پاسخ پرسش چک کردن اعداد متوالی O(n)
def is_permutation(a):
counter = [0] * len(a)
limit = len(a)
for element in a:
if not 1 <= element <= limit or counter[element - 1] != 0:
return False
else:
counter[element - 1] = 1
return True