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allcoding1

allcoding1

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allcoding1 (@allcoding1) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 20 819 obunachidan iborat bo'lib, Taʼlim toifasida 9 647-o'rinni va Hindiston mintaqasida 19 582-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 20 819 obunachiga ega bo‘ldi.

23 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -422 ga, so‘nggi 24 soatda esa -15 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 1.26% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.18% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 263 marta ko‘riladi; birinchi sutkada odatda 246 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 1 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 24 Oktabr, 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.

20 819
Obunachilar
-1524 soatlar
-987 kun
-42230 kun
Postlar arxiv
Repost from allcoding1

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Repost from allcoding1
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def PowerLimit(a, b): current_resource, max_seconds = 1, 0 while a > 0 and b > 0: if a > b: current_resource, a, b = 1, a - 2
def PowerLimit(a, b):   current_resource, max_seconds = 1, 0   while a > 0 and b > 0:     if a > b:       current_resource, a, b = 1, a - 2, b + 1     else:       current_resource, b, a = 2, b - 2, a + 1     max_seconds += 1   max_seconds += a // 2 if current_resource == 1 else b // 2   return max_seconds power limit ✅ Telegram:- @allcoding1

long long OneBlock(int N, vector<int> Arr) {     vector<int> one_indices;     for (int i = 0; i < N; ++i) {         if (Arr[i] == 1) {             one_indices.push_back(i);         }     }     if (one_indices.size() <= 1) {         return 1;     }     long long ways = 1;     for (int i = 1; i < one_indices.size(); ++i) {         int zeros_between_ones = one_indices[i] - one_indices[i - 1] - 1;         ways *= (zeros_between_ones + 1);     }     return ways; } //one block C++ Telegram:- @allcoding1_official