Your job is to figure out the index of which vowel is missing from a given string:

* `A` has an index of 0,
* `E` has an index of 1, 
* `I` has an index of 2, 
* `O` has an index of 3,
* `U` has an index of 4.

**Notes:** There is no need for string validation and every sentence given will contain all vowles but one. Also, you won't need to worry about capitals.


## Examples

```
"John Doe hs seven red pples under his bsket"          =>  0  ; missing: "a"
"Bb Smith sent us six neatly arranged range bicycles"  =>  3  ; missing: "o"
```

Source: codewars | Difficulty: EASY
Tags: String algorithms

Data structures

Starter code: def absent_vowel(x):
	

Solutions:
def absent_vowel(x):
	return ['aeiou'.index(i) for i in 'aeiou' if i not in x][0]


def absent_vowel(x):
	return 'aeiou'.index((set('aeiou') - set(x.lower())).pop())


def absent_vowel(stg):
	return next((i for (i, v) in enumerate('aeiou') if v not in stg))


def absent_vowel(x):
	strings = ['a', 'e', 'i', 'o', 'u']
	for i in strings:
		if i not in x:
			return strings.index(i)


def absent_vowel(x):
	code = {'a': 0, 'e': 1, 'i': 2, 'o': 3, 'u': 4}
	for i in code:
		if i not in x:
			return code[i]


def absent_vowel(x):
	vowels = ['a', 'e', 'i', 'o', 'u']
	for (i, v) in enumerate(vowels):
		if v not in x.lower():
			return i


def absent_vowel(x):
	VOWELS = set('aeiou')
	INDEX = {'a': 0, 'e': 1, 'i': 2, 'o': 3, 'u': 4}
	missing = VOWELS.difference(set(x.lower()))
	return INDEX.get(''.join(missing))


def absent_vowel(x):
	dictionary = {'a': 0, 'e': 1, 'i': 2, 'o': 3, 'u': 4}
	for vow in 'aeiou':
		if vow not in x:
			return dictionary.get(vow)


def absent_vowel(x):
	vowels = ['a', 'e', 'i', 'o', 'u']
	for i in vowels:
		if i not in x.lower():
			return vowels.index(i)