Space complexity of algo

Question

Is the space complexity of this function N^2 as the output is a linked list?

I am learning about space complexity in school and am stuck on this question.

def myHealthcare(record):
    l2=[]
    count=0 # num of records generated and the specific time 
    for i in range(record):
        l=[]
        now = datetime.datetime.now()
        ts = now.strftime("%d/%m/%Y %H:%M:%S") # str timestamp 
        ts=ts +' '+str(count)
        l.append(ts)
        l.append(rand.randint(36,39)) #temp
        l.append(rand.randint(55,100)) #hr
        l.append(rand.randint(55,100)) #Pulse
        l.append(rand.randint(120,121)) #bp
        l.append(rand.randint(11,17)) #respiratory rate
        l.append(rand.randint(93,100)) #oxygen sat
        l.append(round(rand.uniform(7.1,7.6),1)) #pH
        l2.append(l)
        count+=1
    return l2

Can you explain your thought process in more detail? Why does being a linked list mean it's O(n^2)? — John Kugelman, Dec 09 '19 at 19:52

score 1 · Accepted Answer · answered Dec 09 '19 at 19:54

The space complexity of a linked list is not quadratic; each linked list node takes a constant amount of auxiliary memory, so the auxiliary memory used by the whole data structure is O(n) where n is the number of nodes.

However, you are also constructing strings and storing these in memory. The string str(count) is part of a string that gets appended to your list l on each iteration, and the length of this string is O(log n) since count is a number up to n, and it has O(log n) digits when represented as a string. So the overall space complexity of this algorithm is O(n log n) because of that.

Space complexity of algo

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