Iahub got bored, so he invented a game to be played on paper.
 He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
 The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
```
@param {Array} line of the input there are n integers: a1, a2, ..., an. 1 ≤ n ≤ 100. It is guaranteed that each of those n values is either 0 or 1
@return {Integer} the maximal number of 1s that can be obtained after exactly one move
```
Examples :
```
[1, 0, 0, 1, 0, 0]  =>  5
[1, 0, 0, 1]  =>  4

```
Note:
 In the first case, flip the segment from 2 to 6 (i = 2, j = 6). That flip changes the sequence, it becomes: [1 1 1 0 1 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1 1].
 In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.

(c)ll931110 & fchirica

Difficulty: EASY

Starter code:
def flipping_game(num):
	

Solutions:
def flipping_game(num):
	current = 0
	biggest = 0
	for i in num:
		current = max(0, current - (i or -1))
		biggest = max(biggest, current)
	return sum(num) + (biggest or -1)


def flipping_game(num):
	return max([sum(num) + num[i:j + 1].count(0) - num[i:j + 1].count(1) for i in range(len(num)) for j in range(i, len(num))])


def flipping_game(a):
	current = best = 0
	for n in a:
		current = max(0, current - (n or -1))
		best = max(best, current)
	return sum(a) + (best or -1)


def flipping_game(num):
	poss_start = [i for (i, x) in enumerate(num) if x == 0 and (i == 0 or num[i - 1] == 1)]
	poss_end = [i for (i, x) in enumerate(num) if x == 0 and (i == len(num) - 1 or num[i + 1] == 1)]
	if not poss_start or not poss_end:
		return sum(num) - 1
	curmax = 0
	for st in poss_start:
		for en in poss_end:
			curmax = max(curmax, sum((x if i < st or i > en else 1 - x for (i, x) in enumerate(num))))
	return curmax


def flipping_game(num):
	if all(num):
		return len(num) - 1
	x = 0
	m = 0
	for n in num:
		if n == 0:
			x += 1
			m = max(x, m)
		else:
			x = max(x - 1, 0)
	return num.count(1) + m


from itertools import groupby

def flipping_game(lst):
	split = [(k, sum((1 for _ in g))) for (k, g) in groupby(lst)]
	zeroes = [i for (i, t) in enumerate(split) if not t[0]]
	return max((sum(((k ^ 1 if a <= i <= b else k) * n for (i, (k, n)) in enumerate(split))) for (z, a) in enumerate(zeroes) for b in zeroes[z:]), default=sum(lst) - 1)


def flipping_game(num):
	all_pairs = ((i, j) for i in range(len(num)) for j in range(i + 1, len(num) + 1))
	ones = (sum(num[:i]) + j - i - sum(num[i:j]) + sum(num[j:]) for (i, j) in all_pairs)
	return max(ones)


flipping_game = lambda a: sum(a) + max((j - i - 2 * sum(a[i:j]) for j in range(1, len(a) + 1) for i in range(j)))


def flipping_game(a):
	m = 0
	for i in range(len(a)):
		for j in range(i, len(a) + 1):
			x = sum(a[:i]) + a[i:j + 1].count(0) + sum(a[j + 1:])
			if x > m:
				m = x
	return m


def flipping_game(num):
	if 0 not in num:
		return len(num) - 1
	elif num == [0]:
		return 1
	m = -1
	for i in range(len(num) - 1):
		for j in range(i + 1, len(num) + 1):
			c = num[:i].count(1) + num[i:j].count(0) + num[j:].count(1)
			if c > m:
				m = c
	return m