LeetCode #2586 — EASY

Count the Number of Vowel Strings in Range

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 0-indexed array of string words and two integers left and right.

A string is called a vowel string if it starts with a vowel character and ends with a vowel character where vowel characters are 'a', 'e', 'i', 'o', and 'u'.

Return the number of vowel strings words[i] where i belongs to the inclusive range [left, right].

Example 1:

Input: words = ["are","amy","u"], left = 0, right = 2
Output: 2
Explanation: 
- "are" is a vowel string because it starts with 'a' and ends with 'e'.
- "amy" is not a vowel string because it does not end with a vowel.
- "u" is a vowel string because it starts with 'u' and ends with 'u'.
The number of vowel strings in the mentioned range is 2.

Example 2:

Input: words = ["hey","aeo","mu","ooo","artro"], left = 1, right = 4
Output: 3
Explanation: 
- "aeo" is a vowel string because it starts with 'a' and ends with 'o'.
- "mu" is not a vowel string because it does not start with a vowel.
- "ooo" is a vowel string because it starts with 'o' and ends with 'o'.
- "artro" is a vowel string because it starts with 'a' and ends with 'o'.
The number of vowel strings in the mentioned range is 3.

Constraints:

1 <= words.length <= 1000
1 <= words[i].length <= 10
words[i] consists of only lowercase English letters.
0 <= left <= right < words.length

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array of string words and two integers left and right. A string is called a vowel string if it starts with a vowel character and ends with a vowel character where vowel characters are 'a', 'e', 'i', 'o', and 'u'. Return the number of vowel strings words[i] where i belongs to the inclusive range [left, right].

Baseline thinking

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

Pattern signal: Array

Example 1

["are","amy","u"]
0
2

Example 2

["hey","aeo","mu","ooo","artro"]
1
4

Step 02

Core Insight

What unlocks the optimal approach

consider iterating over all strings from left to right and use an if condition to check if the first character and last character are vowels.

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 #2586: Count the Number of Vowel Strings in Range
class Solution {
    public int vowelStrings(String[] words, int left, int right) {
        int ans = 0;
        for (int i = left; i <= right; ++i) {
            var w = words[i];
            if (check(w.charAt(0)) && check(w.charAt(w.length() - 1))) {
                ++ans;
            }
        }
        return ans;
    }

    private boolean check(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
}

// Accepted solution for LeetCode #2586: Count the Number of Vowel Strings in Range
func vowelStrings(words []string, left int, right int) (ans int) {
	check := func(c byte) bool {
		return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'
	}
	for _, w := range words[left : right+1] {
		if check(w[0]) && check(w[len(w)-1]) {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #2586: Count the Number of Vowel Strings in Range
class Solution:
    def vowelStrings(self, words: List[str], left: int, right: int) -> int:
        return sum(
            w[0] in 'aeiou' and w[-1] in 'aeiou' for w in words[left : right + 1]
        )

// Accepted solution for LeetCode #2586: Count the Number of Vowel Strings in Range
impl Solution {
    pub fn vowel_strings(words: Vec<String>, left: i32, right: i32) -> i32 {
        let check =
            |c: u8| -> bool { c == b'a' || c == b'e' || c == b'i' || c == b'o' || c == b'u' };

        let mut ans = 0;
        for i in left..=right {
            let w = words[i as usize].as_bytes();
            if check(w[0]) && check(w[w.len() - 1]) {
                ans += 1;
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #2586: Count the Number of Vowel Strings in Range
function vowelStrings(words: string[], left: number, right: number): number {
    let ans = 0;
    const check: string[] = ['a', 'e', 'i', 'o', 'u'];
    for (let i = left; i <= right; ++i) {
        const w = words[i];
        if (check.includes(w[0]) && check.includes(w.at(-1))) {
            ++ans;
        }
    }
    return ans;
}

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(m)

Space

O(1)

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.