Yes, it is O(1) to pop the last element of a Python list, and O(N) to pop an arbitrary element (since the whole rest of the list has to be shifted).
Here's a great article on how Python lists are stored and manipulated: An Introduction to Python Lists.
Answer from Dan Lenski on Stack OverflowYes, it is O(1) to pop the last element of a Python list, and O(N) to pop an arbitrary element (since the whole rest of the list has to be shifted).
Here's a great article on how Python lists are stored and manipulated: An Introduction to Python Lists.
Pop() for the last element ought to be O(1) since you only need to return the element referred to by the last element in the array and update the index of the last element. I would expect pop() for an arbitrary element to be O(N) and require on average N/2 operations since you would need to move any elements beyond the element you are removing one position up in the array of pointers.
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Your algorithm genuinely does take O(n) time and the "pop in reverse order" algorithm genuinely does take O(n²) time. However, LeetCode isn't reporting that your time complexity is better than 89% of submissions; it is reporting your actual running time is better than 89% of all submissions. The actual running time depends on what inputs the algorithm is tested with; not just the sizes but also the number of duplicates.
It also depends how the running times across multiple test cases are averaged; if most of the test cases are for small inputs where the quadratic solution is faster, then the quadratic solution may come out ahead overall even though its time complexity is higher. @Heap Overflow also points out in the comments that the overhead time of LeetCode's judging system is proportionally large and quite variable compared to the time it takes for the algorithms to run, so the discrepancy could simply be due to random variation in that overhead.
To shed some light on this, I measured running times using timeit. The graph below shows my results; the shapes are exactly what you'd expect given the time complexities, and the crossover point is somewhere between 8000 < n < 9000 on my machine. This is based on sorted lists where each distinct element appears on average twice. The code I used to generate the times is given below.
Timing code:
def linear_solution(nums):
left, right = 0, 0
while right < len(nums)-1:
if nums[right] != nums[right+1]:
nums[left+1]=nums[right+1]
left += 1
right += 1
return left + 1
def quadratic_solution(nums):
prev_obj = []
for i in range(len(nums)-1,-1,-1):
if prev_obj == nums[i]:
nums.pop(i)
prev_obj = nums[i]
return len(nums)
from random import randint
from timeit import timeit
def gen_list(n):
max_n = n // 2
return sorted(randint(0, max_n) for i in range(n))
# I used a step size of 1000 up to 15000, then a step size of 5000 up to 50000
step = 1000
max_n = 15000
reps = 100
print('n', 'linear time (ms)', 'quadratic time (ms)', sep='\t')
for n in range(step, max_n+1, step):
# generate input lists
lsts1 = [ gen_list(n) for i in range(reps) ]
# copy the lists by value, since the algorithms will mutate them
lsts2 = [ list(g) for g in lsts1 ]
# use iterators to supply the input lists one-by-one to timeit
iter1 = iter(lsts1)
iter2 = iter(lsts2)
t1 = timeit(lambda: linear_solution(next(iter1)), number=reps)
t2 = timeit(lambda: quadratic_solution(next(iter2)), number=reps)
# timeit reports the total time in seconds across all reps
print(n, 1000*t1/reps, 1000*t2/reps, sep='\t')
The conclusion is that your algorithm is indeed faster than the quadratic solution for large enough inputs, but the inputs LeetCode is using to measure running times are not "large enough" to overcome the variation in the judging overhead, and the fact that the average includes times measured on smaller inputs where the quadratic algorithm is faster.
Just because the solution is not O(n), you can't assume it to be O(n^2).
It doesn't quite become O(n^2) because he is using pop in reverse order which decreases the time to pop every time, using pop(i) on forward order will consume more time than that on reverse, as the pop searches from reverse and in every loop he is decreasing the number of elements on the back. Try that same solution in non-reverse order, run few times to make sure, you'll see.
Anyway, regarding why his solution is faster, You have an if condition with a lot of variables, he has only used one variable prev_obj, using the reverse order makes it possible to do with just one variable. So the number of basic mathematical operations are more in your case, so with same O(n) complexity each of your n-loops is longer than his.
Just look at your count varible, in every iteration its value is left+1 you could return left+1, just removing that would decrease n amount of count=count+1 you have to do.
I just posted this solution and it is 76% faster
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
a=sorted(set(nums),key=lambda item:item)
for i,v in enumerate(a):
nums[i]=v
return len(a)
and this one gives faster than 90%.
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
a ={k:1 for k in nums} #<--- this is O(n)
for i,v in enumerate(a.keys()): #<--- this is another O(n), but the length is small so O(m)
nums[i]=v
return len(a)
You can say both of them are more than O(n) if you look at the for loop, But since we are working with dublicate members when I am looping over the reduced memebers while your code is looping over all memebers. So the time required to make that unique set/dict is if lesser than time required for you to loop over those extra members and to check for if conditions, then my solution can be faster.
Python's list implementation uses a dynamically resized C array under the hood, removing elements usually requires you to move elements following after up to prevent gaps.
list.pop() with no arguments removes the last element. Accessing that element can be done in constant time. There are no elements following so nothing needs to be shifted.
list.pop(0) removes the first element. All remaining elements have to be shifted up one step, so that takes O(n) linear time.
To add to Martijn's answer, if you want a datastructure that has constant time pops at both ends, look at collections.deque.