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📈 Аналитический обзор Telegram-канала allcoding1_official

Канал allcoding1_official (@allcoding1_official) языкового сегмента Английский является активным участником. Сейчас сообщество объединяет 85 687 подписчиков, занимая 1 509 место в категории Технологии и приложения и 3 512 место в регионе Индия.

📊 Показатели аудитории и динамика

С момента создания невідомо проект демонстрирует стремительный рост, собрав аудиторию из 85 687 подписчиков.

Согласно последним данным от 20 июня, 2026, канал показывает стабильную активность. За последние 30 дней изменение числа участников составило -1 460, а за последние 24 часа — -39, при этом общий охват остаётся высоким.

  • Статус верификации: Не верифицирован
  • Уровень вовлечённости (ER): Средний показатель вовлечённости аудитории составляет 3.36%. В первые 24 часа после публикации контент обычно набирает 0.73% реакций от общего числа подписчиков.
  • Охват публикаций: В среднем каждый пост получает 2 882 просмотров. В течение первых суток публикация набирает 625 просмотров.
  • Реакции и взаимодействия: Аудитория активно поддерживает контент: среднее количество реакций на один пост — 1.
  • Тематические интересы: Контент сосредоточен на ключевых темах, таких как dsa, stack, namaste, javascript, dev.

📝 Описание и контентная политика

Описание канала не предоставлено.

Благодаря высокой частоте обновлений (последние данные получены 21 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 Технологии и приложения.

85 687
Подписчики
-3924 часа
-3267 дней
-1 46030 день
Архив постов
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))

You are given a string S consisting of lowercase latin letters, i.e. {a, b,c, ..z} You can perform the following operation on S any number times Remove two consecutive characters. Find the total number of distinct strings that you can generate. Note: All operations are mutually exclusive. This means that all operations are going to be performed independently on the initial string S. Input Format The first line contains a string, S. denoting the given string. Constraints 1 <= len(S) <= 10^5 Sample Test Cases Case 1 Input aaabcc Output 4 Explanation: S = "aaabcc" we can get the strings "abcc", "aacc", "aaac", and "aaab". Hence, the answer for this case is equal to 4. Case 2 Input aaaaaaaaaa Output 1 Explanation: S = "аaаааааааа" We can get only one string which is "aaaaaaaaaa". Hence, the answer for this case is equal to 1. Case 3 Input: abcdef Output: 5 Explanation: S = "abcdef" We can get the strings "cdef", "adef", "abef", "abcf", and "abcd". Hence, the answer for this case is equal to 5.

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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