Python list.pop(i) time complexity?
Timing code: 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. 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.
Python list.pop(i) time complexity?
I look up online and know that list.pop() has O(1) time complexity but list.pop(i) has O(n) time complexity. While I am writing leetcode, many people use pop(i) in a for loop and they say it is O(n) time complexity and in fact it is faster than my code, which only uses one loop but many lines in that loop. I wonder why this would happen, and should I use pop(i) instead of many lines to avoid it?
class Solution(object): def removeDuplicates(self, nums): """ :type nums: List[int] :rtype: int """ left, right = 0, 0 count = 1 while right < len(nums)-1: if nums[right] == nums[right+1]: right += 1 else: nums[left+1]=nums[right+1] left += 1 right += 1 count += 1 return count and other people's code, faster than 90%: (this guy does not say O(n), but why O(n^2) faster than my O(n)?)
class Solution(object): def removeDuplicates(self, nums): """ :type nums: List[int] :rtype: int """ 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 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.
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 = # 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 - Time complexity of list and tuple lookup, But please add a link to where ever you found the time complexity – Bhargav Rao. Sep 25, 2015 at 10:47 | Show 8 more comments. Browse other questions tagged python list tuples time-complexity or ask your own question. The Overflow Blog Stack Exchange sites are getting prettier faster: Introducing …
What is the time complexity of popping an element and appending the same element to a Python list?
So I know popping an element from an an array is O(n) since you have to shift all the elements over by one. BUT, what if you popped an element AND appended that same element to the back of the array, all in one line. Would this be constant operation O(1) since you don't have to shift anything? Please see my code below with the comment.
def moveZeroes(self, nums: List[int]) -> None: index = 0 zeroes = nums.count(0) counter = 0 while True: if counter == zeroes: break if nums[index] == 0: nums.append(nums.pop(index)) # THIS LINE RIGHT HERE counter += 1 else: index += 1 - When you pop any element using its index, you make at most len(array) shifts to reduce the length by 1.
- You can append an element into an array in O(1) time complexity.
However, when you call nums.append(nums.pop(index)) . It will first do step 1 and then do step 2. So overall, you still do O(n) operations.
What is the time complexity of pop() for the set in Python?, On modern CPython implementations, pop takes amortized constant-ish time (I'll explain further). On Python 2, it's usually the same, but performance can degrade heavily in certain cases. A Python set is based on a hash table, and pop has to find an occupied entry in the table to remove and return. If it …
Why is index faster than pop in python list?
Hello I'm practice coding test on leetcode and I have one question.
"26. Remove Duplicates from Sorted Array" description is as below.
Given a sorted array nums, remove the duplicates in-place such that each element appears only once and returns the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
Solution 01
# 4.4948496559999995 idx = 0 for _ in range(len(nums)-1): if nums[-1-idx] == nums[-2-idx]: nums.pop(-1-idx) else: idx += 1 return len(nums) Solution 02
# 3.689027874 ans = 1 for i in range(len(nums)-1): if nums[i] != nums[i+1]: nums[ans] = nums[i+1] ans += 1 return ans Check Time Complexity
if __name__ == '__main__': from timeit import Timer t = Timer("for t in [[1, 1, 1, 2], [1, 1, 1, 2, 2, 3], [1, 2, 3, 3, 3, 3], []]: Solution().removeDuplicates(t)", "from __main__ import Solution") print(t.timeit()) Solution 01 running time was 4.4948496559999995 and Solution 02 running time was 3.689027874
pop and index have O(1) time complexity but why is solution 02 faster than solution 01 ?
One thing that I am sure about leetcode is that the running time does not equal the time complexity of your algorithm. Sometimes it depends on when you are submitting it and how busy the website is. Once I submitted the same answer three times and every time got a different running time.
What is the time complexity of popping an element from, The time complexity of dict.pop is exactly the same as that of dict.get.list.pop is O(N) because it needs to shift elements, but dict.pop doesn't do that.. That said, dict.pop probably won't improve your lookup time. Deleting a key from a dict needs to leave a DKIX_DUMMY marker in its place, and the lookup routine needs to …
What's the time complexity of list access in Python?
I'm trying to make a python function as fast as I can. Let's suppose I have a prime list and I'm calling primes[i] n times for the same i .
I've the intuition that from a certain value of n, it becomes faser to keep the value of primes[i] in a variable.
I made some tries by comparing the two following implementations, and I can't figure out which one is the fastest. It looks like time access to primes[i] depends on a lot of factor.
1st implementation
while n != 1: p = primes[i] if n % p == 0: n = n // p factorization.append(p) else: i += 1 2nd implementation
while n != 1: if n % primes[i] == 0: n = n // primes[i] factorization.append(primes[i]) else: i += 1 Is there any rule to know from how many calls it becomes interesting to keep the value of an element of a list in a variable ?
Accessing primes[i] is done in constant time, O(1) . What that means is that the time needed to read primes[i] does not increase as the primes becomes bigger and that it does not increase when i becomes bigger. In layman's terms: it's damn fast!
Then again, accessing a local variable p is still faster than accessing primes[i] , because the latter has to look up and call the __getitem__ implementation of the primes object. Therefore caching a value in a local variable instead of looking up a list twice is marginally faster.
On the other hand, caring about marginal speed improvements is meaningless compared to reducing algorithm complexity. For the problem of finding prime numbers, you should focus on finding a smart algorithm rather than on improving built-in-list access times.
import time start = time.time() while n != 1: p = primes[i] if n % p == 0: n = n // p factorization.append(p) else: i += 1 end = time.time() print(end - start) do the same for implementation 2 and compare.
And also, try doing it in google colab or any other external machine for better results.
Complexity Cheat Sheet for Python Operations, The first has a time complexity of O(N) for Python2, O(1) for Python3 and the latter has O(1) which can create a lot of difference in nested statements. Pop: s.pop() O(1) O(N) Python Code for time Complexity plot of Heap Sort. 06, Sep 18. KNN Model Complexity. 04, Sep 20. Writing to an excel sheet using …