您将获得一个包含N个元素的整数数组：d [0],d [1],… d [N-1].
您可以对数组执行AT MOST动作：选择任意两个整数[L,R],并翻转第L和R位之间的所有元素. L和R代表位的最左和最右索引,这些位标记了您决定翻转的段的边界.
在最终的位串中可以获取的最大1位位数(用S表示)是多少？
“翻转”表示将0变换为1,将1变换为0(0-> 1,1-> 0).
输入格式：整数N,下一行包含N位,以空格分隔：d [0] d [1] … d [N-1]
输出：S
限制条件：
1 <= N <= 100000,
d[i] can only be 0 or 1 ,
0 <= L <= R < n ,
输入样例：
8
1 0 0 1 0 0 1 0
样本输出：6
说明：
通过执行以下任一操作,在给定的二进制数组中最多可以得到6个：
Flip [1, 5] ==> 1 1 1 0 1 1 1 0
解决方法:
清理并制作Pythonic
arr1 = [1, 0, 0, 1, 0, 0, 1, 0]
arr2 = [1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1]
arr3 = [0,0,0,1,1,0,1,0,1,1,0,0,1,1,1]
def maximum_ones(arr):
"""
Returns max possible number of ones after flipping a span of bit array
"""
total_one = 0
net = 0
maximum = 0
for bit in arr:
if bit:
total_one += 1
net -= 1
else:
net += 1
maximum = max(maximum, net)
if net < 0:
net = 0
return total_one + maximum
print(maximum_ones(arr1))
print(maximum_ones(arr2))
print(maximum_ones(arr3))
输出：
6
14
11
如果我们想要L和R指数
对此不太确定.它可能可以做得更干净.
arr1 = [1, 0, 0, 1, 0, 0, 1, 0]
arr2_0 = [1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1]
arr2_1 = [1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1]
arr2_2 = [1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1]
arr3 = [0,0,0,1,1,0,1,0,1,1,0,0,1,1,1]
def maximum_ones(arr):
"""
Returns max possible number of ones after flipping a span of bit array
and the (L,R) indices (inclusive) of such a flip
"""
total_one = 0
net = 0
maximum = 0
L = R = 0
started_flipping = False
for i, bit in enumerate(arr):
if bit:
total_one += 1
net -= 1
else:
net += 1
if not started_flipping:
started_flipping = True
L = i
if net > maximum:
maximum = net
R = i
if net < 0:
net = 0
if i < R:
L = i
return (total_one + maximum, (L,R))
print(maximum_ones(arr1))
print(maximum_ones(arr2_0))
print(maximum_ones(arr2_1))
print(maximum_ones(arr2_2))
print(maximum_ones(arr3))
输出：
(6, (1, 5))
(14, (1, 16))
(14, (2, 16))
(14, (3, 16))
(11, (0, 2))
第一次迭代
如果您想了解思维过程的演变,这就是我最初的想法.在这里,我实质上是在纸上翻译出我想出的东西.
本质上,我们遍历数组并开始翻转位(好吧,不是真的),跟踪两个独立数组中的累积翻转零和累积翻转1以及整数计数器中的总翻转比特.如果在给定索引处翻转的零与零之间的差(“净”)降到零以下,我们将在该索引处将累计计数“重置”为零(但没有其他值).在此过程中,我们还跟踪已达到的最大净额以及发生的指数.因此,总数就是我们所见的总数1,加上最大索引处的净值.
arr = [1, 0, 0, 1, 0, 0, 1, 0]
total_one = 0
one_flip = [0 for _ in range(len(arr))]
zero_flip = [0 for _ in range(len(arr))]
# deal with first element of array
if arr[0]:
total_one += 1
else:
zero_flip[0] = 1
maximum = dict(index=0,value=0) #index, value
i = 1
# now deal with the rest
while i < len(arr):
# if element is 1 we definitely increment total_one, else, we definitely flip
if arr[i]:
total_one += 1
one_flip[i] = one_flip[i-1] + 1
zero_flip[i] = zero_flip[i-1]
else:
zero_flip[i] = zero_flip[i-1] + 1
one_flip[i] = one_flip[i-1]
net = zero_flip[i] - one_flip[i]
if net > 0:
if maximum['value'] < net:
maximum['value'] = net
maximum['index'] = i
else: # net == 0, we restart counting our "net"
one_flip[i] = 0
zero_flip[i] = 0
i += 1
maximum_flipped = total_one - one_flip[maximum['index']] + zero_flip[maximum['index']]
结果：
print(total_one, -one_flip[maximum['index']], zero_flip[maximum['index']] )
print(maximum_flipped)
print('________________________________________________')
print(zero_flip, arr, one_flip, sep='\n')
print('maximum index', maximum['index'])
输出：
3 -1 4
6
________________________________________________
[0, 1, 2, 2, 3, 4, 4, 5]
[1, 0, 0, 1, 0, 0, 1, 0]
[0, 0, 0, 1, 1, 1, 2, 2]
maximum index 5
如果arr = [1、0、0、0、1、0、0、1、0、1、0、0、0、0、1、0、0、1]
6 -4 12
14
________________________________________________
[0, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 10, 10, 11, 12, 12]
[1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1]
[0, 0, 0, 0, 1, 1, 1, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5]
maximum index 16
最后,如果arr = [0,0,0,1,1,0,1,0,1,1,1,0,1,1,1,1]
8 0 3
11
________________________________________________
[1, 2, 3, 3, 3, 4, 4, 5, 5, 0, 1, 2, 2, 0, 0]
[0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1]
[0, 0, 0, 1, 2, 2, 3, 3, 4, 0, 0, 0, 1, 0, 0]
maximum index 2
太好了,现在把它拆开,人！