##Background - the Collatz Conjecture:

Imagine you are given a positive integer, `n`, then:

* if `n` is even, calculate: `n / 2`
* if `n` is odd, calculate: `3 * n + 1`

Repeat until your answer is `1`. The Collatz conjecture states that performing this operation repeatedly, you will always eventually reach `1`.

You can try creating Collatz sequences with [this](http://www.codewars.com/kata/5286b2e162056fd0cb000c20) kata. For further information, see the [wiki page](https://en.wikipedia.org/wiki/Collatz_conjecture).

##Now! Your task:

**Given an array of positive integers, return the integer whose Collatz sequence is the longest.**

Example:

```python
longest_collatz([2, 4, 3])==3
```

Explanation: The Collatz sequence for `2` has a length of `1`, the sequence for `4` has a length of `2`, and the sequence for `3` has a length of `7`. So from our array, the integer `3` is the one with the longest Collatz sequence.

Hence, your function should return `3`.

##Note:

There may be more than one answer, i.e. two or more integers produce the longest Collatz sequence, because they happen to have sequences of the same length. **In this case, your function should return the integer that appears first in the array.**

Example:
Given an array: `[2, 5, 32]`, both `5` and `32` have Collatz sequences of length 5. These are also the longest sequences from our array.

In this case, our function returns `5`, because `5` comes before `32` in our array.

Difficulty: EASY
Starter code: def longest_collatz(input_array):
	

Solutions:
AVAILABLE_METHODS = {'even': lambda x: x / 2, 'odd': lambda x: 3 * x + 1}

def generator(x):
	temp = x
	temp_num = 0
	while temp > 1:
		if temp % 2 == 0:
			temp = AVAILABLE_METHODS['even'](temp)
		else:
			temp = AVAILABLE_METHODS['odd'](temp)
		temp_num += 1
		yield temp_num

def longest_collatz(input_array):
	answer_list = [sum([item for item in generator(dig)]) for dig in input_array]
	print(max(answer_list), answer_list)
	return input_array[answer_list.index(max(answer_list))]


def collatz(n, t=1):
	return t if n == 1 else collatz(n / 2 if n % 2 == 0 else 3 * n + 1, t + 1)

def longest_collatz(input_array):
	return sorted(list(zip(list(map(collatz, input_array)), input_array)))[-1][1]


def longest_collatz(arr):
	return max([(gen_s(e), e) for e in arr])[1]

def gen_s(n, c=1):
	return c + gen_s((n // 2, n * 3 + 1)[n % 2], c + 1) if n != 1 else 1


from functools import lru_cache

@lru_cache(maxsize=None)
def collatz(n):
	return n != 1 and 1 + collatz(3 * n + 1 if n & 1 else n >> 1)

def longest_collatz(input_array):
	return max(input_array, key=collatz)


def longest_collatz(a):

	def f(n):
		r = 1
		while n > 1:
			n = n * 3 + 1 if n % 2 else n // 2
			r += 1
		return r
	return max(a, key=f)


def longest_collatz(lst):
	return max(lst, key=seq_len)

def seq_len(n):
	count = 0
	while n > 1:
		n = 3 * n + 1 if n % 2 else n // 2
		count += 1
	return count


def collatz_len(n):
	result = 1
	while n != 1:
		if n % 2:
			n = 3 * n + 1
		else:
			n //= 2
		result += 1
	return result

def longest_collatz(input_array):
	return max(input_array, key=collatz_len)


def longest_collatz(input_array):
	results = {1: 1}
	for n in range(2, max(input_array) + 1):
		count = 0
		x = n
		while x not in results:
			x = x * 3 + 1 if x % 2 else x // 2
			count += 1
		count += results[x]
		results[n] = count
	return max(input_array, key=results.get)


def collatz(n):
	d = {1: 1}
	count = 0
	while n != 1:
		if n in d:
			count += d[n]
			n = 1
		elif n % 2 == 0:
			n /= 2
			count += 1
		else:
			n = (3 * n + 1) / 2
			count += 2
	return count

def longest_collatz(input_array):
	return max(input_array, key=collatz)


def longest_collatz(input_array):

	def collatz_len(n):
		ctr = 0
		while n > 1:
			(ctr, n) = (ctr + 1, n // 2 if n % 2 == 0 else 3 * n + 1)
		return ctr
	return max((n for n in input_array), key=collatz_len)