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📈 Análisis del canal de Telegram allcoding1_official
El canal allcoding1_official (@allcoding1_official) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 79 009 suscriptores, ocupando la posición 1 613 en la categoría Tecnologías y Aplicaciones y el puesto 3 828 en la región India.
📊 Métricas de audiencia y dinámica
Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 79 009 suscriptores.
Según los últimos datos del 25 octubre, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -1 654, y en las últimas 24 horas de -52, conservando un alto alcance.
- Estado de verificación: No verificado
- Tasa de interacción (ER): El promedio de interacción de la audiencia es 1.79%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 0.62% de reacciones respecto al total de suscriptores.
- Alcance de las publicaciones: Cada publicación recibe en promedio 1 414 visualizaciones. En el primer día suele acumular 489 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, dev.
📝 Descripción y política de contenido
No se ha proporcionado la descripción del canal.
Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 26 octubre, 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 Tecnologías y Aplicaciones.
79 009
Suscriptores
-5224 horas
-3677 días
-1 65430 días
Archivo de publicaciones
79 009
import math
AC
def is_nprime(x): if x <2: return False for i in range(2, int(math.sqrt(x)) + 1): if x % i == 0: return x == i*i return True
def count_nearly_primes(L, R): nPrimes = 0 for x in range(L+1, R): if is_nprime(x): nPrimes += 1 return nPrimes
L = int(input()) R = int(input()) result = count_nearly..primes(L, R) print(result)
Count of Nearly Primes
Python3
79 009
def count_arrays(N, K, X):
MOD = 1000000007
if N == 1:
return 1 if X == 1 else 0
a = [0] * (N + 1)
b = [0] * (N + 1)
a[1] = 1
b[1] = 0
for i in range(2, N + 1):
a[i] = (a[i - 1] * (K - 2) + b[i - 1] * (K - 1)) % MOD
b[i] = a[i - 1] % MOD
return b[N]
N = int(input())
K = int(input())
X = int(input())
print(count_arrays(N, K, X))
Number Of Array code
Thoughtwork exam
Python3
Telegram:- @allcoding1_official
79 009
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Repost from allcoding1
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Repost from allcoding1
+8
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79 009
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79 009
Repost from allcoding1
+8
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📌Abdul bari courses
📌Ashok IT
📌Linux
📌Networking
📌Design patterns
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📌Entity framework
📌Node.js
📌ASP. Net
📌Aps. Net cro
📌java
📌JavaScript
📌full stack developer
Tutorials + Books + Courses + Trainings + Workshops + Educational Resources
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🔹Learning language resources English , 🇫🇷
All courses (100 rupees)
Contact:- @meterials_available
proof:- https://t.me/+mP2YTsMihZozNWQ1
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📌All programing courses
📌Abdul bari courses
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Tutorials + Books + Courses + Trainings + Workshops + Educational Resources
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🔹Learning language resources English , 🇫🇷
All courses (100 rupees)
Contact:- @meterials_available
79 009
+8
📌IT learning courses
📌All programing courses
📌Abdul bari courses
📌Ashok IT
Tutorials + Books + Courses + Trainings + Workshops + Educational Resources
🔹Data science
🔹Python
🔹Artificial Intelligence
🔹AWS Certified
🔹Cloud
🔹BIG DATA
🔹Data Analytics
🔹BI
🔹Google Cloud Platform
🔹IT Training
🔹MBA
🔹Machine Learning
🔹Deep Learning
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🔹SPSS
🔹Statistics
🔹Data Base
🔹Learning language resources English , 🇫🇷
All courses (100 rupees)
Contact:- @meterials_available
79 009
#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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