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MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

Kanalga Telegram’da oβ€˜tish

πŸ”₯Guys plz Stop fearing for daily exams πŸ“ πŸ‘¨β€πŸ’» @srksvk is here to help you all at lowest cost possible.πŸ’ͺ πŸŒ€ ” Our Only Aim Is To Let Get Placed To You In A Reputed Company πŸ”₯Effort from our side = πŸ’― πŸ“±Main Channel: @coding_are πŸ“±Tel I'd : @srksvk

Ko'proq ko'rsatish

πŸ“ˆ Telegram kanali MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer analitikasi

MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 13 242 obunachidan iborat bo'lib, TaΚΌlim toifasida 15 362-o'rinni va Hindiston mintaqasida 32 092-o'rinni egallagan.

πŸ“Š Auditoriya koβ€˜rsatkichlari va dinamika

Π½Π΅Π²Ρ–Π΄ΠΎΠΌΠΎ sanasidan buyon loyiha tez oβ€˜sib, 13 242 obunachiga ega boβ€˜ldi.

18 Iyun, 2026 dagi oxirgi ma’lumotlarga koβ€˜ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -138 ga, soβ€˜nggi 24 soatda esa -2 ga oβ€˜zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya oβ€˜rtacha 2.93% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.11% ini tashkil etuvchi reaksiyalarni toβ€˜playdi.
  • Post qamrovi: Har bir post oβ€˜rtacha 388 marta koβ€˜riladi; birinchi sutkada odatda 147 ta koβ€˜rish yigβ€˜iladi.
  • Reaksiyalar va oβ€˜zaro ta’sir: Auditoriya faol: har bir postga oβ€˜rtacha 2 ta reaksiya keladi.
  • Tematik yoβ€˜nalishlar: Kontent placement, gaurntee, suree, capgemini, infosy kabi asosiy mavzularga jamlangan.

πŸ“ Tavsif va kontent siyosati

Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
β€œπŸ”₯Guys plz Stop fearing for daily exams πŸ“ πŸ‘¨β€πŸ’» @srksvk is here to help you all at lowest cost possible.πŸ’ͺ πŸŒ€ ” Our Only Aim Is To Let Get Placed To You In A Reputed Company πŸ”₯Effort from our side = πŸ’― πŸ“±Main Channel: @coding_are πŸ“±Tel I'd : @srks...”

Yuqori yangilanish chastotasi (oxirgi ma’lumot 19 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli boβ€˜lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni TaΚΌlim toifasidagi muhim ta’sir nuqtasiga aylantirishini koβ€˜rsatadi.

13 242
Obunachilar
-224 soatlar
-457 kunlar
-13830 kunlar
Postlar arxiv
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All are correct code with verified ☺️ So do correctly everyone

using System; using System.Collections.Generic; class Program { static void Main() { int N = int.Parse(Console.ReadLine()); var graph = new Dictionary>(); var indegrees = new Dictionary(); for (int i = 0; i < N; i++) { var edge = Console.ReadLine().Split(); string from = edge[0]; string to = edge[1]; if (!graph.ContainsKey(from)) graph[from] = new List(); graph[from].Add(to); if (!indegrees.ContainsKey(from)) indegrees[from] = 0; if (!indegrees.ContainsKey(to)) indegrees[to] = 0; indegrees[to]++; } var words = Console.ReadLine().Split(); var levels = new Dictionary(); var queue = new Queue(); foreach (var node in indegrees.Keys) { if (indegrees[node] == 0) { levels[node] = 1; // Root level is 1 queue.Enqueue(node); } } while (queue.Count > 0) { var current = queue.Dequeue(); int currentLevel = levels[current]; if (graph.ContainsKey(current)) { foreach (var neighbor in graph[current]) { if (!levels.ContainsKey(neighbor)) { levels[neighbor] = currentLevel + 1; queue.Enqueue(neighbor); } indegrees[neighbor]--; if (indegrees[neighbor] == 0 && !levels.ContainsKey(neighbor)) { queue.Enqueue(neighbor); } } } } int totalValue = 0; foreach (var word in words) { if (levels.ContainsKey(word)) { totalValue += levels[word]; } else { totalValue += -1; } } Console.Write(totalValue); } } String Puzzle C+

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int main() { string s; cin >> s; int n = s.length(), res = 0; vector<int> v(n); for (int i = 0; i < n; ++i) cin >> v[i]; int lw = s[0] - '0', lwv = v[0]; for (int i = 1; i < n; ++i) { if (s[i] - '0' == lw) { res += min(lwv, v[i]); lwv = max(lwv, v[i]); } else { lw = s[i] - '0'; lwv = v[i]; } } cout << res; return 0; } Form alternating string Change accordingly before submitting to avoid plagiarism @codeing_are

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