3

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 a or b grows.

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] &lt; 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?

anshur
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2 Answers2

2

You are completely correct in your analysis.

Space. You construct an array $x$ which has length the same as that of $b$, which is $n$. It contains a pair for each element in $b$, so it has size $O(n)$. The other variables do not count, hence space complexity is indeed $O(n)$.

Time. You correctly observed that the sorting of $x$ takes time $O(n \log n)$, and that both iterations take time $O(n)$. Everything else is arithmetic operations, which all run in $O(1)$ time. Hence, something like $O(n) + O(n \log n) + O(n) + O(1)$, which all in all results in a time complexity of $O(n \log n)$.

Well done!

Ainsley H.
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0

The second loop is indeed tricky. Without a deeper look, it is hard to tell its time complexity tightly, because it could be that the break always occurs very early. But you are on the safe side by assuming $O(n)$ for this loop, which, fortunately, is absorbed by the $O(n\log n)$ of the sort.

The space is indeed $O(n)$, due to the filling of $x$ with $n$ elements, given that appends are done in amortized constant time.