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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
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Athletes code Accenture Hackdiva
Athletes code Accenture Hackdiva

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Latin Alpha Numerals Accenture Hackdiva Code
Latin Alpha Numerals Accenture Hackdiva Code

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LI.GCD code in another way of solving
LI.GCD code in another way of solving

LI Gcd Code in python Infosys
LI Gcd Code in python Infosys

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def f(S): n = len(S) lf, rf = {}, {} ls, rs = set(), set() # Initialize the rf map and rs with the entire string S for c in S: rf[c] = rf.get(c, 0) + 1 rs.add(c) mx = 0 # Traverse the string and adjust the ls and rs sets and maps for i in range(n - 1): c = S[i] lf[c] = lf.get(c, 0) + 1 rf[c] -= 1 if rf[c] == 0: rs.remove(c) ls.add(c) cs = len(ls) + len(rs) mx = max(mx, cs) return n - mx // spilit screen @allcoding1_official

public static int solve(int N, int[] A) { int t = 0; for (int num : A) { t += num; } int x = 0; int y = 0; for (int i = 0; i < N; i++) { int res = t - x - A[i]; if (x == res) { y++; } x += A[i]; } return y; } //Equilibrium Point

def GetAnswer(N, A, B, P): dp = [0] * (N + 1) max_d = 0 for i in range(N - 1, -1, -1): max_p = P[i] min_p = P[i] for j in range(1, B + 1): if i + j <= N: max_p = max(max_p, P[i + j - 1]) min_p = min(min_p, P[i + j - 1]) max_d = max(max_d, max_p - min_p) if i + 1 == N: dp[i] = max(dp[i], max_d) else: dp[i] = max(dp[i], max_d, dp[i + 1]) return dp[0] import sys input = sys.stdin.read data = input().split() N = int(data[0]) A = int(data[1]) B = int(data[2]) P = list(map(int, data[3:])) result = GetAnswer(N, A, B, P) print(result) // XOR formaating code