How to encrypt the count of vowel in a string

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i came through a question which states that to define a function vowelEncryption which accepts an argument str representing a string value which contains only alphabets.


.

 the function must encrypt the given string by replacing the consecutive vowels(one or more ) with their count then the function must return the encrypted string.

.
 for example : if the input is "eager" the output should be "2g1r". as we should replace the consecutive vowels with their count the string becomes 2g1r..
.
 example2: input :'kgooooooooooseee' output:'kg10s3'
'
 example3: input: 'ZAbcdeioufghiOAabcd' output: 'Z1bcd4fgh4bcd'

What I have tried:

s=input().strip()
vowels='aeiou'
ctr=0
count=0
for i in s:
    if i not in vowels:
        print(i,end="")
    else:
        count+=1
    print(count,end="")

Posted 15-Aug-21 0:50am

timothee chalamet

Updated 15-Aug-21 1:27am

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Richard MacCutchan · Answer 1 · 2021-08-15T01:27:00

You are only capturing single vowels. You need to use a flag to identify when the substring changes from vowel to consonant and vice versa. Something like:

Python

count=0
vowel = False
for letter in sentence:
    if letter in vowels:
        if vowel == False:
            vowel = True
            count = 1
        else:
            count += 1
    else:
        if vowel == True:
            print(count, end="")
            vowel = False
        print(letter, end='')
print('')

Note: I have not run exhaustive tests on this so you need to try with a number of different words.