My understanding is the following:
Time = With the initial not state is just to check if there are no elements in the list
a. This is done in O(1) time. The first loop enumerates the second list
b with the operation
append() which is done in O(1) time therefore the first loop is O(n). The second loop I am confused if the sorting plays a factor since I know the sort method is O(nlogn) time. Within the second loop I am a bit confused about the operations as far as what it is specifically doing which I wanted to ask about although I do understand the fact that if the i-th element in
a is < c it goes into the if statement.
Space = Overall I think it is O(n) as the only additional space needed was the list
x plus the input space and the length of
x grows as
Given the following Python code:
a = [3, 1, 2] # input length will be size n b = [1, 2, 3] # input length will be size n c = 4 def foo(a, b, c): res = 0 if not a: return res x =  for i, j in enumerate(b): x.append((j, i)) for j, i in sorted(x, reverse=True): if a[i] < c: res += a[i] * j c -= a[i] else: res += c * j break return res
Can someone help go over what the time and space complexity is plus the idea behind it mainly?