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def interleave ( L 1 , L 2 ) : # L 1 and L 2 are lists i = 0 result = [ ]

def interleave ( L1, L2):
# L1 and L2 are lists
i =0
result =[]
while i < min (len ( L1), len ( L2)):
result . append ( L1[ i ])
result . append ( L2[ i ])
i +=1
if len ( L1)> len ( L2):
longer = L1
else :
longer = L2
for j in range (i , len( longer )):
result . append ( longer [ j ])
return result
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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