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allcoding1_official

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📈 نظرة تحليلية على قناة تيليجرام allcoding1_official

تُعد قناة allcoding1_official (@allcoding1_official) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 84 584 مشتركاً، محتلاً المرتبة 1 497 في فئة التكنولوجيات والتطبيقات والمرتبة 3 527 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 84 584 مشتركاً.

بحسب آخر البيانات بتاريخ 10 يوليو, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -1 556، وفي آخر 24 ساعة بمقدار -30، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 2.01‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 0.85‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 1 701 مشاهدة. وخلال اليوم الأول يجمع عادةً 723 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 1.
  • الاهتمامات الموضوعية: يركز المحتوى على مواضيع رئيسية مثل dsa, stack, namaste, javascript, dev.

📝 الوصف وسياسة المحتوى

وصف القناة غير متوفر.

بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 11 يوليو, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التكنولوجيات والتطبيقات.

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-1 55630 أيام
أرشيف المشاركات
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Guys ❤️ Once check a your gmail's Update Accenture exam date

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🎯RDA Labs Off Campus Drive | CTC: INR 5.5 LPA * Job Role : Associate Engineer * Qualification : B.E/B.Tech * Experience : Freshers * Package : 5.5 LPA https://www.joboffersadda.com/2022/04/rda-labs-off-campus-drive-ctc-inr-55-lpa.html https://www.joboffersadda.com/2022/04/rda-labs-off-campus-drive-ctc-inr-55-lpa.html TELEGRAM :-@allcoding1

IBM TELEGRAM :-@allcoding1
IBM TELEGRAM :-@allcoding1

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Penalty calculations C++ language
Penalty calculations C++ language

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#include using namespace std; int vowel(string s){ int n = s.length(); vector arr; for (int i = 0; i < n; i++) { if (i ==
#include <bits/stdc++.h> using namespace std; int vowel(string s){ int n = s.length(); vector<int> arr; for (int i = 0; i < n; i++) { if (i == 0){ arr.push_back(n); } else{ arr.push_back((n - i) + arr[i - 1] - i); } } int sum = 0; for (int i = 0; i < n; i++) if (s[i] == 'a' s[i] == 'e' s[i] == 'i' s[i] == 'o' s[i] == 'u'){ sum += arr[i]; } return sum; } int main(){ string s = "abc"; cout << vowel(s) << endl; } EPAM VOWEL Code C++ TELEGRAM :-@allcoding1

#include using namespace std; int vowel(string s){ int n = s.length(); vector arr; for (int i = 0; i &lt; n; i++) { if (i ==
#include <bits/stdc++.h> using namespace std; int vowel(string s){ int n = s.length(); vector<int> arr; for (int i = 0; i < n; i++) { if (i == 0){ arr.push_back(n); } else{ arr.push_back((n - i) + arr[i - 1] - i); } } int sum = 0; for (int i = 0; i < n; i++) if (s[i] == 'a' s[i] == 'e' s[i] == 'i' s[i] == 'o' s[i] == 'u'){ sum += arr[i]; } return sum; } int main(){ string s = "abc"; cout << vowel(s) << endl; } EPAM CODE Telegram👇 @allcoding1 @allcoding1 @allcoding1 @allcoding1 @allcoding1 @allcoding1 ⚠️ Share post in ur college and telegram Group's

from bisect import bisect_left, insort_left n, d = map(int, input().split()) t = list(map(int, input().split())) noti = 0 las
from bisect import bisect_left, insort_left n, d = map(int, input().split()) t = list(map(int, input().split())) noti = 0 lastd = sorted(t[:d]) def med(): return lastd[d//2] if d % 2 == 1 else ((lastd[d//2] + lastd[d//2-1])/2) for i in range(d, n): if t[i] >= 2*med(): noti += 1 del lastd[bisect_left(lastd,t[i-d])] insort_left(lastd, t[i]) print(noti) EPAM Bank Notification Code Telegram👇 @allcoding1 @allcoding1 @allcoding1 @allcoding1 @allcoding1 @allcoding1 ⚠️ Share post in ur college and telegram Group's

#include using namespace std; int vowel(string s){ int n = s.length(); vector arr; for (int i = 0; i &lt; n; i++) { if (i ==
#include <bits/stdc++.h> using namespace std; int vowel(string s){ int n = s.length(); vector<int> arr; for (int i = 0; i < n; i++) { if (i == 0){ arr.push_back(n); } else{ arr.push_back((n - i) + arr[i - 1] - i); } } int sum = 0; for (int i = 0; i < n; i++) if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i'|| s[i] == 'o' || s[i] == 'u'){ sum += arr[i]; } return sum; } int main(){ string s = "abc"; cout << vowel(s) << endl; } EPAM Telegram - https://t.me/allcoding1

Number search epam Python https://t.me/allcoding1
Number search epam Python https://t.me/allcoding1

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Bike Register code Java Telegram:-@allcoding1
Bike Register code Java Telegram:-@allcoding1

Love letter Code Python Telegram:-@allcoding1
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Love letter Code Python Telegram:-@allcoding1

Two Postive Integers Python 3 Telegram:-@allcoding1
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Two Postive Integers Python 3 Telegram:-@allcoding1