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📈 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 21 533 名订阅者,在 教育 类别中位列第 9 159,并在 印度 地区排名第 19 101

📊 受众指标与增长动态

невідомо 创建以来,项目保持高速增长,吸引了 21 533 名订阅者。

根据 01 九月, 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 -367,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 7.16%。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 1 543 次浏览,首日通常累积 270 次浏览。
  • 互动与反馈: 受众积极参与,单帖平均反应数为 0
  • 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 02 九月, 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。

21 533
订阅者
-1424 小时
-797 天
-36730 天
帖子存档
photo content
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def PowerLimit(a, b): current_resource, max_seconds = 1, 0 while a > 0 and b > 0: if a > b: current_resource, a, b = 1, a - 2
def PowerLimit(a, b):   current_resource, max_seconds = 1, 0   while a > 0 and b > 0:     if a > b:       current_resource, a, b = 1, a - 2, b + 1     else:       current_resource, b, a = 2, b - 2, a + 1     max_seconds += 1   max_seconds += a // 2 if current_resource == 1 else b // 2   return max_seconds power limit ✅ Telegram:- @allcoding1

long long OneBlock(int N, vector<int> Arr) {     vector<int> one_indices;     for (int i = 0; i < N; ++i) {         if (Arr[i] == 1) {             one_indices.push_back(i);         }     }     if (one_indices.size() <= 1) {         return 1;     }     long long ways = 1;     for (int i = 1; i < one_indices.size(); ++i) {         int zeros_between_ones = one_indices[i] - one_indices[i - 1] - 1;         ways *= (zeros_between_ones + 1);     }     return ways; } //one block C++ Telegram:- @allcoding1_official

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def count(a, X): &nbsp;&nbsp;&nbsp; a.sort() &nbsp;&nbsp;&nbsp; ans = 0 &nbsp;&nbsp;&nbsp; n = len(a) &nbsp;&nbsp;&nbsp; for
def count(a, X):     a.sort()     ans = 0     n = len(a)     for i in range(n):         if (i < n // 2):             ans += max(0, a[i] - X)         elif (i == n // 2):             ans += abs(X - a[i])         else:             ans += max(0, X - a[i]);     return ans IBM Telegram:- @allcoding1

#include using namespace std; void dfs(int node, vector&amp; vis, vector&gt;&amp; adj) { &nbsp;&nbsp;&nbsp; vis[node] = 1; &n
#include<bits/stdc++.h> using namespace std; void dfs(int node, vector<int>& vis, vector<vector<int>>& adj) {     vis[node] = 1;     for(auto it : adj[node]) {         if(!vis[it]) {             dfs(it, vis, adj);         }     } } int main() {     int n, m;     cin >> n >> m;     vector<vector<int>> adj(n);     for(int i = 0; i < n; i++) {         for(int j = 0; j < m; j++) {             int x;             cin >> x;             if(x == 1) {                 adj[i].push_back(j);                 adj[j].push_back(i);             }         }     }     vector<int> vis(n, 0);     int cc = 0;     for(int i = 0; i < n; i++) {         if(!vis[i]) {             dfs(i, vis, adj);             cc++;         }     }     cout << cc << endl;     return 0; } Telegram:- @allcoding1

int solve(vector&amp; nums) { &nbsp;&nbsp;&nbsp; vector s(nums); &nbsp;&nbsp;&nbsp; sort(s.begin(), s.end()); &nbsp;&nbsp;&nb
int solve(vector<int>& nums) {     vector<int> s(nums);     sort(s.begin(), s.end());     int i = 0, j = nums.size() - 1;     while (i < nums.size() && nums[i] == s[i]) {         i++;     }     while (j > i && nums[j] == s[j]) {         j--;     }     return j - i + 1; } Arrange the heights Apple ✅ Telegram:- @allcoding1

#include <bits/stdc++.h> #define ll long long using namespace std; ll solve(ll k, ll s) {     auto comb = [](ll n) {         return n * (n - 1) / 2;     };     if (k > 3 * s) return 0;     ll ans = comb(k + 2);     if (k > s) ans -= 3 * comb(k - s + 1);     if (k - 2 >= 2 * s) ans += 3 * comb(k - 2 * s);     return ans; } int main() {     ll n, s;     cin >> n >> s;     cout << solve(s,n) << endl;     return 0; } Distribute Car Toy Service Now ✅ Telegram:- @allcoding1

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Valid user ✅ Telegram:- @allcoding1
Valid user ✅ Telegram:- @allcoding1