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Firibgarlarga uchmang! Telemetrio bunday kanallarni topadi va belgilaydi 👉 Belgini ko‘rmoqchi bo‘lsangiz, obuna bo‘ling

allcoding1_official

allcoding1_official

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📈 Telegram kanali allcoding1_official analitikasi

allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 78 877 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 622-o'rinni va Hindiston mintaqasida 3 839-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 78 877 obunachiga ega bo‘ldi.

27 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 611 ga, so‘nggi 24 soatda esa -26 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 2.10% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.61% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 1 659 marta ko‘riladi; birinchi sutkada odatda 478 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 0 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 28 Oktabr, 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.

78 877
Obunachilar
-2624 soatlar
-3827 kun
-1 61130 kun
Postlar arxiv
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Number of valid Words IBM
Number of valid Words IBM

IBM
IBM

Min cost of project IBM
Min cost of project IBM

Minimum operation IBM
Minimum operation IBM

Count Faults IBM
Count Faults IBM

Aspiring Programmer code IBM
Aspiring Programmer code IBM

def max_units(boxes, units_per_box, truck_size): products = list(zip(units_per_box, boxes)) products.sort(reverse=True) max_u
def max_units(boxes, units_per_box, truck_size):     products = list(zip(units_per_box, boxes))     products.sort(reverse=True)         max_units = 0     for units, boxes in products:         if truck_size >= boxes:             max_units += units * boxes             truck_size -= boxes         else:             max_units += units * truck_size             break     return max_units IBM