Your job at E-Corp is both boring and difficult. It isn't made any easier by the fact that everyone constantly wants to have a meeting with you, and that the meeting rooms are always taken!

In this kata, you will be given an array. Each value represents a meeting room. Your job? Find the **first** empty one and return its index (N.B. There may be more than one empty room in some test cases). 

'X' --> busy
'O' --> empty

If all rooms are busy, return 'None available!'.


More in this series:

The Office I - Outed
The Office II - Boredeom Score
The Office III - Broken Photocopier
The Office V - Find a Chair

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Data structures

Starter code: def meeting(rooms):
	

Solutions:
def meeting(rooms):
	try:
		return rooms.index('O')
	except ValueError:
		return 'None available!'


def meeting(rooms):
	return rooms.index('O') if 'O' in rooms else 'None available!'


def meeting(rooms):
	return next((i for (i, r) in enumerate(rooms) if r == 'O'), 'None available!')


def meeting(rooms):
	if 'O' not in rooms:
		return 'None available!'
	return rooms.index('O')


def meeting(rooms):
	for (num, status) in enumerate(rooms):
		if status == 'O':
			return num
	return 'None available!'


def meeting(rooms):
	return [o for (o, v) in enumerate(rooms) if v == 'O'][0] if 'O' in rooms else 'None available!'


def meeting(rooms):
	tally = 0
	for i in rooms:
		if i == 'O':
			return tally
		else:
			tally += 1
	return 'None available!'


def meeting(rooms):
	for x in rooms:
		if x is 'O':
			return rooms.index(x)
	return 'None available!'