LeetCode #3775 — MEDIUM

Reverse Words With Same Vowel Count

Move from brute-force thinking to an efficient approach using two pointers strategy.

The Problem

Problem Statement

You are given a string s consisting of lowercase English words, each separated by a single space.

Determine how many vowels appear in the first word. Then, reverse each following word that has the same vowel count. Leave all remaining words unchanged.

Return the resulting string.

Vowels are 'a', 'e', 'i', 'o', and 'u'.

Example 1:

Input: s = "cat and mice"

Output: "cat dna mice"

Explanation:

The first word "cat" has 1 vowel.
"and" has 1 vowel, so it is reversed to form "dna".
"mice" has 2 vowels, so it remains unchanged.
Thus, the resulting string is "cat dna mice".

Example 2:

Input: s = "book is nice"

Output: "book is ecin"

Explanation:

The first word "book" has 2 vowels.
"is" has 1 vowel, so it remains unchanged.
"nice" has 2 vowels, so it is reversed to form "ecin".
Thus, the resulting string is "book is ecin".

Example 3:

Input: s = "banana healthy"

Output: "banana healthy"

Explanation:

The first word "banana" has 3 vowels.
"healthy" has 2 vowels, so it remains unchanged.
Thus, the resulting string is "banana healthy".

Constraints:

1 <= s.length <= 10⁵
s consists of lowercase English letters and spaces.
Words in s are separated by a single space.
s does not contain leading or trailing spaces.

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting of lowercase English words, each separated by a single space. Determine how many vowels appear in the first word. Then, reverse each following word that has the same vowel count. Leave all remaining words unchanged. Return the resulting string. Vowels are 'a', 'e', 'i', 'o', and 'u'.

Baseline thinking

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

Pattern signal: Two Pointers

Example 1

"cat and mice"

Example 2

"book is nice"

Example 3

"banana healthy"

Step 02

Core Insight

What unlocks the optimal approach

Simulate as described

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 #3775: Reverse Words With Same Vowel Count
class Solution {
    public String reverseWords(String s) {
        String[] words = s.split("\\s+");

        int cnt = calc(words[0]);
        List<String> ans = new ArrayList<>();
        ans.add(words[0]);

        for (int i = 1; i < words.length; i++) {
            String w = words[i];
            if (calc(w) == cnt) {
                ans.add(new StringBuilder(w).reverse().toString());
            } else {
                ans.add(w);
            }
        }
        return String.join(" ", ans);
    }

    private int calc(String w) {
        int res = 0;
        for (char c : w.toCharArray()) {
            if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
                res++;
            }
        }
        return res;
    }
}

// Accepted solution for LeetCode #3775: Reverse Words With Same Vowel Count
func reverseWords(s string) string {
	words := strings.Fields(s)

	calc := func(w string) int {
		cnt := 0
		for _, c := range w {
			switch c {
			case 'a', 'e', 'i', 'o', 'u':
				cnt++
			}
		}
		return cnt
	}

	cnt := calc(words[0])
	var ans []string
	ans = append(ans, words[0])

	for i := 1; i < len(words); i++ {
		w := words[i]
		if calc(w) == cnt {
			b := []rune(w)
			for l, r := 0, len(b)-1; l < r; l, r = l+1, r-1 {
				b[l], b[r] = b[r], b[l]
			}
			w = string(b)
		}
		ans = append(ans, w)
	}

	return strings.Join(ans, " ")
}

# Accepted solution for LeetCode #3775: Reverse Words With Same Vowel Count
class Solution:
    def reverseWords(self, s: str) -> str:
        def calc(w: str) -> int:
            return sum(c in "aeiou" for c in w)

        words = s.split()
        cnt = calc(words[0])
        ans = [words[0]]
        for w in words[1:]:
            if calc(w) == cnt:
                ans.append(w[::-1])
            else:
                ans.append(w)
        return " ".join(ans)

// Accepted solution for LeetCode #3775: Reverse Words With Same Vowel Count
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3775: Reverse Words With Same Vowel Count
// class Solution {
//     public String reverseWords(String s) {
//         String[] words = s.split("\\s+");
// 
//         int cnt = calc(words[0]);
//         List<String> ans = new ArrayList<>();
//         ans.add(words[0]);
// 
//         for (int i = 1; i < words.length; i++) {
//             String w = words[i];
//             if (calc(w) == cnt) {
//                 ans.add(new StringBuilder(w).reverse().toString());
//             } else {
//                 ans.add(w);
//             }
//         }
//         return String.join(" ", ans);
//     }
// 
//     private int calc(String w) {
//         int res = 0;
//         for (char c : w.toCharArray()) {
//             if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
//                 res++;
//             }
//         }
//         return res;
//     }
// }

// Accepted solution for LeetCode #3775: Reverse Words With Same Vowel Count
function reverseWords(s: string): string {
    const words = s.split(/\s+/);

    const calc = (w: string): number => {
        let cnt = 0;
        for (const c of w) {
            if ('aeiou'.includes(c)) cnt++;
        }
        return cnt;
    };

    const cnt = calc(words[0]);
    const ans: string[] = [words[0]];

    for (let i = 1; i < words.length; i++) {
        let w = words[i];
        if (calc(w) === cnt) {
            w = w.split('').reverse().join('');
        }
        ans.push(w);
    }

    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)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

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.

TWO POINTERS

O(n) time

O(1) space

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.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Moving both pointers on every comparison

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.