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allcoding1

allcoding1

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📈 Аналитический обзор Telegram-канала allcoding1

Канал allcoding1 (@allcoding1) языкового сегмента Английский является активным участником. Сейчас сообщество объединяет 22 590 подписчиков, занимая 8 822 место в категории Образование и 19 518 место в регионе Индия.

📊 Показатели аудитории и динамика

С момента создания невідомо проект демонстрирует стремительный рост, собрав аудиторию из 22 590 подписчиков.

Согласно последним данным от 12 июня, 2026, канал показывает стабильную активность. За последние 30 дней изменение числа участников составило -437, а за последние 24 часа — -6, при этом общий охват остаётся высоким.

  • Статус верификации: Не верифицирован
  • Уровень вовлечённости (ER): Средний показатель вовлечённости аудитории составляет 5.99%. В первые 24 часа после публикации контент обычно набирает 1.25% реакций от общего числа подписчиков.
  • Охват публикаций: В среднем каждый пост получает 1 353 просмотров. В течение первых суток публикация набирает 283 просмотров.
  • Реакции и взаимодействия: Аудитория активно поддерживает контент: среднее количество реакций на один пост — 2.
  • Тематические интересы: Контент сосредоточен на ключевых темах, таких как dsa, stack, namaste, javascript, learning.

📝 Описание и контентная политика

Описание канала не предоставлено.

Благодаря высокой частоте обновлений (последние данные получены 13 июня, 2026) канал поддерживает актуальность и высокий уровень охвата публикаций. Аналитика показывает, что аудитория активно взаимодействует с контентом, что делает его важной точкой влияния в категории Образование.

22 590
Подписчики
-624 часа
-967 дней
-43730 день
Архив постов
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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