Find Longest Substring of All Vowels in Order in Python

Suppose we have a string s with only English vowels, we have to find the length of the longest beautiful substring of s. If we cannot find such substring then, return 0. A string is said to be beautiful if it satisfies the following conditions −

• Each of the 5 vowels must appear at least once in it.

• Letters must be sorted in alphabetical sequence

So, if the input is like s = "aaioaaaaeiiouuooaauu", then the output will be 10 because the substring is "aaaaeiiouu" which is beautiful.

To solve this, we will follow these steps −

• vowels := a list of all vowels ['a', 'e', 'i', 'o', 'u']

• l := 0, r := 0, longest := 0

• while l < size of s, do

  • valid := True

  • for each vowel in vowels, do

    • valid := valid is true and (r < size of s and s[r] is same as vowel)

    • while r < size of s and s[r] is same as vowel, do

      • r := r + 1

  • if valid is true, then

    • longest := maximum of longest and (r - l)

  • l := r

• return longest

Example

Let us see the following implementation to get better understanding −

def solve(s):
   vowels = ['a', 'e', 'i', 'o', 'u']
   l, r, longest = 0, 0, 0
   while (l < len(s)):
      valid = True
      for vowel in vowels:
         valid &= (r < len(s) and s[r] == vowel)
         while (r < len(s) and s[r] == vowel):
            r += 1
      if (valid):
         longest = max(longest, r - l)
      l = r
   return longest

s = "aaioaaaaeiiouuooaauu"
print(solve(s))

Input

"aaioaaaaeiiouuooaauu"

Output

10