LeetCode #2559 — MEDIUM

Count Vowel Strings in Ranges

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

The Problem

Problem Statement

You are given a 0-indexed array of strings words and a 2D array of integers queries.

Each query queries[i] = [l_i, r_i] asks us to find the number of strings present at the indices ranging from l_i to r_i (both inclusive) of words that start and end with a vowel.

Return an array ans of size queries.length, where ans[i] is the answer to the i^th query.

Note that the vowel letters are 'a', 'e', 'i', 'o', and 'u'.

Example 1:

Input: words = ["aba","bcb","ece","aa","e"], queries = [[0,2],[1,4],[1,1]]
Output: [2,3,0]
Explanation: The strings starting and ending with a vowel are "aba", "ece", "aa" and "e".
The answer to the query [0,2] is 2 (strings "aba" and "ece").
to query [1,4] is 3 (strings "ece", "aa", "e").
to query [1,1] is 0.
We return [2,3,0].

Example 2:

Input: words = ["a","e","i"], queries = [[0,2],[0,1],[2,2]]
Output: [3,2,1]
Explanation: Every string satisfies the conditions, so we return [3,2,1].

Constraints:

1 <= words.length <= 10⁵
1 <= words[i].length <= 40
words[i] consists only of lowercase English letters.
sum(words[i].length) <= 3 * 10⁵
1 <= queries.length <= 10⁵
0 <= l_i <= r_i < words.length

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array of strings words and a 2D array of integers queries. Each query queries[i] = [li, ri] asks us to find the number of strings present at the indices ranging from li to ri (both inclusive) of words that start and end with a vowel. Return an array ans of size queries.length, where ans[i] is the answer to the ith query. Note that the vowel letters 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: Array

Example 1

["aba","bcb","ece","aa","e"]
[[0,2],[1,4],[1,1]]

Example 2

["a","e","i"]
[[0,2],[0,1],[2,2]]

Core Insight

What unlocks the optimal approach

Precompute the prefix sum of strings that start and end with vowels.
Use unordered_set to store vowels.
Check if the first and last characters of the string are present in the vowels set.
Subtract prefix sum for range [l-1, r] to find the number of strings starting and ending with 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 #2559: Count Vowel Strings in Ranges
class Solution {
    private List<Integer> nums = new ArrayList<>();

    public int[] vowelStrings(String[] words, int[][] queries) {
        Set<Character> vowels = Set.of('a', 'e', 'i', 'o', 'u');
        for (int i = 0; i < words.length; ++i) {
            char a = words[i].charAt(0), b = words[i].charAt(words[i].length() - 1);
            if (vowels.contains(a) && vowels.contains(b)) {
                nums.add(i);
            }
        }
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            int l = queries[i][0], r = queries[i][1];
            ans[i] = search(r + 1) - search(l);
        }
        return ans;
    }

    private int search(int x) {
        int l = 0, r = nums.size();
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums.get(mid) >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

// Accepted solution for LeetCode #2559: Count Vowel Strings in Ranges
func vowelStrings(words []string, queries [][]int) []int {
	vowels := map[byte]bool{'a': true, 'e': true, 'i': true, 'o': true, 'u': true}
	nums := []int{}
	for i, w := range words {
		if vowels[w[0]] && vowels[w[len(w)-1]] {
			nums = append(nums, i)
		}
	}
	ans := make([]int, len(queries))
	for i, q := range queries {
		l, r := q[0], q[1]
		ans[i] = sort.SearchInts(nums, r+1) - sort.SearchInts(nums, l)
	}
	return ans
}

# Accepted solution for LeetCode #2559: Count Vowel Strings in Ranges
class Solution:
    def vowelStrings(self, words: List[str], queries: List[List[int]]) -> List[int]:
        vowels = set("aeiou")
        nums = [i for i, w in enumerate(words) if w[0] in vowels and w[-1] in vowels]
        return [bisect_right(nums, r) - bisect_left(nums, l) for l, r in queries]

// Accepted solution for LeetCode #2559: Count Vowel Strings in Ranges
// 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 #2559: Count Vowel Strings in Ranges
// class Solution {
//     private List<Integer> nums = new ArrayList<>();
// 
//     public int[] vowelStrings(String[] words, int[][] queries) {
//         Set<Character> vowels = Set.of('a', 'e', 'i', 'o', 'u');
//         for (int i = 0; i < words.length; ++i) {
//             char a = words[i].charAt(0), b = words[i].charAt(words[i].length() - 1);
//             if (vowels.contains(a) && vowels.contains(b)) {
//                 nums.add(i);
//             }
//         }
//         int m = queries.length;
//         int[] ans = new int[m];
//         for (int i = 0; i < m; ++i) {
//             int l = queries[i][0], r = queries[i][1];
//             ans[i] = search(r + 1) - search(l);
//         }
//         return ans;
//     }
// 
//     private int search(int x) {
//         int l = 0, r = nums.size();
//         while (l < r) {
//             int mid = (l + r) >> 1;
//             if (nums.get(mid) >= x) {
//                 r = mid;
//             } else {
//                 l = mid + 1;
//             }
//         }
//         return l;
//     }
// }

// Accepted solution for LeetCode #2559: Count Vowel Strings in Ranges
function vowelStrings(words: string[], queries: number[][]): number[] {
    const vowels = new Set(['a', 'e', 'i', 'o', 'u']);
    const nums: number[] = [];
    for (let i = 0; i < words.length; ++i) {
        if (vowels.has(words[i][0]) && vowels.has(words[i][words[i].length - 1])) {
            nums.push(i);
        }
    }
    const search = (x: number): number => {
        let l = 0,
            r = nums.length;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    return queries.map(([l, r]) => search(r + 1) - search(l));
}

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 + m × 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.