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allcoding1

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 22 590 suscriptores, ocupando la posición 8 822 en la categoría Educación y el puesto 19 518 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 22 590 suscriptores.

Según los últimos datos del 12 junio, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -437, y en las últimas 24 horas de -6, conservando un alto alcance.

  • Estado de verificación: No verificado
  • Tasa de interacción (ER): El promedio de interacción de la audiencia es 5.99%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 1.25% de reacciones respecto al total de suscriptores.
  • Alcance de las publicaciones: Cada publicación recibe en promedio 1 353 visualizaciones. En el primer día suele acumular 283 visualizaciones.
  • Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 2.
  • 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 13 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.

22 590
Suscriptores
-624 horas
-967 días
-43730 días
Archivo de publicaciones
def max_sum_of_distinct_characters(S): n = len(S) left_chars = set() right_chars = set() left_count = [0] * n right_count = [0] * n for i in range(n): left_chars.add(S[i]) left_count[i] = len(left_chars) for i in range(n-1, -1, -1): right_chars.add(S[i]) right_count[i] = len(right_char) max_sum = 0 for i in range(n-1): max_sum = max(max_sum, left_count[i] + right_count[i+1]) return n- max_sum Split String code Python 3 All passed

import sys det solve(N, A) for i in range(N): if A[i]=0; A[1] ps=0 M-1 pm={0:-1) for i in range(N): ps+=A[i] if ps in pm: m=max(m,i-pm[ps]) else: pm[ps]=i return m def main(): Nint(sys.stdin.readline().strip()) A-[] for_ in range(N): A.append(int(sys.stdin.readline().strip())) result = solve(N, A) print(result) Largest Subarray with equal number Infosys

def solve(N, A): unique_sums = set() for start in range(N): current_sum = 0 for end in range(start, N): current_sum += A[end] unique_sums.add(current_sum) print(len(unique_sums))

Minimum substring ..
Minimum substring ..

def count_distinct_strings(S): distinct_strings = set() for i in range(len(S) - 1): new_string = S[:i] + S[i+2:] distinct_strings.add(new_string) return len(distinct_strings) # Read input string S = input().strip() # Get the number of distinct strings that can be generated result = count_distinct_strings(S) print(result)

def minimum_unique_sum(A): N = len(A) A.sort() total = A[0] for i in range(1, N): if A[i] <= A[i-1]: A[i] = A[i-1] + 1 total += A[i] return total # Input format N = int(input()) A = [] for i in range(N): A.append(int(input())) # Output result = minimum_unique_sum(A) print(result)

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def split_string_cost(S): # Length of the string S len_S = len(S) # To store the cost of the split parts max_cost = 0 # Set to keep track of distinct characters in the first part distinct_chars_A = set() # List to keep track of the cost for the second part from each split position cost_B = [0] * len_S # Set to keep track of distinct characters in the second part distinct_chars_B = set() # Calculate cost for second part from the end for i in range(len_S - 1, -1, -1): distinct_chars_B.add(S[i]) cost_B[i] = len(distinct_chars_B) # Calculate maximum sum of cost for parts A and B for i in range(len_S - 1): distinct_chars_A.add(S[i]) cost_A = len(distinct_chars_A) cost = cost_A + cost_B[i + 1] max_cost = max(max_cost, cost) # Calculate the result as |S| - X result = len_S - max_cost return result # Example usage S = "aaabbb" print(split_string_cost(S)) # Output: 3

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