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allcoding1

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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 22 575 名订阅者,在 教育 类别中位列第 8 822,并在 印度 地区排名第 19 518

📊 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 22 575 名订阅者。

根据 12 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 -437,过去 24 小时变化为 -6,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 5.99%。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 1 353 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 13 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。

22 575
订阅者
-624 小时
-967
-43730
帖子存档
📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Work
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📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT 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 , 🇫🇷 All courses (100 rupees) Contact:- @meterials_available

int solve(vector tc) {     int n = tc.size();     vector p(n), s(n);     p[0] =
int solve(vector tc) {     int n = tc.size();     vector p(n), s(n);     p[0] = tc[0];     for (int i = 1; i < n; ++i) {         p[i] = p[i - 1] + tc[i];     }     s[n - 1] = tc[n - 1];     for (int i = n - 2; i >= 0; --i) {         s[i] = s[i + 1] + tc[i];     }     int m = max(p[0], s[0]);     for (int i = 1; i < n; ++i) {         m = max(m, max(p[i], s[i]));     }     return m; } Amazon

#include #include #include using namespace std; long getkRepValue(string user_history, long k) { &nbsp;&nbsp;&nbsp; long n =
#include <iostream> #include <string> #include <unordered_map> using namespace std; long getkRepValue(string user_history, long k) {     long n = user_history.size();     unordered_map<char, long> count;     long left = 0, right = 0, ans = 0;     while (right < n) {         count[user_history[right]]++;         while (count[user_history[right]] >= k && left <= right) {             ans += n - right;             count[user_history[left]]--;             left++;         }         right++;     }     return ans; }  Machine learning Amazon

📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Work
+8
📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT 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 , 🇫🇷 All courses (100 rupees) Contact:- @meterials_available

🎯Infosys Recruitment Drive 2024 | Hiring For Process Executive , Salary Up to ₹3.5 LPA Degree:- Any Bachelor's Batch:- 2021, 2022, 2023 & 2024 Apply now:- www.allcoding1.com

class Message(object): def init(self, message: str, sender: int, receiver: int) -> None: self.message = message self.sender =
class Message(object):     def init(self, message: str, sender: int, receiver: int) -> None:         self.message = message         self.sender = sender         self.receiver = receiver     def str(self) -> str:         return self.message     def eq(self, other: object) -> bool:         if not isinstance(other, Message):             return False         return self.message == other.message Message Objects

#include <iostream> #include <sstream> #include <string> #include <vector> std::string stemmer(const std::string& text) {     std::stringstream ss(text);     std::string word;     std::vector<std::string> stemmed_words;     while (ss >> word) {         if (word.size() > 2 && (word.substr(word.size() - 2) == "ed" word.substr(word.size() - 2) == "ly" word.substr(word.size() - 3) == "ing")) {             word = word.substr(0, word.size() - 2);         }         if (word.size() > 8) {             word = word.substr(0, 8);         }         stemmed_words.push_back(word);     }     std::string result;     for (const std::string& stemmed_word : stemmed_words) {         result += stemmed_word + " ";     }     result.pop_back();     return result; } int main() {     std::string text = "an extremely dangerous dog is barking";     std::cout << stemmer(text) << std::endl;  // Output: "an extreme dangerou dog is bark"     return 0; } Suffix stripping stemmer Telegram:- @allcoding1

Paid Promotion:- @Priya_i

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heights = [int(i) for i in input().split()] freq_dict = {} for h in heights: if h in freq_dict: freq_dict[h] += 1 else: freq_
heights = [int(i) for i in input().split()] freq_dict = {} for h in heights: if h in freq_dict:   freq_dict[h] += 1 else:   freq_dict[h] = 1 for h in sorted(freq_dict): print(h, freq_dict[h])

#include <iostream> #include <string> #include <vector> using namespace std; int numDecodings(string msg) {     int MOD = 1000000007;     int n = msg.size();         vector<long long> dp(n + 1, 0);     dp[0] = 1;         if (msg[0] == '0')         dp[1] = 0;     else if (msg[0] == '*')         dp[1] = 9;     else         dp[1] = 1;         for (int i = 2; i <= n; ++i) {         if (msg[i - 1] == '0') {                         if (msg[i - 2] == '1' || msg[i - 2] == '2')                 dp[i] += dp[i - 2];             else if (msg[i - 2] == '*')                 dp[i] += 2 * dp[i - 2];         } else if (msg[i - 1] >= '1' && msg[i - 1] <= '9') {                   dp[i] += dp[i - 1];                         if (msg[i - 2] == '1' || (msg[i - 2] == '2' && msg[i - 1] <= '6'))                 dp[i] += dp[i - 2];             else if (msg[i - 2] == '*') {                                 if (msg[i - 1] <= '6')                     dp[i] += 2 * dp[i - 2];                 else                     dp[i] += dp[i - 2];             }         } else if (msg[i - 1] == '*') {             dp[i] += 9 * dp[i - 1];                         if (msg[i - 2] == '1')                 dp[i] += 9 * dp[i - 2];             else if (msg[i - 2] == '2')                 dp[i] += 6 * dp[i - 2];             else if (msg[i - 2] == '*')                 dp[i] += 15 * dp[i - 2];         }                 dp[i] %= MOD;     }         return dp[n]; } Number of ways decode

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import re a = int(input().strip()) b = set() c = r'\b[A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\.[A-Z|a-z]{2,}\b' for _ in range(a): li
import re a = int(input().strip()) b = set() c = r'\b[A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\.[A-Z|a-z]{2,}\b' for _ in range(a):     line = input().strip()     b.update(re.findall(c, line)) print(';'.join(sorted(b)))

int solve(int k, vector&amp; s) { &nbsp;&nbsp;&nbsp; sort(s.rbegin(), s.rend()); &nbsp;&nbsp;&nbsp; int c = 0; &nbsp;&nbsp;&n
int solve(int k, vector<int>& s) {     sort(s.rbegin(), s.rend());     int c = 0;     for (int i = 0; i < s.size(); i++) {         if (i < k && s[i] > 0) {             c++;         } else if (s[i] == s[i - 1] && s[i] > 0) {             c++;         } else {             break;         }     }     return c; } Competitive Gaming Accenture exam Telegram:- @allcoding1

#include <iostream> #include <vector> #include <algorithm> using namespace std; int findLIS(vector<int>& s) {     vector<int> tails;     for (int x : s) {         auto it = lower_bound(tails.begin(), tails.end(), x);         if (it == tails.end()) {             tails.push_back(x);         } else {             *it = x;         }     }     return tails.size(); }   Swiggy LIS