This question will be really annoying due to the fact it is for a class and we have a lot of restrictions on our code.
The objective is to make a function to see if two lists (random ordered) have the same elements in them. So, if a=[2,5,4]
and b=[4,2,5]
a==b
would be true
. Now the restrictions are that we cannot use any built-in functions except len()
. So I can't use anything like set()
or the like. I also am not allowed to edit the lists, so I could not check to see if items are in both lists then delete them as I go if they are in both lists until it is empty.
With all these restrictions, I'm running out of ideas. Please help.
Is recursivity allowed ? That way, you don't have to modify existing lists in place. Obviously, not very efficient, but given your requirements this shouldn't really be an issue here...
def are_items_equal(a, b):
# First the guard clause: if the first list is empty,
# return True if the second list is empty too, False otherwise
if not a:
return not b
# There is now at least 1 item in list a
# Perform a linear search in the b list to find a[0]
# (could have used a "for" loop, but I assumed here this was
# forbidden too)
ib = 0;
while ib < len(b):
if a[0] == b[ib]:
# At this point, we know that at index `ib` in list `b`
# there is the same item as in `a[0]`
# Both list match if the "rest" of those two lists match too
# Check that by performing a recursive call to are_items_equal
# (discarding the pair matched at this step)
return are_items_equal(a[1:], b[:ib]+b[ib+1:])
ib += 1
# You only reach this point when `a[0]` was not found
# in the `b` list.
return False
Testing:
test_case = (
([2,5,4], [4,2,5]),
([2, 2, 5, 4], [4, 5, 2, 5]),
([2,2,5,4], [4,2,5]),
([2,2,5,4],[4,2,5,2]),
)
for a,b in test_case:
print(are_items_equal(a, b), a, b)
Producing:
True [2, 5, 4] [4, 2, 5]
False [2, 2, 5, 4] [4, 5, 2, 5]
False [2, 2, 5, 4] [4, 2, 5]
True [2, 2, 5, 4] [4, 2, 5, 2]