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ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

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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 14 133 مشتركاً، محتلاً المرتبة 14 094 في فئة التعليم والمرتبة 28 066 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 14 133 مشتركاً.

بحسب آخر البيانات بتاريخ 25 سبتمبر, 2026، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -111، وفي آخر 24 ساعة بمقدار -3، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 3.69‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً 1.55‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 522 مشاهدة. وخلال اليوم الأول يجمع عادةً 219 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 25 سبتمبر, 2026) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التعليم.

14 133
المشتركون
-324 ساعات
-157 أيام
-11130 أيام
أرشيف المشاركات
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from collections import deque, defaultdict def Solve(n, r, t, k, s, e): graph = defaultdict(list) for u, v in r: graph[u].append(v) graph[v].append(u) visited_t = set(t) queue_t = deque([(th, 0) for th in t]) while queue_t: current_t, distance_t = queue_t.popleft() if distance_t < k: for neighbor in graph[current_t]: if neighbor not in visited_t: visited_t.add(neighbor) queue_t.append((neighbor, distance_t + 1)) visited_s = set() queue_s = deque([(s, 0)]) while queue_s: current_s, distance_s = queue_s.popleft() if current_s in visited_s or current_s in visited_t: continue if current_s == e: return distance_s visited_s.add(current_s) for neighbor in graph[current_s]: if neighbor not in visited_s and neighbor not in visited_t: queue_s.append((neighbor, distance_s + 1)) return -1 Uber ✅

typedef long long ll; ll solution(int n, vector a1, vector a2) { vector left(n + 1, 0), right(n + 2, 0), prefixSum(n + 1, 0), maxL(n + 1, 0), maxM(n + 1, LLONG_MIN); for (int i = 1; i <= n; i++) { left[i] = i == 1 ? a1[0] : max(left[i - 1] + (ll)a1[i - 1], (ll)a1[i - 1]); } ll res = LLONG_MIN; for (int i = 1; i <= n; i++) { res = max(res, left[i]); } for (int i = n; i >= 1; i--) { right[i] = i == n ? (ll)a1[n - 1] : max(right[i + 1] + (ll)a1[i - 1], (ll)a1[i - 1]); } for (int i = 1; i <= n; i++) { prefixSum[i] = prefixSum[i - 1] + (ll)a2[i - 1]; } maxL[1] = left[0] - prefixSum[0]; maxM[1] = maxL[1]; for (int i = 2; i <= n; i++) { maxL[i] = left[i - 1] - prefixSum[i - 1]; maxM[i] = max(maxM[i - 1], maxL[i]); } for (int j = 1; j <= n; j++) { res = max(res, maxM[j] + prefixSum[j] + right[j + 1]); } return res; } Tichnas sir and bonds mam Uber ✅

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#include <iostream> #include <string> #include <unordered_map> #include <queue> using namespace std; string reorganizeString(string s) {     unordered_map<char, int> freq_map;     for (char c : s) {         freq_map[c]++;     }     priority_queue<pair<int, char>> max_heap;     for (auto &[ch, freq] : freq_map) {         max_heap.push({freq, ch});     }     string res;     while (max_heap.size() >= 2) {         auto [freq1, char1] = max_heap.top(); max_heap.pop();         auto [freq2, char2] = max_heap.top(); max_heap.pop();         res += char1;         res += char2;         if (--freq1 > 0) max_heap.push({freq1, char1});         if (--freq2 > 0) max_heap.push({freq2, char2});     }     if (!max_heap.empty()) {         auto [freq, ch] = max_heap.top();         if (freq > 1) return "";         res += ch;     }     return res; } int main() {     string input;     cout << "Enter a string: ";     cin >> input;     string result = reorganizeString(input);     if (result.empty()) {         cout << "The string cannot be reorganized to avoid adjacent repeating characters." << endl;     } else {         cout << "Reorganized string: " << result << endl;     }     return 0; } IBM ✅

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