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allcoding1

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📈 نظرة تحليلية على قناة تيليجرام allcoding1

تُعد قناة allcoding1 (@allcoding1) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 21 584 مشتركاً، محتلاً المرتبة 9 062 في فئة التعليم والمرتبة 19 025 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 21 584 مشتركاً.

بحسب آخر البيانات بتاريخ 29 أغسطس, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -360، وفي آخر 24 ساعة بمقدار -8، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 5.78‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 1.10‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 1 247 مشاهدة. وخلال اليوم الأول يجمع عادةً 238 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 30 أغسطس, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التعليم.

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public static int findMaximumGreatness(int[] arr) { int n = arr.length; Arrays.sort(arr); int greatness = 0; int j = 0; boole
public static int findMaximumGreatness(int[] arr) { int n = arr.length; Arrays.sort(arr); int greatness = 0; int j = 0; boolean[] used = new boolean[n]; for (int i = 0; i < n; i++) { while (j < n && (used[j] || arr[j] <= arr[i])) { j++; } if (j < n && arr[j] > arr[i]) { used[j] = true; greatness++; j++; } } return greatness; }

public static int findEarliestMeetingTime(List&gt; schedules, int length) { int DAY_END = 24 * 60; List allMeetings = new Arr
public static int findEarliestMeetingTime(List<List<int[]>> schedules, int length) { int DAY_END = 24 * 60; List<int[]> allMeetings = new ArrayList<>(); for (List<int[]> employeeMeetings : schedules) { allMeetings.addAll(employeeMeetings); } Collections.sort(allMeetings, (a, b) -> Integer.compare(a[0], b[0])); int earliestAvailable = 0; for (int[] meeting : allMeetings) { if (earliestAvailable + length <= meeting[0]) { return earliestAvailable; } earliestAvailable = Math.max(earliestAvailable, meeting[1]); } if (earliestAvailable + length <= DAY_END) { return earliestAvailable; } return -1; }

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MountBlue1-10.pdf1.86 KB

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