercase English letters