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def find _ triples ( L ) : # L is a list if len ( L ) = = 0 : return 0 L

def find_triples ( L ):
# L is a list
if len ( L )==0:
return 0
L = sorted ( L )
triples =0
repeats =1
prev = L [0]
for i in range (1, len( L )):
cur = L [ i ]
if cur != prev :
repeats =1
else :
repeats +=1
if repeats ==3:
triples +=1
prev = cur
return triples
With regard to the function above:
What is its big-O time complexity? Very briefly justify your answer.
If you used any variables, such as N, in your time complexity, define them in term of the functions
parameters.

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