英文:
Binary Array Function Solution
问题
以下是翻译好的部分：
一个接受二进制数组作为输入的函数，找出在只能对子数组进行一次操作的情况下，数组中存在的最大零的数量，该操作可以将子数组的元素翻转，即将0变为1，将1变为0。
有人能否用Python编写程序，考虑到所有边界情况。
我尝试过使用ChatGPT和其他人工智能，但无法解决一些隐藏的测试案例，我也想不出可能失败的测试案例。
一些示例测试案例如下：
输入: [ 0, 1, 0, 0, 1, 1, 0 ]
输出: 6
输入: [ 0, 0, 0, 0, 0 ]
输出: 5
我需要帮助创建一个接受二进制数组输入并返回输出的完整函数。
英文:
A function that takes a binary array as input and finds out the maximum no. of zeroes present in the array after applying only one operation in which we can flip the elements of a subarray i.e. changing 0 to 1 and vice versa.
Can someone write the program in python considering all the edge cases.
I tried using chatgpt and other AI but could solve some of the hidden test cases and I can't think of a test cases which might be failing.
Some example test cases are:
Input : [ 0, 1, 0, 0, 1, 1, 0 ]
Output : 6
Input : [ 0, 0, 0, 0, 0 ]
Output : 5
I need help in creating the complete function which takes the binary array input and return the output.
答案1
得分: 1
the trick is to find the longest subarray containing ones and flipping its values then counting the zeros.
def foo(list1):
# 先寻找包含1的最长子数组
start_index = 0 # 最长子数组的起始索引
length = 0 # 包含的元素数量
for i in range(len(list1)):
if list1[i] == 1: # 找到一个1，检查后面有多少个1
j = i
temp_length = 0
while j < len(list1) and list1[j] == 1:
j += 1
temp_length += 1
if temp_length > length:
start_index = i
length = temp_length
# 翻转该子数组中的值为0
list1[start_index:start_index + length] = [0] * length
# 返回0的计数
return list1.count(0)
英文:
the trick is to find the longest subarray containing ones and flipping its values then counting the zeros.
def foo(list1):
# first search for the longest subarray containing ones
start_index=0 # start index of the longest subarray
length=0 # number of elements it contains
for i in range(len(list1)):
if list1[i]==1: # we have found a one, check how many ones come after it
j=i
temp_length=0
while j<len(list1) and list1[j]==1:
j+=1
temp_length+=1
if temp_length>length:
start_index=i
length=temp_length
# flip the values in that subarray to zeros
list1[start_index:start_index+length]=[0]*length
# return the count of zeros
return list1.count(0)
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