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allcoding1

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allcoding1 (@allcoding1) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 22 409 obunachidan iborat bo'lib, Taʼlim toifasida 8 808-o'rinni va Hindiston mintaqasida 18 944-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 22 409 obunachiga ega bo‘ldi.

27 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -383 ga, so‘nggi 24 soatda esa -21 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 4.45% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.42% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 997 marta ko‘riladi; birinchi sutkada odatda 318 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 28 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.

22 409
Obunachilar
-2124 soatlar
-927 kunlar
-38330 kunlar
Postlar arxiv
A Telegram:- @allcoding1
A Telegram:- @allcoding1

A Telegram:- @allcoding1
A Telegram:- @allcoding1

A Telegram:- @allcoding1
A Telegram:- @allcoding1

Y Telegram:- @allcoding1
Y Telegram:- @allcoding1

A Telegram:- @allcoding1
A Telegram:- @allcoding1

A Telegram:- @allcoding1
A Telegram:- @allcoding1

Kth max code in java Telegram:- @allcoding1
Kth max code in java Telegram:- @allcoding1

Maximum multiple code in python 3 Telegram:-@allcoding1
Maximum multiple code in python 3 Telegram:-@allcoding1

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#include int main() { int n, reverse = 0, remainder; scanf("%d", &n); int p=n; while (n != 0) { remainder = n % 10; rever
#include <stdio.h> int main() { int n, reverse = 0, remainder; scanf("%d", &n); int p=n; while (n != 0) { remainder = n % 10; reverse = reverse * 10 + remainder; n /= 10; } int c=reverse -p; int d=abs(c); printf(" %d", d); return 0; } Code in c Telegram:- @allcoding1

int diff_primes_in_ranges(int input1, int intput2) { int ans = 0; int i, j, flag; for (i = input1; i &lt;= intput2; i++) { if
int diff_primes_in_ranges(int input1, int intput2) { int ans = 0; int i, j, flag; for (i = input1; i <= intput2; i++) { if (i == 1 || i == 0) continue; flag = 1; for (j = 2; j <= i / 2; ++j) { if (i % j == 0) { flag = 0; break; } } if (flag == 1) ans++; } return ans; } C++✅ KPIT Telegram:- @allcoding1

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Python Telegram:-@allcoding1
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Python Telegram:-@allcoding1

Python Telegram:-@allcoding1
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Python Telegram:-@allcoding1

🎯Global Edge Hiring | 2022 graduates CTC - 4LPA Direct Link :- https://www.allcoding1.com/2022/08/global-edge-hiring-2022-graduatesctc.html Telegram :- @allcoding1