【TUST“码蹄杯”编程之星】「journey：[小程序] 4.21 每日一题」

答案为2。
显然，答案应为连续0数量与2中的较小值

ip = "000001111000110101010000111010111100010100110"

blocks = []
pchar = None
for char in ip:
    if char != pchar:
        blocks.append(char)
        pchar = char
block = sum(1 for block in blocks if block == '0')

has_zero = '0' in ip
has_one = '1' in ip

if not has_zero:
    min_mex = 0
elif not has_one:
    min_mex = block
else:
    min_mex = min(block, 2)

print(min_mex)