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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 22 561 名订阅者,在 教育 类别中位列第 8 836,并在 印度 地区排名第 19 517

📊 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 22 561 名订阅者。

根据 13 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 -442,过去 24 小时变化为 -20,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 6.17%。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 1 394 次浏览,首日通常累积 283 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 14 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。

22 561
订阅者
-2024 小时
-897
-44230
帖子存档
#include <iostream> #include <vector> #include <unordered_map> using namespace std; const int MOD = 1e9 + 7; vector<vector<int>> tree; vector<int> a; unordered_map<int, int> countMap; bool checkPalindrome(unordered_map<int, int>& countMap) { int oddCount = 0; for (auto& it : countMap) { if (it.second % 2 != 0) oddCount++; if (oddCount > 1) return false; } return true; } int dfs(int node) { int ans = 0; countMap[a[node]]++; if (checkPalindrome(countMap)) { ans = 1; } for (auto& child : tree[node]) { ans += dfs(child); ans %= MOD; } countMap[a[node]]--; // Backtrack to remove the current node's count return ans; } int main() { int n; cin >> n; tree.resize(n + 1); a.resize(n + 1); vector<int> par(n + 1); for (int i = 2; i <= n; i++) { cin >> par[i]; tree[par[i]].push_back(i); } for (int i = 1; i <= n; i++) { cin >> a[i]; } int ans = dfs(1); cout << ans << endl; return 0; } c++ Palindromic subtrees

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Maximum Count Hackwithinfy
Maximum Count Hackwithinfy

Gas Station Hackwithinfy
Gas Station Hackwithinfy

Removable subarrays Hackwithinfy
Removable subarrays Hackwithinfy

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Repost from allcoding1
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vector babuswiggyjayega(vector a, vector b) { &nbsp;&nbsp;&nbsp; ll n1 = a.size(), n2 = b.size(); &nbsp;&nbsp;&nbsp; ll i = 0
vector<ll> babuswiggyjayega(vector<ll> a, vector<ll> b) {     ll n1 = a.size(), n2 = b.size();     ll i = 0, j = 0, k = 0;     vector<ll> golu(n1 + n2);     while (i < n1 && j < n2)     {         if (a[i] < b[j])             golu[k++] = a[i++];         else             golu[k++] = b[j++];     }     while (i < n1)         golu[k++] = a[i++];     while (j < n2)         golu[k++] = b[j++];     return golu; } Merge 2 Arrays Telegram:- @allcoding1

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