If in the first list an element is repeated, and in the second list this element also appears, this might be counted as 1 coincidence
Example:
>>> comptaCoincidencies(['verd', 'groc', 'blau', 'vermell'], ['marro', 'blau', 'blanc', 'negre']) 
0 
>>> comptaCoincidencies(['verd', 'groc', 'blau', 'vermell'], ['blanc', 'negre', 'verd', 'groc']) 
2
>>> comptaCoincidencies(['verd', 'groc', 'blau', 'vermell'], ['vermell', 'blau', 'groc', 'verd']) 
4
>>> comptaCoincidencies(['verd', 'verd', 'verd', 'verd'], ['vermell', 'blau', 'groc', 'verd']) 
1
I have done this:
def comptaCoincidencies(l1, l2):
    """
    Donades dues llistes, retorna el nombre d'elements coincidents entre le llistes
    >>> comptaCoincidencies(["verd", "verd"], ["blau", "blau"])
    0
    >>> comptaCoincidencies(["verd", "vermell"], ["verd", "blau"])
    1
    >>> comptaCoincidencies(["verd", "blau"], ["blau", "verd"])
    2
    """
    comptador = 0
    for i in range(0, len(l1)):
        if l1[i] in l2:
            comptador = comptador + 1
    return comptador
 
    