现已上线！2025 年 Telegram 研究 — 年度关键洞察 
allcoding1
前往频道在 Telegram
📈 Telegram 频道 allcoding1 的分析概览
频道 allcoding1 (@allcoding1) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 22 558 名订阅者，在 教育 类别中位列第 8 854，并在 印度 地区排名第 19 507 位。
📊 受众指标与增长动态
自 невідомо 创建以来，项目保持高速增长，吸引了 22 558 名订阅者。
根据 14 六月, 2026 的最新数据，频道保持稳定运转。过去 30 天订阅人数变化为 -445，过去 24 小时变化为 -14，整体触达仍然可观。
- 认证状态： 未认证
- 互动率 (ER)： 平均受众互动率为 6.31%。内容发布后 24 小时内通常能获得 1.25% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 1 423 次浏览，首日通常累积 282 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 2。
- 主题关注点： 内容集中在 dsa, stack, namaste, javascript, learning 等核心主题上。
📝 描述与内容策略
尚未提供频道描述。
凭借高频更新（最新数据采集于 15 六月, 2026），频道始终保持新鲜度与高覆盖。分析显示受众积极互动，使其成为 教育 类别中的关键影响点。
22 558
订阅者
-1424 小时
-947 天
-44530 天
帖子存档
22 558
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@allcoding1_official
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22 558
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Bootcamp 2.0 Course
@allcoding1_official
> What you’ll learn:
• HTML and DOM
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• Projects- HTML & CSS
• JavaScript Refresher
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> Download link:
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22 558
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22 558
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22 558
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22 558
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22 558
#include <iostream>
#include <vector>
using namespace std;
class SinglyLinkedListNode {
public:
string data;
SinglyLinkedListNode* next;
SinglyLinkedListNode(string value) {
data = value;
next = nullptr;
}
};
SinglyLinkedListNode* getShoppingCart(SinglyLinkedListNode* head, vector<vector<string>> queries) {
for (auto& query : queries) {
string action = query[0];
string item_name = query[1];
if (action == "PUSH_HEAD") {
SinglyLinkedListNode* new_node = new SinglyLinkedListNode(item_name);
new_node->next = head;
head = new_node;
} else if (action == "PUSH_TAIL") {
SinglyLinkedListNode* new_node = new SinglyLinkedListNode(item_name);
if (head == nullptr) {
head = new_node;
} else {
SinglyLinkedListNode* current = head;
while (current->next != nullptr) {
current = current->next;
}
current->next = new_node;
}
} else if (action == "POP_HEAD") {
if (head != nullptr) {
SinglyLinkedListNode* temp = head;
head = head->next;
delete temp;
}
}
}
return head;
}
Amazon
22 558
#include <iostream>
#include <vector>
using namespace std;
long makePowerNondecreasing(vector<int>& power) {
int n = power.size();
long adjustments = 0;
for (int i = 1; i < n; ++i) {
if (power[i] < power[i - 1]) {
adjustments += power[i - 1] - power[i];
}
}
return adjustments;
}
Amazon
Telegram:- @allcoding1
