التعقيد المساحي

def reversed_copy(a):          # ذاكرة إضافية O(n)    result = []    for i in range(len(a) - 1, -1, -1):        result.append(a[i])    return result def reverse_in_place(a):       # ذاكرة إضافية O(1)    i, j = 0, len(a) - 1    while i < j:        a[i], a[j] = a[j], a[i]        i += 1        j -= 1 nums = [1, 2, 3, 4, 5]print(reversed_copy(nums), nums)reverse_in_place(nums)print(nums)

import tracemalloc def reversed_copy(a):    return [a[i] for i in range(len(a) - 1, -1, -1)] def reverse_in_place(a):    i, j = 0, len(a) - 1    while i < j:        a[i], a[j] = a[j], a[i]        i += 1; j -= 1 data = list(range(100_000))for f in (reversed_copy, reverse_in_place):    tracemalloc.start()    f(data)    peak = tracemalloc.get_traced_memory()[1]    tracemalloc.stop()    print(f.__name__, "←", peak // 1024, "KB")

import sys def total_rec(a, i=0):    if i == len(a):        return 0    return a[i] + total_rec(a, i + 1) def total_loop(a):    s = 0    for x in a:        s += x    return s print(total_loop(list(range(5000))))print("حد العمق في بايثون:", sys.getrecursionlimit())print(total_rec(list(range(500))))

import sys squares_list = [x * x for x in range(1_000_000)]   # مليون قيمة محفوظةsquares_gen = (x * x for x in range(1_000_000))    # تُحسب عند الطلب print(sys.getsizeof(squares_list) // 1024, "KB")print(sys.getsizeof(squares_gen), "بايت")print(sum(squares_gen) == sum(squares_list))

def is_prime(n):                 # بلا ذاكرة: O(√n) لكل سؤال    if n < 2:        return False    k = 2    while k * k <= n:        if n % k == 0:            return False        k += 1    return True # جدول محسوب مسبقًا (غربال إراتوستينس): ذاكرة O(N)، وكل سؤال بعده O(1)N = 100table = [True] * (N + 1)table[0] = table[1] = Falsefor i in range(2, int(N ** 0.5) + 1):    if table[i]:        for j in range(i * i, N + 1, i):            table[j] = False print(is_prime(97), table[97])print([n for n in range(N + 1) if table[n]][:10])