Given an array containing only zeros and ones, find the index of the zero that, if converted to one, will make the longest sequence of ones.

For instance, given the array:

```
[1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1]
```

replacing the zero at index 10 (counting from 0) forms a sequence of 9 ones:

```
[1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1]
                  '------------^------------'
```


Your task is to complete the function that determines where to replace a zero with a one to make the maximum length subsequence.


**Notes:**
- If there are multiple results, return the last one:

 `[1, 1, 0, 1, 1, 0, 1, 1] ==> 5`


- The array will always contain only zeros and ones.


Can you do this in one pass?

Difficulty: EASY
Starter code: def replace_zero(arr):
	

Solutions:
def replace_zero(arr):
	(m, im, i, lst) = (0, -1, -1, ''.join(map(str, arr)).split('0'))
	for (a, b) in zip(lst, lst[1:]):
		i += len(a) + 1
		candidate = len(a) + len(b) + 1
		if m <= candidate:
			(im, m) = (i, candidate)
	return im


def replace_zero(arr):
	ret = [0, 0]
	arr = [None] + arr + [None]
	for (i, e) in enumerate(arr):
		if type(arr[i]) == int and (not e) and any((arr[i + 1], arr[i - 1])):
			mx = sm(arr[:i]) + sm(arr[i + 1:][::-1])
			ret = [ret, [mx, i - 1]][mx >= ret[0]]
	return ret[-1]

def sm(arr, c=0):
	i = len(arr) - 1
	while arr[i]:
		i -= 1
		c += 1
	return c


def replace_zero(lst):
	z = [-1] + [i for (i, n) in enumerate(lst) if not n] + [len(lst)]
	return max(((z[i + 1] - z[i - 1], z[i]) for i in range(1, len(z) - 1)))[1]


import itertools

def replace_zero(arr):
	xs = list(itertools.chain.from_iterable(([length if key else 0] * length for (key, grp) in itertools.groupby(arr) for (_, length) in [(key, sum((1 for _ in grp)))])))
	return max(((0, i) if x else (1 + (xs[i - 1] if i > 0 else 0) + (xs[i + 1] if i < len(xs) - 1 else 0), i) for (i, x) in enumerate(xs)))[1]


def replace_zero(arr):
	(best_len, best_idx) = (0, 0)
	(curr_len, curr_idx) = (0, None)
	for (i, e) in enumerate(arr):
		if e == 1:
			curr_len += 1
		else:
			if curr_len >= best_len:
				(best_len, best_idx) = (curr_len, curr_idx)
			if curr_idx is None:
				curr_len += 1
			elif i - curr_idx < 2:
				curr_len = 1
			else:
				curr_len = i - curr_idx
			curr_idx = i
	if curr_len >= best_len:
		(best_len, best_idx) = (curr_len, curr_idx)
	return best_idx


def replace_zero(arr):
	labels = []
	max = 1
	ans = -1
	for i in range(len(arr)):
		if arr[i] == 0:
			labels.append(i)
	for j in range(len(labels)):
		arr[labels[j]] = 1
		count = 1
		end = 0
		k = labels[j]
		while end == 0 and k != 0:
			if arr[k - 1]:
				count = count + 1
				k = k - 1
			else:
				end = 1
		end = 0
		k = labels[j]
		while end == 0 and k != len(arr) - 1:
			if arr[k + 1]:
				count = count + 1
				k = k + 1
			else:
				end = 1
		arr[labels[j]] = 0
		if count >= max:
			ans = labels[j]
			max = count
	return ans


def replace_zero(arr):
	(max_idx, count) = (None, 0)
	left = idx = right = 0
	for (i, n) in enumerate(arr + [0]):
		if n:
			right += 1
		else:
			if left + right >= count:
				(max_idx, count) = (idx, left + right)
			(left, idx, right) = (right, i, 0)
	return max_idx


def replace_zero(a):
	li1 = []
	for (i, j) in enumerate(a):
		if j == 0:
			temp = a[:]
			(t, t1, li, temp[i]) = (i, i + 1, [], 1)
			while t >= 0 and temp[t] == 1:
				li.append(temp[t])
				t -= 1
			while t1 <= len(a) - 1 and temp[t1] == 1:
				li.append(temp[t1])
				t1 += 1
			li1.append([len(li), i])
	return sorted(li1, key=lambda x: (-x[0], -x[1]))[0][1]


import itertools as it

def iter_len(iterator):
	return len(list(iterator))

def window(iterable, length):
	iterators = [it.islice(iterator, idx, None) for (idx, iterator) in enumerate(it.tee(iterable, length))]
	return zip(*iterators)

def replace_zero(arr):

	def accumulate_group(total, group):
		(_, prev_length, prev_begin) = total
		(key, length, _) = group
		return (key, length, prev_begin + prev_length)
	groups = ((key, iter_len(group), 0) for (key, group) in it.groupby(arr))
	empty = [(None, 0, 0)]
	indexed_groups = it.accumulate(it.chain(empty, groups, empty), accumulate_group)
	is_zero = lambda group: group[0] == 0
	is_one = lambda group: group[0] == 1
	group_len = lambda group: group[1]
	first_pos = lambda group: group[2]
	last_pos = lambda group: group[2] + group[1] - 1
	revelant = lambda triple: is_zero(triple[1])
	triples = filter(revelant, window(indexed_groups, 3))

	def solutions(triples):
		for (left, this, right) in triples:
			if group_len(this) == 1:
				yield (group_len(left) + 1 + group_len(right), first_pos(this))
			else:
				yield (group_len(left) + 1, first_pos(this))
				yield (group_len(right) + 1, last_pos(this))
	(length, positions) = max(solutions(triples))
	return positions


def replace_zero(arr):
	z = [0] + [i for i in range(len(arr)) if not arr[i]] + [len(arr)]
	sums = sorted([[sum(arr[z[i]:z[i + 2]]), z[i + 1]] for i in range(len(z) - 2)])
	return sums[-1][1]