What is the brute force approach and complexity for Reverse Vowels of a String?

Copy vowels to a list, then refill the vowel slots from the end of the list with time complexity O(N) and space complexity O(N).

What is the optimal approach and complexity for Reverse Vowels of a String?

Each side skips non-vowels; swap when both stop on a vowel with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Reverse Vowels of a String?

Look for key signals in the prompt: "reverse" indicates Reversing = swap first with last, then move inward → opposite ends.; "only the vowels" indicates Each pointer must skip the characters that don't count, on its own.; "everything else stays where it is" indicates Skipped characters never need another look..

Reverse Vowels of a String

Easy·LeetCode #345 Pro OnlySlight Variation

LeetCode

Problem Statement

You are given a string s.

Reverse the order of only the vowels in it, and leave every other character exactly where it is. Return the new string.

The vowels are a, e, i, o, u, in lowercase or uppercase.

Input

A string s of printable ASCII characters.

Output

The string with its vowels in reverse order and all other characters unchanged.

Example 1

Input: s = "IceCreAm"

Output: "AceCreIm"

The vowels, in order, are I, e, e, A. Reversed they are A, e, e, I. Put them back into the vowel slots: "AceCreIm".

Example 2

Input: s = "leetcode"

Output: "leotcede"

Vowels e, e, o, e become e, o, e, e.

Constraints

• 1 ≤ s.length ≤ 3 · 10^5
• s consists of printable ASCII characters.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“reverse”

→ Reversing = swap first with last, then move inward → opposite ends.

“only the vowels”

→ Each pointer must skip the characters that don't count, on its own.

“everything else stays where it is”

→ Skipped characters never need another look.

Approach Complexity Ladder

Brute Force — collect the vowels, write them back reversedO(N)O(N)

Copy vowels to a list, then refill the vowel slots from the end of the list

Optimal — two pointers that skip consonantsO(N)O(1)

Each side skips non-vowels; swap when both stop on a vowel

Brute Force Intuition

Walk through s and copy every vowel into a separate string. Then walk through s again. Every time you hit a vowel slot, write in the next vowel from the end of the list.

Recognition rule for this sub-pattern:

Start at both ends of a sorted (or symmetric) structure. Each step, move the side that cannot do better.

Remember this · Two Pointers

Trigger

Sorted array, pairs/triplets with a target, comparing from both ends, compacting an array in place.

Think

If I compare the elements at two positions, can I safely discard one side?

Key question

Which pointer moves, and why can the skipped elements never be the answer?

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