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AMAZON EXAM SOLUTIONS

AMAZON 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

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📈 Análisis del canal de Telegram AMAZON EXAM SOLUTIONS

El canal AMAZON EXAM SOLUTIONS (@coding_are) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 14 368 suscriptores, ocupando la posición 14 155 en la categoría Educación y el puesto 29 163 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 14 368 suscriptores.

Según los últimos datos del 02 agosto, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de 1 172, y en las últimas 24 horas de 177, conservando un alto alcance.

  • Estado de verificación: No verificado
  • Tasa de interacción (ER): El promedio de interacción de la audiencia es 7.73%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 2.73% de reacciones respecto al total de suscriptores.
  • Alcance de las publicaciones: Cada publicación recibe en promedio 1 111 visualizaciones. En el primer día suele acumular 392 visualizaciones.
  • Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 3.
  • 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 03 agosto, 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.

14 368
Suscriptores
+17724 horas
+9637 días
+1 17230 días
Archivo de publicaciones
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2nd linkdin code✅

public static List getMinSum(List arr, List query, int k) { List result = new ArrayList<>(); int n = arr.size(); for (int z : query) { int minSum = Integer.MAX_VALUE; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { int x = arr.get(i); int y = arr.get(j); if (x == 0 && y == 0) continue; if (x != 0 && z % x == 0) { int a = z / x, b = 0; if (a >= 0) minSum = Math.min(minSum, a + b); } if (y != 0 && z % y == 0) { int a = 0, b = z / y; if (b >= 0) minSum = Math.min(minSum, a + b); } if (x != 0) { for (int a = 0; a <= k; a++) { int rem = z - a * x; if (rem < 0) break; if (y != 0 && rem % y == 0) { int b = rem / y; if (b >= 0 && a + b <= k) { minSum = Math.min(minSum, a + b); } } } } } } result.add(minSum <= k ? minSum : -1); } return result; }

Linkdin first' code✅

import java.util.*; class Result { public static int findMinOperations(List<Integer> arr) { int n = arr.size(); int operations = 0; int rootCount = 0; for (int x : arr) { if (x == -1) rootCount++; } if (rootCount == 0) { arr.set(0, -1); operations++; } else if (rootCount > 1) { boolean found = false; for (int i = 0; i < n; i++) { if (arr.get(i) == -1) { if (!found) { found = true; } else { arr.set(i, 1); operations++; } } } } for (int i = 0; i < n; i++) { int parent = arr.get(i); if (parent != -1 && (parent < 1 || parent > n)) { arr.set(i, 1); operations++; } } boolean[] visited = new boolean[n]; boolean[] inStack = new boolean[n]; for (int i = 0; i < n; i++) { if (!visited[i]) { operations += dfsFixCycle(arr, visited, inStack, i); } } return operations; } private static int dfsFixCycle(List<Integer> arr, boolean[] visited, boolean[] inStack, int node) { int operations = 0; while (node != -1 && !visited[node]) { visited[node] = true; inStack[node] = true; int parent = arr.get(node) == -1 ? -1 : arr.get(node) - 1; if (parent != -1) { if (!visited[parent]) { node = parent; } else if (inStack[parent]) { arr.set(node, -1); operations++; break; } else { break; } } else { break; } } Arrays.fill(inStack, false); return operations; } } public class Solution { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); List<Integer> arr = new ArrayList<>(); for (int i = 0; i < n; i++) { arr.add(sc.nextInt()); } System.out.println(Result.findMinOperations(arr)); } }