#1839

Medium

auto_awesomeSliding window with running state updates

LeetCode Problem Workspace

Longest Substring Of All Vowels in Order

Find the longest substring of all vowels in order within a given string of vowels using sliding window technique.

Problem Statement

You are given a string of vowels. A string is considered beautiful if it contains the vowels 'a', 'e', 'i', 'o', and 'u' in that exact order. The task is to return the length of the longest beautiful substring within the given string of vowels. If no such substring exists, return 0.

For example, "aeiou" and "aaaaaaeiiiioou" are considered beautiful, but "uaeio" or "aeoiu" are not. Your goal is to find the longest beautiful substring, or determine that no such substring exists.

Examples

Example 1

Input: word = "aeiaaioaaaaeiiiiouuuooaauuaeiu"

Output: 13

The longest beautiful substring in word is "aaaaeiiiiouuu" of length 13.

Example 2

Input: word = "aeeeiiiioooauuuaeiou"

Output: 5

The longest beautiful substring in word is "aeiou" of length 5.

Example 3

Input: word = "a"

Output: 0

There is no beautiful substring, so return 0.

Constraints

1 <= word.length <= 5 * 105
word consists of characters 'a', 'e', 'i', 'o', and 'u'.

Solution Approach

Sliding Window Approach

Use a sliding window to expand through the string while ensuring the sequence of vowels remains in order. Start by finding an 'a' and expand the window to include subsequent vowels, making sure they follow the order 'a', 'e', 'i', 'o', 'u'.

Efficient State Updates

Maintain the current state of the window using a simple pointer or variable to track which vowel you are expecting next. This avoids redundant checks and recalculations when expanding the window.

Handle Edge Cases

Consider cases where the string is too short, or when there are no complete sequences of all five vowels. Also, handle substrings that contain repetitions of a single vowel.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the final implementation but is generally O(n), where n is the length of the string, because each character is processed once. The space complexity is O(1) since we are only maintaining pointers and state variables.

What Interviewers Usually Probe

Look for a solution that efficiently handles the sliding window and state updates.
Check for the candidate's ability to handle edge cases, such as strings without all vowels in order.
Evaluate the candidate's approach to maintaining a running state to track the expected vowels.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the solution with unnecessary recalculations of vowel positions.
Ignoring edge cases such as strings shorter than five characters or without the full set of vowels.
Failing to handle the order of vowels correctly, especially when vowels repeat.

Follow-up variants

Modify the solution to return the actual substring rather than its length.
Consider variations where the string contains additional characters beyond vowels.
Adapt the solution to work with a larger set of characters or more complex vowel sequences.

FAQ

What is the sliding window technique?

The sliding window technique involves expanding and contracting a window over the input to find the solution, while maintaining an efficient state for the computation.

How do I know if my substring is beautiful?

A substring is beautiful if it contains the vowels 'a', 'e', 'i', 'o', and 'u' in the correct order, without any characters in between or out of order.

How do I optimize my solution for large strings?

By using the sliding window approach, you can reduce the need for repetitive checks and ensure the solution scales linearly with the string length.

What is the expected time complexity for this problem?

The expected time complexity is O(n), where n is the length of the input string, as each character is processed at most once.

Can I modify the approach for a different sequence of characters?

Yes, you can adapt the sliding window technique for different sequences by adjusting the conditions for the order of characters within the window.

terminal

Solution

Solution 1: Two Pointers + Simulation

We can first transform the string `word`. For example, for `word="aaaeiouu"`, we can transform it into data items `('a', 3)`, `('e', 1)`, `('i', 1)`, `('o', 1)`, `('u', 2)` and store them in an array `arr`. Each data item's first element represents a vowel, and the second element represents the number of times the vowel appears consecutively. This transformation can be implemented using two pointers.

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class Solution:
    def longestBeautifulSubstring(self, word: str) -> int:
        arr = []
        n = len(word)
        i = 0
        while i < n:
            j = i
            while j < n and word[j] == word[i]:
                j += 1
            arr.append((word[i], j - i))
            i = j
        ans = 0
        for i in range(len(arr) - 4):
            a, b, c, d, e = arr[i : i + 5]
            if a[0] + b[0] + c[0] + d[0] + e[0] == "aeiou":
                ans = max(ans, a[1] + b[1] + c[1] + d[1] + e[1])
        return ans

Continue Practicing

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