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allcoding1_official

allcoding1_official

前往频道在 Telegram

📈 Telegram 频道 allcoding1_official 的分析概览

频道 allcoding1_official (@allcoding1_official) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 78 714 名订阅者,在 技术与应用 类别中位列第 1 622,并在 印度 地区排名第 3 757 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 78 714 名订阅者。

根据 30 十月, 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 -1 600,过去 24 小时变化为 -27,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 1.38%。内容发布后 24 小时内通常能获得 0.62% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 1 091 次浏览,首日通常累积 486 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, dev 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 31 十月, 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 技术与应用 类别中的关键影响点。

78 714
订阅者
-2724 小时
-3217 天
-1 60030 天
帖子存档
5
5

15
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Nvidia exam Ans

function ArithGeoll(arr) { let arith = true; let d = Math.abs(arr[0]- arr[1]); for (let i = 2; i < arr.length; i++) { if (
function ArithGeoll(arr) { let arith = true; let d = Math.abs(arr[0]- arr[1]); for (let i = 2; i < arr.length; i++) { if (Math.abs(arr[i] - arr[i - 1]) !== d) { arith = false; } } if (arith) { return "Arithmetic"; } let geo = true; d = arr[1]/ arr[0]; for (let i = 2; i < arr.length; i++) if (Math.abs(arr[i] / arr[i- { 1]) !== d) { geo = false; } } if (geo) { return "Geometric"; } if (larith && !geo) { return "-1"; } }

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function ArithGeoll(arr) { let arith = true; let d = Math.abs(arr[0]- arr[1]); for (let i = 2; i < arr.length; i++) { if (Math.abs(arr[i] - arr[i - 1]) !== d) { arith = false; } } if (arith) { return "Arithmetic"; } let geo = true; d = arr[1]/ arr[0]; for (let i = 2; i < arr.length; i++) if (Math.abs(arr[i] / arr[i- { 1]) !== d) { geo = false; } }

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500 TB Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intel
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500 TB Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources ( English🏴󐁧󐁢󐁥󐁮󐁧󐁿 , French🇨🇵 , German🇩🇪 ) ₹300 Contact:- @meterials_available

struct Request { string id; string start_time; string end_time; }; bool compare(Request a, Request b) { return a.start_time &
struct Request { string id; string start_time; string end_time; }; bool compare(Request a, Request b) { return a.start_time < b.start_time; } int solve(vector requests) { sort(requests.begin(), requests.end(), compare); int max_concurrent = 0; int current_concurrent = 0; string current_end_time = "00:00:00"; for (Request request : requests) { if (request.start_time >= current_end_time) { current_concurrent--; } else { current_concurrent++; if (current_concurrent > max_concurrent) { max_concurrent = current_concurrent; } } if (request.end_time > current_end_time) { current_end_time = request.end_time; } } return max_concurrent; }

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500 TB Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intel
+2
500 TB Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources ( English🏴󐁧󐁢󐁥󐁮󐁧󐁿 , French🇨🇵 , German🇩🇪 ) ₹300 Contact:- @meterials_available

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#define vv(a) vector<vector<int>>(a) int solve(int n, int m, vector<vector<int>> &a) { vv(right)(n, vector<int>(m, 0)); vv(down)(n, vector<int>(m, 0)); vv(left)(n, vector<int>(m, 0)); for (int i = 0; i < n; i++) { int cnt = 0; for (int j = m - 1; j >= 0; j--) { right[i][j] = cnt; if (a[i][j] == 1) cnt++; } } for (int i = 0; i < n; i++) { int cnt = 0; for (int j = 0; j < m; j++) { left[i][j] = cnt; if (a[i][j] == 1) cnt++; } } for (int j = 0; j < m; j++) { int cnt = 0; for (int i = n - 1; i >= 0; i--) { down[i][j] = cnt; if (a[i][j] == 1) cnt++; } } int ans = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (a[i][j] == 1) { ans += (right[i][j] * down[i][j]); ans += (left[i][j] * down[i][j]); } } } return ans; } C++ Telegram:- @allcoding1_official

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