Your code doesn't work becuase slicing as in divide(alist[:middle]) creates a new list, so alist after the first recursion doesn't refer to temp anymore.
It is more common to return results rather than trying to create side effects in the calling arguments, e.g.:
def divide(alist):
# when the list have only one element, it should return the 0
if len(alist) == 1:
return [0]
middle = len(alist) // 2
return divide(alist[:middle]) + divide(alist[middle:])
print(divide(temp))
# [0, 0, 0, 0, 0, 0]
Obviously, this is relatively nonsense but I'm assuming you are just setting up the structure to do something specific.
If you really wanted to do this with side effects (not recommended!!!) then you would need to keep a left and right index and use that to eventually assign the [0], e.g.:
def divide(alist, left, right):
middle = (right - left) // 2
# when the list have only one element, it should return the 0
if middle == 0:
alist[left:right] = [0]
else:
divide(alist, left, left+middle)
divide(alist, left+middle, right)
temp = [1, 2, 3, 4, 5, 6]
divide(temp, 0, len(temp))
print(temp)
# [0, 0, 0, 0, 0, 0]
If you want to keep the original calling signature then you can use an _inner() function to handle the recursion, which would allow you to default the args but in reality you could just return _inner(0, len(alist)) without them:
def divide(alist):
def _inner(left=0, right=len(alist)): # default args based on comment @EricDuminil
middle = (right - left) // 2
# when the list have only one element, it should return the 0
if middle == 0:
alist[left:right] = [0]
else:
_inner(left, left+middle)
_inner(left+middle, right)
return _inner()
temp = [1, 2, 3, 4, 5, 6]
divide(temp)
print(temp)
# [0, 0, 0, 0, 0, 0]
