Question

I need to use recursion to find the number of vowels in a string. So if hello is entered I want it to return 2.

The problem I’m having is going to the next character in the string.

 def recVowelCount(i):
    count = 0
    if i in 'aeiou':
        count +=1
        reVowelCount(i)
    else:
        reVowelCount(i)
        return count

Answer 1

Here’s one way to do it using recursion 🙂

def recVowelCount(i, chars_to_find='aeiou'):
    if not chars_to_find:
        return 0
    return i.count(chars_to_find[0]) + recVowelCount(i, chars_to_find[1:])

Now, the problem in your code is that

if i in 'aeiou':

would be asking if 'hello' in 'aeiou':, which isn’t very useful. You need to check if i[0] in 'aeiou' where i[0] will be each letter of "hello" each time the function is called recursively

Start with the simple case. What happens if the input string is empty? You’d just return 0 right?

def recVowelCount(i):
    if not i:
        return 0

So we’re half done. Now you need to think about what happens in the case the i isn’t empty. If the first character is a vowel, we’ll count 1 and then pass the rest of the string into the function recursively

def recVowelCount(i):
    if not i:
        return 0
    if i[0] in 'aeiou':
        count = 1
    else:
        count = 0
    return count + recVowelCount(i[1:])

ok.. that can be refactored a little

def recVowelCount(i):
    if not i:
        return 0
    count = 'aeiou'.count(i[0])
    return count + recVowelCount(i[1:])

and finally

def recVowelCount(i):
    if not i:
        return 0
    return 'aeiou'.count(i[0]) + recVowelCount(i[1:])