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📈 Аналитический обзор Telegram-канала allcoding1_official

Канал allcoding1_official (@allcoding1_official) языкового сегмента Английский является активным участником. Сейчас сообщество объединяет 85 678 подписчиков, занимая 1 508 место в категории Технологии и приложения и 3 501 место в регионе Индия.

📊 Показатели аудитории и динамика

С момента создания невідомо проект демонстрирует стремительный рост, собрав аудиторию из 85 678 подписчиков.

Согласно последним данным от 21 июня, 2026, канал показывает стабильную активность. За последние 30 дней изменение числа участников составило -1 434, а за последние 24 часа — -33, при этом общий охват остаётся высоким.

  • Статус верификации: Не верифицирован
  • Уровень вовлечённости (ER): Средний показатель вовлечённости аудитории составляет 3.09%. В первые 24 часа после публикации контент обычно набирает 0.73% реакций от общего числа подписчиков.
  • Охват публикаций: В среднем каждый пост получает 2 645 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 22 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.

85 678
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-1 43430 день
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If what's promotions👇 👇👇 @Priya_i

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Repost from allcoding1
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Repost from allcoding1
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#include <iostream> #include <vector> #include <climits> #include <algorithm> using namespace std; int minimum_energy_cost_recursive(int index,int idxy, int left_consecutive, int right_consecutive, int N, int X, int Y, int El, int Er, vector<int>& weights) {     // Base case: if we have reached the end of bags     if (index > idxy) {         return 0;     }         // Calculate the cost of picking from the left and the right     int cost_left = weights[index] * X + (left_consecutive ? El : 0) +                     minimum_energy_cost_recursive(index + 1,idxy, 1, 0, N, X, Y, El, Er, weights);     int cost_right = weights[idxy] * Y + (right_consecutive ? Er : 0) +                      minimum_energy_cost_recursive(index ,idxy-1, 0, 1, N, X, Y, El, Er, weights);         // Return the minimum of both choices     return min(cost_left, cost_right); } int main() {     int N, X, Y, El, Er;     cin >> N >> X >> Y >> El >> Er;         vector<int> weights(N);     for (int i = 0; i < N; ++i) {         cin >> weights[i];     }         cout << minimum_energy_cost_recursive(0,N-1, 0, 0, N, X, Y, El, Er, weights) << endl;         return 0; } Amazon Hackon exam Telegram:-

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