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لا تقع ضحية للمخادعين! تيليمتريو يكتشف ويُميّز هذه القنوات 👉 إذا كنت تريد رؤية العلامة، اشترك 👈

ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

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🔥Guys plz Stop fearing for daily exams 📝 👨‍💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srksvk

إظهار المزيد

📈 نظرة تحليلية على قناة تيليجرام ACCENTURE EXAM SOLUTIONS

تُعد قناة ACCENTURE EXAM SOLUTIONS (@coding_are) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 14 042 مشتركاً، محتلاً المرتبة 14 513 في فئة التعليم والمرتبة 28 436 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 14 042 مشتركاً.

بحسب آخر البيانات بتاريخ 13 أكتوبر, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -124، وفي آخر 24 ساعة بمقدار -6، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 3.63‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 1.83‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 510 مشاهدة. وخلال اليوم الأول يجمع عادةً 257 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 2.
  • الاهتمامات الموضوعية: يركز المحتوى على مواضيع رئيسية مثل placement, gaurntee, suree, capgemini, infosy.

📝 الوصف وسياسة المحتوى

يصف المؤلف القناة بأنها مساحة للتعبير عن الآراء الذاتية:
“🔥Guys plz Stop fearing for daily exams 📝 👨‍💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srks...”

بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 14 أكتوبر, 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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int f(int i, int j, int flag, vector& arr, int sum, vector>>& dp) {     if (i >= j) return 0;     if (dp[i][j][flag] != -1) return dp[i][j][flag];     int firsttwo = 0;     int lasttwo = 0;     int firstlast = 0;     if (i + 1 < arr.size() && arr[i] + arr[i + 1] == sum) {         firsttwo = 1 + f(i + 2, j, flag, arr, sum, dp);     }     if (arr[i] + arr[j] == sum) {         firstlast = 1 + f(i + 1, j - 1, flag, arr, sum, dp);     }     if (j >= 0 && arr[j] + arr[j - 1] == sum) {         lasttwo = 1 + f(i, j - 2, flag, arr, sum, dp);     }     return dp[i][j][flag] = max({firsttwo, lasttwo, firstlast}); } int solution(vector& arr) {     int n = arr.size();     vector>> dp(n, vector>(n, vector(3, -1)));     int sum1 = arr[0] + arr[n - 1];     int sum2 = arr[0] + arr[1];     int sum3 = arr[n - 1] + arr[n - 2];     int ans1 = f(0, n - 1, 0, arr, sum1, dp);     int ans2 = f(0, n - 1, 1, arr, sum2, dp);     int ans3 = f(0, n - 1, 2, arr, sum3, dp);     int res = max(ans1, max(ans2, ans3));     return res; }

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ACCENTURE EXAM SOLUTIONS - إحصائيات وتحليلات قناة تيليجرام @coding_are