LeetCode #2062 — EASY

Count Vowel Substrings of a String

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

The Problem

Problem Statement

A substring is a contiguous (non-empty) sequence of characters within a string.

A vowel substring is a substring that only consists of vowels ('a', 'e', 'i', 'o', and 'u') and has all five vowels present in it.

Given a string word, return the number of vowel substrings in word.

Example 1:

Input: word = "aeiouu"
Output: 2
Explanation: The vowel substrings of word are as follows (underlined):
- "aeiouu"
- "aeiouu"

Example 2:

Input: word = "unicornarihan"
Output: 0
Explanation: Not all 5 vowels are present, so there are no vowel substrings.

Example 3:

Input: word = "cuaieuouac"
Output: 7
Explanation: The vowel substrings of word are as follows (underlined):
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"
- "cuaieuouac"

Constraints:

1 <= word.length <= 100
word consists of lowercase English letters only.

Step 01

Brute Force Baseline

Problem summary: A substring is a contiguous (non-empty) sequence of characters within a string. A vowel substring is a substring that only consists of vowels ('a', 'e', 'i', 'o', and 'u') and has all five vowels present in it. Given a string word, return the number of vowel substrings in word.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"aeiouu"

Example 2

"unicornarihan"

Example 3

"cuaieuouac"

Core Insight

What unlocks the optimal approach

While generating substrings starting at any index, do you need to continue generating larger substrings if you encounter a consonant?
Can you store the count of characters to avoid generating substrings altogether?

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 #2062: Count Vowel Substrings of a String
class Solution {
    public int countVowelSubstrings(String word) {
        int n = word.length();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            Set<Character> t = new HashSet<>();
            for (int j = i; j < n; ++j) {
                char c = word.charAt(j);
                if (!isVowel(c)) {
                    break;
                }
                t.add(c);
                if (t.size() == 5) {
                    ++ans;
                }
            }
        }
        return ans;
    }

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

// Accepted solution for LeetCode #2062: Count Vowel Substrings of a String
func countVowelSubstrings(word string) int {
	ans, n := 0, len(word)
	for i := range word {
		t := map[byte]bool{}
		for j := i; j < n; j++ {
			c := word[j]
			if !(c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
				break
			}
			t[c] = true
			if len(t) == 5 {
				ans++
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #2062: Count Vowel Substrings of a String
class Solution:
    def countVowelSubstrings(self, word: str) -> int:
        s = set("aeiou")
        ans, n = 0, len(word)
        for i in range(n):
            t = set()
            for c in word[i:]:
                if c not in s:
                    break
                t.add(c)
                ans += len(t) == 5
        return ans

// Accepted solution for LeetCode #2062: Count Vowel Substrings of a String
struct Solution;

impl Solution {
    fn count_vowel_substrings(word: String) -> i32 {
        let s: Vec<char> = word.chars().collect();
        let n = s.len();
        let mut res = 0;
        for i in 0..n {
            for j in i..n {
                if Solution::is_vowel(&s, i, j) {
                    res += 1;
                }
            }
        }
        res
    }

    fn is_vowel(s: &[char], start: usize, end: usize) -> bool {
        let mut ca = 0;
        let mut ce = 0;
        let mut ci = 0;
        let mut co = 0;
        let mut cu = 0;

        for i in start..=end {
            match s[i] {
                'a' => {
                    ca += 1;
                }
                'e' => {
                    ce += 1;
                }
                'i' => {
                    ci += 1;
                }
                'o' => {
                    co += 1;
                }
                'u' => {
                    cu += 1;
                }
                _ => {
                    return false;
                }
            }
        }
        ca >= 1 && ce >= 1 && ci >= 1 && co >= 1 && cu >= 1
    }
}

#[test]
fn test() {
    let word = "aeiouu".to_string();
    let res = 2;
    assert_eq!(Solution::count_vowel_substrings(word), res);
    let word = "unicornarihan".to_string();
    let res = 0;
    assert_eq!(Solution::count_vowel_substrings(word), res);
    let word = "cuaieuouac".to_string();
    let res = 7;
    assert_eq!(Solution::count_vowel_substrings(word), res);
}

// Accepted solution for LeetCode #2062: Count Vowel Substrings of a String
function countVowelSubstrings(word: string): number {
    let ans = 0;
    const n = word.length;
    for (let i = 0; i < n; ++i) {
        const t = new Set<string>();
        for (let j = i; j < n; ++j) {
            const c = word[j];
            if (!(c === 'a' || c === 'e' || c === 'i' || c === 'o' || c === 'u')) {
                break;
            }
            t.add(c);
            if (t.size === 5) {
                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(n^2)

Space

O(C)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.