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allcoding1

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📈 Análisis del canal de Telegram allcoding1

El canal allcoding1 (@allcoding1) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 21 578 suscriptores, ocupando la posición 9 062 en la categoría Educación y el puesto 19 025 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 21 578 suscriptores.

Según los últimos datos del 29 agosto, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -360, y en las últimas 24 horas de -8, conservando un alto alcance.

  • Estado de verificación: No verificado
  • Tasa de interacción (ER): El promedio de interacción de la audiencia es 5.78%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 1.10% de reacciones respecto al total de suscriptores.
  • Alcance de las publicaciones: Cada publicación recibe en promedio 1 247 visualizaciones. En el primer día suele acumular 238 visualizaciones.
  • Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 0.
  • Intereses temáticos: El contenido se centra en temas clave como dsa, stack, namaste, javascript, learning.

📝 Descripción y política de contenido

No se ha proporcionado la descripción del canal.

Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 30 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.

21 578
Suscriptores
-824 horas
-787 días
-36030 días
Archivo de publicaciones
#include <iostream> #include <vector> #include <climits> #include <algorithm> using namespace std; int minimum_energy_cost_recursive(int index,int idxy, int left_consecutive, int right_consecutive, int N, int X, int Y, int El, int Er, vector<int>& weights) {     // Base case: if we have reached the end of bags     if (index > idxy) {         return 0;     }         // Calculate the cost of picking from the left and the right     int cost_left = weights[index] * X + (left_consecutive ? El : 0) +                     minimum_energy_cost_recursive(index + 1,idxy, 1, 0, N, X, Y, El, Er, weights);     int cost_right = weights[idxy] * Y + (right_consecutive ? Er : 0) +                      minimum_energy_cost_recursive(index ,idxy-1, 0, 1, N, X, Y, El, Er, weights);         // Return the minimum of both choices     return min(cost_left, cost_right); } int main() {     int N, X, Y, El, Er;     cin >> N >> X >> Y >> El >> Er;         vector<int> weights(N);     for (int i = 0; i < N; ++i) {         cin >> weights[i];     }         cout << minimum_energy_cost_recursive(0,N-1, 0, 0, N, X, Y, El, Er, weights) << endl;         return 0; } Amazon Hackon exam

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UI_UX Design Tips (1).pdf7.77 MB