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allcoding1

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allcoding1 (@allcoding1) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 22 535 obunachidan iborat bo'lib, Taʼlim toifasida 8 857-o'rinni va Hindiston mintaqasida 19 476-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 22 535 obunachiga ega bo‘ldi.

15 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -434 ga, so‘nggi 24 soatda esa -13 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 6.43% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.32% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 1 450 marta ko‘riladi; birinchi sutkada odatda 297 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.

22 535
Obunachilar
-1324 soatlar
-887 kunlar
-43430 kunlar
Postlar arxiv
Merge String code def MergeStrings(str1, str2):     if str1 is None and str2 is None:         return None     if str1 is None:         return str2     if str2 is None:         return str1         len1 = len(str1)     len2 = len(str2)     merged = [''] * (len1 + len2)         for i in range(min(len1, len2)):         if str1[i] <= str2[i]:             merged[i] = str1[i]             merged[-(i+1)] = str2[i]         else:             merged[i] = str2[i]             merged[-(i+1)] = str1[i]         if len1 > len2:         for i in range(len2, len1):             merged[i] = str1[i]     elif len2 > len1:         for i in range(len1, len2):             merged[i] = str2[i]         return ''.join(merged) Python join this channel 👇 @allcoding1

int LetteredNumberSum(char* str, int len) { &nbsp;&nbsp;&nbsp; if (str == nullptr) { &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs
int LetteredNumberSum(char* str, int len) {     if (str == nullptr) {         return 0;     }     int sum = 0;     for (int i = 0; i < len; i++) {         char letter = str[i];         int ans = 0; //@allcoding1         switch (letter) {             case 'A':                 ans = 1;                 break;             case 'B':                 ans = 10;                 break;             case 'C':                 ans = 100;                 break;             case 'D':                 ans = 1000;                 break;             case 'E':                 ans = 10000;                 break;             case 'F':                 ans = 100000;                 break;             case 'G':                 ans = 1000000;                 break;             default:                 ans = 0;                 break;         }         sum += ans;     }     return sum; } C++ join this channel 👇 @allcoding1

networking security and clouds

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Pescudo codes ans

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Critical reasoning

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🎯Amazon is hiring for Data Associate I ML Role Education qualifications:- Any  Graduate /Graduate/ Post Graduates Job Role : Data Associate I ML Location: work from home Apply Now:- http://www.allcoding1.com Telegram:- @allcoding1

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Y88B

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360 WDBY
360 WDBY

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21)2012 22)S 23)AAA
21)2012 22)S 23)AAA