Speedcubing is the hobby involving solving a variety of twisty puzzles, the most famous being the 3x3x3 puzzle or Rubik's Cube, as quickly as possible.

In the majority of Speedcubing competitions, a Cuber solves a scrambled cube 5 times, and their result is found by taking the average of the middle 3 solves (ie. the slowest and fastest times are disregarded, and an average is taken of the remaining times).

In some competitions, the unlikely event of a tie situation is resolved by comparing Cuber's fastest times.

Write a function ```cube_times(times)``` that, given an array of floats ```times``` returns an array / tuple with the Cuber's result (to 2 decimal places) AND their fastest solve.

For example:
```python
cube_times([9.5, 7.6, 11.4, 10.5, 8.1]) = (9.37, 7.6)
# Because (9.5 + 10.5 + 8.1) / 3 = 9.37 and 7.6 was the fastest solve.
```

Note: ```times``` will always be a valid array of 5 positive floats (representing seconds)

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def cube_times(times):
	

Solutions:
def cube_times(times):
	return (round((sum(times) - (min(times) + max(times))) / 3, 2), min(times))


def cube_times(times):
	times = sorted(times)
	return (round(sum(times[1:-1]) / 3, 2), times[0])


from statistics import mean

def cube_times(times):
	return (round(mean(sorted(times)[1:4]), 2), min(times))


def cube_times(times):
	return (round(sum(sorted(times)[1:4]) / 3, 2), min(times))


def cube_times(times):
	record_time = min(times)
	slowest_time = max(times)
	average = 0
	for time in times:
		if time == record_time or time == slowest_time:
			pass
		else:
			average += time
	rounded_time = round(average / 3, 2)
	return (rounded_time, record_time)


def cube_times(arr):
	return (round(sum(sorted(arr)[1:4]) / 3, 2), min(arr))


def cube_times(times):
	times.sort()
	a = [0, 0]
	a[0] = round((float(times[1]) + float(times[2]) + float(times[3])) / 3, 2)
	a[1] = times[0]
	return tuple(a)


def cube_times(times):
	times.sort()
	mid = times[1:4]
	return (round(sum(mid) / len(mid), 2), times[0])