import collections


def five_sum(A, B, C, D, E):
    AB = collections.Counter(a + b for a in A for b in B) # O (N ^ 2) time O(N) space
    CD = collections.Counter(c + d for c in C for d in D) # O (N ^ 2) time O(N) space

    # O (N ^ 3)
    for ab in AB: 
        for cd in CD:
            for e in E:
                if ab + cd + e == 2018:
                    return True
    return False


A = [0, 2000]
B = [0, 10]
C = [0, 10]
D = [0, -1]
E = [0, -1]

print(five_sum(A, B, C, D, E))

Technically, it would be three separate blocks for time complexity analysis:

# This block is O(N ^ 2)
for element in array 1
    for element in array 2
        generate dictionary 1 # O(N) space
# This block is O(N ^ 2)
for element in array 3
    for element in array 4
        generate dictionary 2 # O(N) space

# This block is O(N ^ 3)
for element in dictionary 1
    for element in dictionary 2
        for element in array 5
            if statement
Then for the space complexity, it would be two dictionaries, each one is an order of N, then the memory would be O(N).