```if-not:swift
Create a method that takes an array/list as an input, and outputs the index at which the sole odd number is located.

This method should work with arrays with negative numbers. If there are no odd numbers in the array, then the method should output `-1`.
```
```if:swift
reate a function `oddOne` that takes an `[Int]` as input and outputs the index at which the sole odd number is located.

This method should work with arrays with negative numbers. If there are no odd numbers in the array, then the method should output `nil`.
```
Examples:
```python
odd_one([2,4,6,7,10]) # => 3
odd_one([2,16,98,10,13,78]) # => 4
odd_one([4,-8,98,-12,-7,90,100]) # => 4
odd_one([2,4,6,8]) # => -1
```

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def odd_one(arr):
	

Solutions:
def odd_one(arr):
	for i in range(len(arr)):
		if arr[i] % 2 != 0:
			return i
	return -1


def odd_one(arr):
	return next((i for (i, v) in enumerate(arr) if v & 1), -1)


def odd_one(arr):
	for i in arr:
		if i % 2:
			return arr.index(i)
	return -1


def odd_one(arr):
	ad = 0
	for i in range(0, len(arr)):
		if arr[i] % 2 == 1:
			ad += 1
			return i
		else:
			pass
	if ad == 0:
		return -1


def odd_one(lst):
	return next((i for (i, n) in enumerate(lst) if n % 2 == 1), -1)


def odd_one(arr):
	arr = list(map(lambda x: x % 2 == 0, arr))
	if arr.count(True) == 1:
		return arr.index(True)
	elif arr.count(False) == 1:
		return arr.index(False)
	else:
		return -1


def odd_one(arr):
	for num in arr:
		if num % 2 != 0:
			return arr.index(num)
			break
	else:
		return -1


def odd_one(arr):
	for (index, value) in enumerate(arr):
		if value % 2 != 0:
			break
	return -1 if value % 2 == 0 else index


def odd_one(arr):
	try:
		return next((i for (i, x) in enumerate(arr) if x % 2))
	except:
		return -1