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📈 Telegram kanali allcoding1_official analitikasi

allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 85 483 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 507-o'rinni va Hindiston mintaqasida 3 480-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 85 483 obunachiga ega bo‘ldi.

24 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 369 ga, so‘nggi 24 soatda esa -35 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 2.60% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.55% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 2 222 marta ko‘riladi; birinchi sutkada odatda 474 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 3 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, dev kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Kanal uchun tavsif kiritilmagan.

Yuqori yangilanish chastotasi (oxirgi ma’lumot 25 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

85 483
Obunachilar
-3524 soatlar
-2687 kunlar
-1 36930 kunlar
Postlar arxiv
def check_first_and_last_char(Str): if Str[0].isdigit() and not Str[-1].isdigit(): print("Yes") elif Str[-1].isdigit() and no
def check_first_and_last_char(Str): if Str[0].isdigit() and not Str[-1].isdigit(): print("Yes") elif Str[-1].isdigit() and not Str[0].isdigit(): print("No") else: print("Invalid input") Str = input() check_first_and_last_char(Str)

Deloitte exam send Questions 👇

def check_digit_position(s): if s[0].isdigit() and not s[-1].isdigit(): return "Yes" elif not s[0].isdigit() and s[-1].isdigi
def check_digit_position(s): if s[0].isdigit() and not s[-1].isdigit(): return "Yes" elif not s[0].isdigit() and s[-1].isdigit(): return "No" else: return "Invalid input" input_str = input().strip() print(check_digit_position(input_str))

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Deloitte exam Ans

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