I'm practicing on LeetCode and came across this problem. I'm just learning algorithms and data structures in python, so don't know much about how some built-in methods are implemented and so my initial attempt was (python code)
return [num for num in Counter(nums) if Counter(nums)[num] == 1]
but it would take the entire 12.82 seconds to run on this testcase. On the other hand, there is a very similar one liner in discussions
return [c[0] for c in Counter(nums).items() if c[1] == 1]
which utilizes the same idea but runs much faster, the aforementioned testcase took only 0.0022 seconds. Why is there such a difference in runtime?
The issue with this code:
return [num for num in Counter(nums) if Counter(nums)[num] == 1]
is that for each num in Counter(nums), you have to create a new Counter(nums) to determine if the if condition is true or false. By using Counter(nums).items() in the second version:
return [c[0] for c in Counter(nums).items() if c[1] == 1]
You have access to the number and its count already so there is no need to re-compute the count for each number.
Note I would write the comprehension as
return [num for num, count in Counter(nums).items() if count == 1]
to make it a bit more obvious how it works.