LeetCode #2785 — MEDIUM

Sort Vowels in a String

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

Given a 0-indexed string s, permute s to get a new string t such that:

All consonants remain in their original places. More formally, if there is an index i with 0 <= i < s.length such that s[i] is a consonant, then t[i] = s[i].
The vowels must be sorted in the nondecreasing order of their ASCII values. More formally, for pairs of indices i, j with 0 <= i < j < s.length such that s[i] and s[j] are vowels, then t[i] must not have a higher ASCII value than t[j].

Return the resulting string.

The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in lowercase or uppercase. Consonants comprise all letters that are not vowels.

Example 1:

Input: s = "lEetcOde"
Output: "lEOtcede"
Explanation: 'E', 'O', and 'e' are the vowels in s; 'l', 't', 'c', and 'd' are all consonants. The vowels are sorted according to their ASCII values, and the consonants remain in the same places.

Example 2:

Input: s = "lYmpH"
Output: "lYmpH"
Explanation: There are no vowels in s (all characters in s are consonants), so we return "lYmpH".

Constraints:

1 <= s.length <= 10⁵
s consists only of letters of the English alphabet in uppercase and lowercase.

Step 01

Brute Force Baseline

Problem summary: Given a 0-indexed string s, permute s to get a new string t such that: All consonants remain in their original places. More formally, if there is an index i with 0 <= i < s.length such that s[i] is a consonant, then t[i] = s[i]. The vowels must be sorted in the nondecreasing order of their ASCII values. More formally, for pairs of indices i, j with 0 <= i < j < s.length such that s[i] and s[j] are vowels, then t[i] must not have a higher ASCII value than t[j]. Return the resulting string. The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in lowercase or uppercase. Consonants comprise all letters that are not vowels.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"lEetcOde"

Example 2

"lYmpH"

Core Insight

What unlocks the optimal approach

Add all the vowels in an array and sort the array.
Replace characters in string s if it's a vowel from the new array.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2785: Sort Vowels in a String
class Solution {
    public String sortVowels(String s) {
        List<Character> vs = new ArrayList<>();
        char[] cs = s.toCharArray();
        for (char c : cs) {
            char d = Character.toLowerCase(c);
            if (d == 'a' || d == 'e' || d == 'i' || d == 'o' || d == 'u') {
                vs.add(c);
            }
        }
        Collections.sort(vs);
        for (int i = 0, j = 0; i < cs.length; ++i) {
            char d = Character.toLowerCase(cs[i]);
            if (d == 'a' || d == 'e' || d == 'i' || d == 'o' || d == 'u') {
                cs[i] = vs.get(j++);
            }
        }
        return String.valueOf(cs);
    }
}

// Accepted solution for LeetCode #2785: Sort Vowels in a String
func sortVowels(s string) string {
	cs := []byte(s)
	vs := []byte{}
	for _, c := range cs {
		d := c | 32
		if d == 'a' || d == 'e' || d == 'i' || d == 'o' || d == 'u' {
			vs = append(vs, c)
		}
	}
	sort.Slice(vs, func(i, j int) bool { return vs[i] < vs[j] })
	j := 0
	for i, c := range cs {
		d := c | 32
		if d == 'a' || d == 'e' || d == 'i' || d == 'o' || d == 'u' {
			cs[i] = vs[j]
			j++
		}
	}
	return string(cs)
}

# Accepted solution for LeetCode #2785: Sort Vowels in a String
class Solution:
    def sortVowels(self, s: str) -> str:
        vs = [c for c in s if c.lower() in "aeiou"]
        vs.sort()
        cs = list(s)
        j = 0
        for i, c in enumerate(cs):
            if c.lower() in "aeiou":
                cs[i] = vs[j]
                j += 1
        return "".join(cs)

// Accepted solution for LeetCode #2785: Sort Vowels in a String
impl Solution {
    pub fn sort_vowels(s: String) -> String {
        fn is_vowel(c: char) -> bool {
            matches!(c.to_ascii_lowercase(), 'a' | 'e' | 'i' | 'o' | 'u')
        }

        let mut vs: Vec<char> = s.chars().filter(|&c| is_vowel(c)).collect();
        vs.sort_unstable();

        let mut cs: Vec<char> = s.chars().collect();
        let mut j = 0;

        for (i, c) in cs.clone().into_iter().enumerate() {
            if is_vowel(c) {
                cs[i] = vs[j];
                j += 1;
            }
        }

        cs.into_iter().collect()
    }
}

// Accepted solution for LeetCode #2785: Sort Vowels in a String
function sortVowels(s: string): string {
    const vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'];
    const vs = s
        .split('')
        .filter(c => vowels.includes(c))
        .sort();
    const ans: string[] = [];
    let j = 0;
    for (const c of s) {
        ans.push(vowels.includes(c) ? vs[j++] : c);
    }
    return ans.join('');
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.