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allcoding1

allcoding1

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📈 نظرة تحليلية على قناة تيليجرام allcoding1

تُعد قناة allcoding1 (@allcoding1) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 21 589 مشتركاً، محتلاً المرتبة 9 055 في فئة التعليم والمرتبة 19 081 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 21 589 مشتركاً.

بحسب آخر البيانات بتاريخ 28 أغسطس, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -363، وفي آخر 24 ساعة بمقدار -12، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 4.52‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 1.19‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 977 مشاهدة. وخلال اليوم الأول يجمع عادةً 257 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 0.
  • الاهتمامات الموضوعية: يركز المحتوى على مواضيع رئيسية مثل dsa, stack, namaste, javascript, learning.

📝 الوصف وسياسة المحتوى

وصف القناة غير متوفر.

بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 29 أغسطس, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التعليم.

21 589
المشتركون
-1224 ساعات
-767 أيام
-36330 أيام
أرشيف المشاركات
#include int main() { int n; scanf("%d", &n); char str[n + 1]; scanf("%s", str); int count = 0; char thirdLastConsonant;
#include<stdio.h> int main() { int n; scanf("%d", &n); char str[n + 1]; scanf("%s", str); int count = 0; char thirdLastConsonant; for(int i = n - 1; i >= 0; i--) { if (str[i] != 'a' && str[i] != 'e' && str[i] != 'i' && str[i] != 'o' && str[i] != 'u' && str[i] != 'A' && str[i] != 'E' && str[i] != 'I' && str[i] != 'O' && str[i] != 'U') { count++; if (count == 3) { thirdLastConsonant = str[i]; break; } } } printf("%c\n", thirdLastConsonant); return 0; }

import java.util.Scanner; public class ChangeCase { public static void main(String[] args) { Scanner scanner = new Scanner(Sy
import java.util.Scanner; public class ChangeCase { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); char c = scanner.next().charAt(0); if (Character.isLowerCase(c)) { System.out.println(Character.toUpperCase(c)); } else { System.out.println(Character.toLowerCase(c)); } scanner.close(); } }

1. In order to open a new window in incognito mode of Chrome, you should use: - Answer: B. Ctrl + Shift + N 2. Q05. While configuring your email ID in MS Outlook, which of the following protocols should be used for sending emails? - Answer: D. SMTP 3. Q06. Which of the following network devices operates at the physical layer of the OSI model? - Answer: D. Hub 4. Q07. Which of the following is an example of a lossy compression file? - Answer: A. JPEG 5. Q08. Wireless mouse and keyboard communicate through which of the following? - Answer: A. Electromagnetic signals 6. Q09. MS Outlook will alert you if you forget to: - Answer: B. Attach a file after writing the word "attached" in the email body 7. Q10. If you are receiving all the emails of a client in the SPAM folder, which of the following actions should you take to receive them in the INBOX? - Answer: A. All of the mentioned options 8. Q11. In order to change the slideshow view to normal view in PowerPoint, which of the following can be used? - Answer: B. ESC 9. Q12. In MS Excel, Alt+= is used for which of the following purposes? - Answer: B. AutoSum 10. Q13. Performance information of which of the following is shown in the task manager? - Answer: C. All 1, 2 & 3 (CPU, Memory, GPU) 11. Q14. Ping (command line utility) is used to check which of the following? - Answer: A. Only 1 & 2 (Connectivity between host and server/host, The round-trip time of the packet) 12. Q15. Which of the following memory types in a computer is static as well as non-volatile? - Answer: A. ROM

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