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📈 Análisis del canal de Telegram MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer

El canal MTHREE exam help ! Infosys exam help ! Cognizant exam help ! Amazon exam answer (@coding_are) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 13 252 suscriptores, ocupando la posición 15 363 en la categoría Educación y el puesto 32 287 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 13 252 suscriptores.

Según los últimos datos del 15 junio, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de 92, y en las últimas 24 horas de -11, conservando un alto alcance.

  • Estado de verificación: No verificado
  • Tasa de interacción (ER): El promedio de interacción de la audiencia es 2.94%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 1.28% de reacciones respecto al total de suscriptores.
  • Alcance de las publicaciones: Cada publicación recibe en promedio 390 visualizaciones. En el primer día suele acumular 170 visualizaciones.
  • Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 1.
  • Intereses temáticos: El contenido se centra en temas clave como placement, gaurntee, suree, capgemini, infosy.

📝 Descripción y política de contenido

El autor describe el recurso como un espacio para expresar opiniones subjetivas:
🔥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...

Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 16 junio, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Educación.

13 252
Suscriptores
-1124 horas
-487 días
+9230 días
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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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