After watching a live coding interview experience I decided to develop the solution for the problem. #python #pyton3 #codinginterview #coding #interview #test

join_python.py

"""
Write a function that perform as an SQL-join.

"""


BRAND = [
    {'id': 0, 'name': 'Ferrari', 'year': 1998},
    {'id': 1, 'name': 'Red Bull'},
    {'id': 2, 'name': 'Mclaren'}
]

ENGINE = [
    {'id': 2, 'engine': 'Mercedes', 'year': 1998}
]


# Big O complexity: 
# best case: O(Nx1) -> O(N)
# worst case: O(N^3)
def join_python(left_data: list, right_data: list, key: str) -> list:
    cache = {}
    seek_at = set()
    out_data = []
    for obj in left_data:
        if key in obj:
            indexable_key = obj[key]
            seek_at.add(indexable_key)
            cache[ indexable_key ] = obj

    for row in right_data:
        if key in row and row[key] in seek_at:  # This line could be O(N)
            cache[ row[key] ].update(row)
            out_data.append( cache[row[key]] )  # This bould be O(N)
            
    return out_data

if __name__ == '__main__':


    out = join_python(
        left_data=BRAND,
        right_data=ENGINE,
        # key='id',
        key='year'
    )
    print(out)

"""
out: [{'id': 2, 'name': 'Ferrari', 'year': 1998, 'engine': 'Mercedes'}
"""