Longest Ordered Subsequence of Vowels

Given a string consisting of only vowels, find the longest subsequence in the given string such that it consists of all five vowels and is a sequence of one or more a’s, followed by one or more e’s, followed by one or more i’s, followed by one or more o’s and followed by one or more u’s.

If there is more than one longest subsequence, print any one.

Examples:

Input :  str = "aeiaaioooaauuaeiou"
Output :  {a, a, a, a, a, a, e, i, o, u}
There are two possible outputs in this case:
{a, a, a, a, a, a, e, i, o, u} and,
{a, e, i, i, o, o, o, u, u, u}
each of length 10

Input : str = "aaauuiieeou"
Output : No subsequence possible

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
We loop through all the characters in the string recursively and follow the given conditions:

1. If the subsequence is empty, we include the vowel at the current index only if it is ‘a’. Otherwise, we move on to the next index.
2. If the vowel at the current index is same as the last vowel included in the subsequence, we include it.
3. If the vowel at the current index is the next possible vowel (i.e a–> e–> i–> o–> u ) after the last vowel included in the subsequence, we have two options: either include it or move on to the next index. Hence we choose the one which gives the longest subsequence.
4. If none of the above conditions is satisfied, we move on to the next index (to avoid invalid ordering of vowels in the subsequence).
5. If we have reached the end of the string, we check if the current subsequence is valid or not. If it is valid (i.e if it contains all the vowels), we return it, else we return an empty list.

Below is the implementation of the above approach:

 # Python3 program to find the longest subsequence # of vowels in the specified order    vowels = ['a', 'e', 'i', 'o', 'u']    # Mapping values for vowels  mapping = {'a': 0, 'e': 1, 'i': 2, 'o': 3, 'u': 4}    # Function to check if given subsequence # contains all the vowels or not def isValidSequence(subList):            for vowel in vowels:         if vowel not in subList:             return False                    return True    # Function to find the longest subsequence of vowels # in the given string in specified order def longestSubsequence(string, subList, index):            # If we have reached the end of the string,      # return the subsequence     # if it is valid, else return an empty list     if index == len(string):         if isValidSequence(subList) == True:             return subList         else:             return []                else:         # If there is no vowel in the subsequence yet,          # add vowel at current index if it is 'a',          # else move on to the next character          # in the string         if len(subList) == 0:                            if string[index] != 'a':                 return longestSubsequence(string, subList, index + 1)             else:                 return longestSubsequence(string, subList + \                             [string[index]], index + 1)                    # If the last vowel in the subsequence until          # now is same as the vowel at current index,          # add it to the subsequence         elif mapping[subList[-1]] == mapping[string[index]]:             return longestSubsequence(string, subList + \                             [string[index]], index + 1)                    # If the vowel at the current index comes          # right after the last vowel         # in the subsequence, we have two options:          # either to add the vowel in          # the subsequence, or move on to next character.         # We choose the one which gives the longest subsequence.         elif (mapping[subList[-1]] + 1) == mapping[string[index]]:                            sub1 = longestSubsequence(string, subList + \                                 [string[index]], index + 1)             sub2 = longestSubsequence(string, subList, index + 1)                            if len(sub1) > len(sub2):                 return sub1             else:                 return sub2                            else:             return longestSubsequence(string, subList, index + 1)    # Driver Code if __name__ == "__main__":                string = "aeiaaioooauuaeiou"            subsequence = longestSubsequence(string, [], 0)     if len(subsequence) == 0:         print("No subsequence possible")     else:         print(subsequence)

Output:

['a', 'e', 'i', 'i', 'o', 'o', 'o', 'u', 'u', 'u']

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