Say I have a list (not necessarily sorted):
lst = [1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 8, 9, 9, 10, 10, 10]
I need a function that returns count of unique elements in a list, in this case 10.
I am not allowed to use dict or set.
I thought about doing something like:
num = len(lst)
for e in lst:
for i in lst:
if e == i:
num -= 1
But clearly, it does not work. Thanks!
Try this:
Any solution that does a double nested loop over a list runs in O(n^2) time, the following runs in O(n log(n)) time. Though there might be a way to come with O(n) time solution without using a hash table, I don't see it.
def count_number_unique(my_list):
prev = None
unique_count = 0
for ele in sorted(my_list):
if ele != prev:
unique_count += 1
prev = ele
return unique_count
Result:
In [20]: lst = [1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 8, 9, 9, 10, 10, 10]
In [21]: count_number_unique(lst)
Out[21]: 10
In [22]: count_number_unique([1,2,1,9,1,2])
Out[22]: 3
This function should return: 10