I am trying to write a simple function to find if 0,0,1 occurs in a list, in that order.
It should return True
or False
.
The list can contain any number of numbers.
For the function ZeroZeroOne
examples would be as follows:
>> ZeroZeroOne( [0,0,1] )
>> True
>> ZeroZeroOne( [1,0,0] )
>> False
# there are 2s in between but the following does have 0,0,1 occurring and in correct order
>> ZeroZeroOne( [0,2,2,2,2,0,1] )
>> True
I have this function:
def ZeroZeroOne(nums):
FoundIt = False
#quick return if defo not possible
if (nums.count(0) < 2) and (nums.count(1) == 0):
return FoundIt
n = len(nums)
for x in range(n-2):
if nums[x] == 0:
for i,z in enumerate(nums[(x+1):]):
if z==0 and z!=1:
for j,q in enumerate(nums[(i+1):]):
if q==1 and q!=0:
FoundIt=True
return FoundIt
Why does the function return True for this list [0, 1, 0, 2, 1]
?
Moreover....
This function seems overly-complex for a seemingly simple problem.
You can achieve this with a single loop - O(n) time complexity. Since it is for this specific case. Try the code below.
def ZeroZeroOne(nums):
found_pattern = []
for num in nums:
if num == 1:
found_pattern.append(1)
if len(found_pattern) == 3:
return True
else:
found_pattern = []
elif num == 0 and len(found_pattern) < 2:
found_pattern.append(0)
return False
print(ZeroZeroOne([0, 0, 1]))
print(ZeroZeroOne([0, 1, 0, 2, 1]))
print(ZeroZeroOne([0, 2, 0, 1]))
print(ZeroZeroOne([0, 0, 0, 1]))
print(ZeroZeroOne([0, 2, 2, 2, 2, 0, 1]))
But I think you can generalize this as well if required. Probably you need to look in to how grep works and modify it for your use case if you want a generic approach.