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allcoding1_official

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allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 85 467 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 502-o'rinni va Hindiston mintaqasida 3 471-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 85 467 obunachiga ega bo‘ldi.

25 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 422 ga, so‘nggi 24 soatda esa -71 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 2.42% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.88% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 2 071 marta ko‘riladi; birinchi sutkada odatda 749 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 4 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 26 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.

85 467
Obunachilar
-7124 soatlar
-3077 kunlar
-1 42230 kunlar
Postlar arxiv
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Linux full course ₹500 Contact:- @meterials_available

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Oracle
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Oracle

Linux full course ₹500
Linux full course ₹500

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Meesho extended there exam date And calling students to virtual interview based on there exam dates Company name : MEESHO Rol
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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 ✅

long long OneBlock(int N, vector Arr) {     vector 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

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