solutions brute force (timeout)
def intersection(a, b):
result = []
for item in b:
if item in a:
result.append(item)
return result
n = length of array a, m = length of array b Time: O(n*m) Space: O(min(n,m)) using set (pass)
def intersection(a, b):
set_a = set(a)
return [ item for item in b if item in set_a ]
n = length of array a, m = length of array b
Time: O(n+m) Space: O(n)