CSES - Two lists

You are given two lists AA and BB both containing the numbers 1n1 \dots n in some order. Your task is to count how many of the numbers 1n1 \dots n occur earlier on the list AA than on the list BB.

In this task, nn can be large and an efficient algorithm is required. The time complexity should be O(n)O(n).

In a file twolists.py, implement a function count that returns the desired count.

def count(a, b):
    # TODO

if __name__ == "__main__":
    print(count([2,3,4,1], [1,2,3,4])) # 3
    print(count([1,2,3,4], [1,2,3,4])) # 0
    print(count([4,7,3,1,6,2,5], [5,6,1,2,4,3,7])) # 3
    print(count([5,4,9,1,8,3,2,6,7], [6,2,8,4,9,1,5,7,3])) # 5

Explanation: In the first test, the numbers 2, 3 and 4 occur earlier on the list AA than on the list BB.