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allcoding1

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allcoding1 (@allcoding1) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 22 543 obunachidan iborat bo'lib, Taʼlim toifasida 8 854-o'rinni va Hindiston mintaqasida 19 507-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 22 543 obunachiga ega bo‘ldi.

14 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -445 ga, so‘nggi 24 soatda esa -14 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 6.31% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.25% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 1 423 marta ko‘riladi; birinchi sutkada odatda 282 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, learning kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Kanal uchun tavsif kiritilmagan.

Yuqori yangilanish chastotasi (oxirgi ma’lumot 16 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

22 543
Obunachilar
-1424 soatlar
-947 kunlar
-44530 kunlar
Postlar arxiv
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)) @allcoding1

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)) Telegram:- @allcoding1

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Deloitte exam Ans

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Python 2nd code Deloitte exam Telegram:- @allcoding1

1st code Python Deloitte exam Telegram:- @allcoding1
1st code Python Deloitte exam Telegram:- @allcoding1

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Repost from allcoding1

Repost from allcoding1
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