Build confidence with an intuition-first walkthrough focused on two pointers fundamentals.
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, more than once.
Example 1:
Input: s = "IceCreAm"
Output: "AceCreIm"
Explanation:
The vowels in s are ['I', 'e', 'e', 'A']. On reversing the vowels, s becomes "AceCreIm".
Example 2:
Input: s = "leetcode"
Output: "leotcede"
Constraints:
1 <= s.length <= 3 * 105s consist of printable ASCII characters.Problem summary: 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, more than once.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers
"IceCreAm"
"leetcode"
reverse-string)remove-vowels-from-a-string)faulty-keyboard)sort-vowels-in-a-string)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #345: Reverse Vowels of a String
class Solution {
public String reverseVowels(String s) {
boolean[] vowels = new boolean[128];
for (char c : "aeiouAEIOU".toCharArray()) {
vowels[c] = true;
}
char[] cs = s.toCharArray();
int i = 0, j = cs.length - 1;
while (i < j) {
while (i < j && !vowels[cs[i]]) {
++i;
}
while (i < j && !vowels[cs[j]]) {
--j;
}
if (i < j) {
char t = cs[i];
cs[i] = cs[j];
cs[j] = t;
++i;
--j;
}
}
return String.valueOf(cs);
}
}
// Accepted solution for LeetCode #345: Reverse Vowels of a String
func reverseVowels(s string) string {
vowels := [128]bool{}
for _, c := range "aeiouAEIOU" {
vowels[c] = true
}
cs := []byte(s)
i, j := 0, len(cs)-1
for i < j {
for i < j && !vowels[cs[i]] {
i++
}
for i < j && !vowels[cs[j]] {
j--
}
if i < j {
cs[i], cs[j] = cs[j], cs[i]
i, j = i+1, j-1
}
}
return string(cs)
}
# Accepted solution for LeetCode #345: Reverse Vowels of a String
class Solution:
def reverseVowels(self, s: str) -> str:
vowels = "aeiouAEIOU"
i, j = 0, len(s) - 1
cs = list(s)
while i < j:
while i < j and cs[i] not in vowels:
i += 1
while i < j and cs[j] not in vowels:
j -= 1
if i < j:
cs[i], cs[j] = cs[j], cs[i]
i, j = i + 1, j - 1
return "".join(cs)
// Accepted solution for LeetCode #345: Reverse Vowels of a String
impl Solution {
pub fn reverse_vowels(s: String) -> String {
let vowel = String::from("aeiouAEIOU");
let mut data: Vec<char> = s.chars().collect();
let (mut i, mut j) = (0, s.len() - 1);
while i < j {
while i < j && !vowel.contains(data[i]) {
i += 1;
}
while i < j && !vowel.contains(data[j]) {
j -= 1;
}
if i < j {
data.swap(i, j);
i += 1;
j -= 1;
}
}
data.iter().collect()
}
}
// Accepted solution for LeetCode #345: Reverse Vowels of a String
function reverseVowels(s: string): string {
const vowels = new Set(['a', 'e', 'i', 'o', 'u']);
const cs = s.split('');
for (let i = 0, j = cs.length - 1; i < j; ++i, --j) {
while (i < j && !vowels.has(cs[i].toLowerCase())) {
++i;
}
while (i < j && !vowels.has(cs[j].toLowerCase())) {
--j;
}
[cs[i], cs[j]] = [cs[j], cs[i]];
}
return cs.join('');
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.