How do you solve Sort Vowels in a String on LeetCode?

Sort Vowels in a String (LeetCode #2785) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Vowels can be sorted independently of consonants, allowing for efficient rearrangement.

What is the time complexity of Sort Vowels in a String?

The optimal solution for Sort Vowels in a String runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to sorting the vowels, while the space complexity is O(n) for storing the vowels and the result.

What is the brute force solution for Sort Vowels in a String?

The brute force approach for Sort Vowels in a String is Brute Force, with O(n²) time complexity and O(n) space complexity. In this approach, we will extract all the vowels from the string, sort them, and then replace the vowels in the original string with the sorted vowels. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Sort Vowels in a String?

Sort Vowels in a String uses the following concepts: String, Sorting. The recommended patterns to study are: Sorting, Two Pointers.

What are the interview tips for Sort Vowels in a String?

Always clarify the problem requirements before jumping into coding. Think about edge cases, such as strings with no vowels. Explain your thought process clearly while coding to demonstrate your understanding.

What are common mistakes when solving Sort Vowels in a String?

Not handling both uppercase and lowercase vowels correctly. Forgetting to maintain the original positions of consonants.

#2785

Sort Vowels in a String

Medium

StringSortingSortingTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(n)	O(n)

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Intuition

Time O(n log n)Space O(n)

This approach utilizes a single pass to collect and sort the vowels while maintaining the positions of consonants. This is more efficient than the brute force method as it reduces unnecessary iterations.

⚙️

Algorithm

3 steps

1Step 1: Traverse the string and collect all vowels into a list.
2Step 2: Sort the list of vowels.
3Step 3: Create a new list to build the result string, replacing vowels from the sorted list while keeping consonants in their original positions.

solution.py9 lines

1def sortVowels(s):
2    vowels = sorted([c for c in s if c in 'aeiouAEIOU'])
3    result = list(s)
4    vowel_index = 0
5    for i in range(len(s)):
6        if s[i] in 'aeiouAEIOU':
7            result[i] = vowels[vowel_index]
8            vowel_index += 1
9    return ''.join(result)

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the vowels, while the space complexity is O(n) for storing the vowels and the result.

1Vowels can be sorted independently of consonants, allowing for efficient rearrangement.
2Maintaining the original positions of consonants is crucial for the problem's requirements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.