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allcoding1_official

allcoding1_official

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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 85 605 suscriptores, ocupando la posición 1 508 en la categoría Tecnologías y Aplicaciones y el puesto 3 489 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 85 605 suscriptores.

Según los últimos datos del 22 junio, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -1 409, y en las últimas 24 horas de -59, conservando un alto alcance.

  • Estado de verificación: No verificado
  • Tasa de interacción (ER): El promedio de interacción de la audiencia es 2.98%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 0.69% de reacciones respecto al total de suscriptores.
  • Alcance de las publicaciones: Cada publicación recibe en promedio 2 554 visualizaciones. En el primer día suele acumular 589 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 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 23 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 Tecnologías y Aplicaciones.

85 605
Suscriptores
-5924 horas
-2677 días
-1 40930 días
Archivo de publicaciones
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Infosys Exam Send a Questions

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Infosys Exam codes

def solve(S: str) -> int: def cost(ch): pos=[i for c in enumerate(S) if c==ch] k=len(pos) if k<=1: return 0 q=[pos[i]-i for i in range(k)] m=q[k//2] return sum(abs(x-m) for x in q) return min(cost('0'),cost('1'))

import sys input = sys.stdin.readline def solve(N: int, V: list) -> int: if N == 1: return V[0] A = [[0] * N for _ in range(N)] B = [[0] * N for _ in range(N)] for gap in range(1, N): for l in range(N - gap): r = l + gap best = -10**30 for step in (1, 2): nl = l + step if nl >= r: best = max(best, 0) else: best = max(best, V[nl] + B[nl][r]) A[l][r] = best worst = 10**30 for step in (1, 2): nr = r - step if nr <= l: worst = min(worst, 0) else: worst = min(worst, A[l][nr]) B[l][r] = worst return V[0] + A[0][N - 1] if name == "main": data = list(map(int, sys.stdin.read().split())) N = data[0] V = data[1:1 - N] print(solve(N, V))