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