```python # a = "abc", bo'sh emas, noyob (m) # s = "dfaga[bfkac]ebf" => "aceb" (n) O(n) # s = "cfaafb" => "cfaafb" # s = "affb" => "" from collections import Counter def min_s(a: str, s: str) -> str: need = set(a) window = Counter() res = "" c = 0 left = 0 best_len = float('inf') for right, char in enumerate(s): if char in need and \ window[char] += 1 if window[char]: c += 1 ```
```python from collections import Counter def min_s(a: str, s: str) -> str: need = set(a) window = Counter() res = "" c = 0 left = 0 best_len = float('inf') for right, char in enumerate(s): if char in need and \ window[char] += 1 if window[char]: c += 1 while c == len(need): if right - left + 1< best_len: res = s[left:right + 1] best_len = right - left + 1 if s[left] in need: window[s[left]] -= 1 if window[s[left]] == 0: c -= 1 left += 1 return res ``` Vazifa `min_s(a: str, s: str) -> str` ni amalga oshirish, `s` ichida `a` satridagi barcha belgilarni o'z ichiga olgan eng kichik bo'lakni topish uchun.
from file import ATM, SDK import pytest BILLS = {5000, 1000, 500, 100, 50} @pytest.mark.parametrize( "bills_count_const,expected", [ ({5000: 0, 1000: 0, 500: 0, 100: 0, 50: 0}, False), ({5000: 100, 1000: 100, 500: 100, 100: 100, 50: 100}, True) ] ) def test_false(bills_count_const, expected): atm = ATM(SDK()) atm.bills_count = bills_count_const amount = 1000 res = atm.withdraw(amount) assert res == expected return
""" Berilgan butun sonlar ketma-ketligi. Ketma-ketlikdagi elementlar juftligining minimal mumkin bo'lgan ko'paytmasini topish kerak (juftlik - ikkita istalgan element, ketma-ket bo'lishi shart emas). Masalan, 9 4 2 5 3 sonlar ketma-ketligi uchun javob 6 bo'ladi. """ def find_min_product(arr: list[int]) -> int: ...
from collections import Counter def min_s(a: str, s: str) -> str: need = set(a) window = Counter() res = "" c = 0 left = 0 best_len = float('inf') for right, char in enumerate(s): if char in need: window[char] += 1 if window[char] == 1: c += 1 while c == len(need): if right - left + 1 < best_len: res = s[left:right + 1] best_len = right - left + 1 if s[left] in need: window[s[left]] -= 1 if window[s[left]] == 0: c -= 1 left += 1 return res
Ma'lumotlar bazasida saqlash uchun eng katta ma'lumot hajmi nima edi?
Chap harfni (masalan, A harfi) chap tomondan oynasini qisqartirishda alfavitni qamrab olishini yo'qotmasdan olib tashlash mumkinligini qanday bilish mumkin?
Ko'p ipli yoki asinxronlik bilan dasturlar ishlab chiqishda tajribangiz bormi?
Kvadrat operatsiyasini olib tashlagandan so'ng algoritmning yakuniy murakkabligi nima? Alfavit m ning o'lchami vaqt murakkabligiga hech qanday ta'sir qilmayaptimi?
Qaysi ish formatini ko'rib chiqyapsiz?
""" Shaharlar juftligi to'plami berilgan: - har bir juft shahar orasida xodim to'g'ridan-to'g'ri parvoz qilgan; - parvoz yo'nalishi haqida ma'lumot yo'qotilgan; - shuningdek, parvozlar tartibi ham yo'qotilgan. Barcha parvozlar bitta sayohatga tegishli ekanligi ma'lum. Har bir keyingi parvoz avvalgi tugagan shahar bilan boshlanadi. Hech qanday shahar xodim tomonidan ikki marta tashrif buyurilmagan. Sayohat boshlanish shahrining oxirgi nuqtadan farqi ham bor. Shaharlarni marshrut bo'yicha tartibda chiqarib bering. Ikki mumkin bo'lgan javoblar mavjud, istalgan biri mos keladi. Misollar: [("Moskva", "Belgrad")] -> ["Moskva", "Belgrad"] [(("Moskva", "Belgrad"), ("Moskva", "Erevan")) -> ["Erevan", "Moskva", "Belgrad"] """ Flight = tuple[str, str] def get_route(flights: list[Flight]) -> list[str]: ...
Ushbu vazifada grafikni tasvirlash uchun qanday ma'lumotlar tuzilmasi ishlatilishi mumkin?
# a = "abc", bo'sh emas, noyob # s = "dfagabfkacebf" => "aceb" # s = ""
Ishingizda qanday ma'lumotlar bazalari yoki boshqa saqlash joylaridan foydalanasiz?
Bu butun sonlar ketma-ketligi. Ketma-ketlikdagi elementlar juftligining minimal mumkin bo'lgan ko'paytmasini topish kerak (juftlik - ikkita element, ketma-ket bo'lishi shart emas). Masalan, 9 4 2 5 3 sonlar ketma-ketligi uchun javob 6 bo'ladi.
Eng yuqori yuklangan xizmat bir soniyada qancha so'rovni qayta ishlagan?
Oldin [kompaniya]da bo'sh ish o'rinlarini ko'rib chiqqanmisiz, rekruterlar bilan bog'langanmisiz?
