How do you approach Reverse Vowels of a String?

You only need to touch the vowels - everything else stays put.

What is the time complexity of Reverse Vowels of a String?

Time complexity: O(n) - each character is examined at most once. Space complexity: O(n) - the character array (strings are immutable).

What are common mistakes on Reverse Vowels of a String?

Forgetting the input has to be sorted. If it is not, the pointer moves mean nothing and the answer only looks right on the sample. Moving both pointers on a match when only one should move, which steps over valid pairs. Choosing left < right where left <= right is meant. The first never examines the middle element.

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Reverse Vowels of a String

EasyAlgorithms

Description

Given a string `s`, reverse only all the **vowels** in the string and return it. The vowels are `'a'`, `'e'`, `'i'`, `'o'`, and `'u'`, and they can appear in both lower and upper cases. Every other character stays in its original position.

Example 1

Input: s = "hello"

Output: "holle"

Explanation: The vowels 'e' and 'o' swap positions.

Example 2

Input: s = "leetcode"

Output: "leotcede"

Example 3

Input: s = "aA"

Output: "Aa"

Explanation: Uppercase vowels count too.

Constraints

- 1 <= s.length <= 3 * 10^5 - s consists of printable ASCII characters.

Reverse Vowels of a String: approach, complexity and pitfalls

An easy two pointers problem, graded against 7 test cases (4 of them hidden).

The pattern: Two Pointers

Two indices walking a sorted array or string, usually from opposite ends.

Reach for it when you see: A sorted input, a palindrome check, or a pair/triplet that must satisfy a condition.

More Two Pointersproblems →

If you are stuck

Show the 3-step hint ladder

You only need to touch the vowels - everything else stays put.
Use two pointers from both ends; advance each one until it sits on a vowel.
Swap the two vowels, move both pointers inward, and repeat until the pointers cross.

The optimal approach

Show the approach: Two Pointers with Skipping

Convert the string to a character array (strings are immutable in both JS and Python). Walk a left pointer forward and a right pointer backward; each pointer skips non-vowels. When both land on vowels, swap them and move both inward. Join the array back into a string at the end.

Time: O(n) - each character is examined at most once
Space: O(n) - the character array (strings are immutable)

The full reference solution in every supported language stays in the editor above - reveal it there once you have had a real attempt.

Common Two Pointers mistakes

These apply to the pattern as a whole, not just this problem.

Forgetting the input has to be sorted. If it is not, the pointer moves mean nothing and the answer only looks right on the sample.
Moving both pointers on a match when only one should move, which steps over valid pairs.
Choosing `left < right` where `left <= right` is meant. The first never examines the middle element.