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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