Count vowel-only substrings with all vowels

Given a lowercase English string s, return the number of contiguous substrings that (1) consist only of vowels from the set {a, e, i, o, u}, and (2) contain each of the five vowels at least once. Constraints: 1 <= |s| <= 200000. Design an O(n) time solution using a sliding-window/two-pointer approach with O(1) extra space. Explain how you handle consonants that break windows, repeated vowels, and long runs of vowels. Provide code and analyze time and space complexity.

What a Strong Answer Covers

• A clear algorithm with the right data structures and enough pseudocode or code-level detail to implement it.
• A correctness argument that explains why the algorithm covers all required cases.
• Time and space complexity, plus at least one alternative approach when relevant.
• Focused tests for normal cases, edge cases, and failure modes.

Follow-up Questions

• How would the approach change if the input were streaming or too large for memory?
• What invariants would you assert in production code?
• Which tests would catch off-by-one, duplicate, or tie-breaking bugs?

Count vowel-only substrings with all vowels

Given a lowercase English string s, return the number of contiguous substrings that satisfy BOTH conditions:

1. The substring consists only of vowels from the set {a, e, i, o, u} (no consonants).
2. The substring contains each of the five vowels at least once.

A consonant anywhere inside a candidate substring disqualifies it, so consonants act as hard breaks that split s into maximal vowel-only segments. Within each such segment you count the substrings that include all five vowels.

Example 1

Input: s = "aeiou"
Output: 1

The only valid substring is "aeiou" itself.

Example 2

Input: s = "aeiouu"
Output: 2

Valid substrings: "aeiou" and "aeiouu".

Example 3

Input: s = "aeioub"
Output: 1

The trailing consonant b cannot be part of any valid substring; only "aeiou" qualifies.

Constraints:

• 1 <= |s| <= 200000
• s contains only lowercase English letters.
• Aim for O(n) time using a sliding window / two-pointer approach with O(1) extra space (the vowel-count table is fixed size 5).

Hints

1. Consonants can never appear in a valid substring, so first split s into maximal runs of vowels and solve each run independently.
2. Within one vowel run, for a fixed right end, count the number of valid left starts. As right moves forward, the smallest valid left only moves forward too (monotonic) — that is the two-pointer invariant.
3. Track how many DISTINCT vowels the current window has using a size-5 frequency table. While all 5 are present, advance left as far as possible; the count of valid starts for this right is (left - segmentStart).
4. Use a 64-bit accumulator: with up to 200000 characters the answer can exceed 32-bit range.

Solution

Solution Alignment

The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.

Problem Restatement

Given a lowercase English string s, return the number of contiguous substrings that ( 1) consist only of vowels from the set {a, e, i, o, u}, and ( 2) contain each of the five vowels at least once. Constraints: 1 <= |s| <= 200000. Design an O(n) time solution using a sliding-window/two-pointer approach with O(1) extra space. Explain how you handle consonants that break windows, repeated vowels, and long runs of vowels. Provide code and analyze time and space complexity.

Recommended Approach

Choose traversal based on the required output. DFS is natural for subtree computations, reconstruction, and range pruning; BFS is natural for level order or side views. Keep per-depth or per-position state when the output depends on columns, rows, or depths.

Correctness

The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.

Complexity

Most tree traversals are O(n) time and O(h) recursion stack for DFS or O(w) queue space for BFS, where h is height and w is maximum width.

Edge Cases and Tests

Empty tree, one node, skewed tree, duplicate values when reconstruction assumes uniqueness, deep recursion, and tie-breaking for same row/column nodes.