现已上线！2025 年 Telegram 研究 — 年度关键洞察 
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 85 483 名订阅者，在 技术与应用 类别中位列第 1 507，并在 印度 地区排名第 3 480 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 85 483 名订阅者。
根据 24 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 -1 369，过去 24 小时变化为 -35，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 2.60%。内容发布后 24 小时内通常能获得 0.55% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 2 222 次浏览，首日通常累积 474 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 3。
- 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 25 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 技术与应用 类别中的关键影响点。
85 483
订阅者
-3524 小时
-2687 天
-1 36930 天
帖子存档
85 483
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🎯SAS Off Campus Drive 2024 – Associate Software Engineer – Freshers | Rs 4-8 LPA
Job Role : Associate Software Developer
Qualification : B.E/B/Tech/M.E. / M.Tech
Experience : Freshers
Job Location : Pune
Package : Rs 4-8 LPA
Apply Now:- https://www.allcoding1.com/2024/02/sas-off-campus-drive-2024-associate_9.html?m=1
Telegram:- @allcoding1_official
85 483
🎯Atos Is Hiring
Degree:- Any Graduate
Batch:- 2019, 2020, 2021, 2022, 2023 & 2024
Apply Now:- https://www.allcoding1.com/2024/02/atos-off-campus-drive-for-any-graduate.html
Telegram:- @allcoding1_official
85 483
+2500 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
85 483
🎯Atos Is Hiring
Degree:- Any Graduate
Batch:- 2019, 2020, 2021, 2022, 2023 & 2024
Apply Now:- https://www.allcoding1.com/2024/02/atos-off-campus-drive-for-any-graduate.html
Telegram:- @allcoding1_official
85 483
long getMaxProfit(vector p, vector pr) {
int n = p.size();
long m=-1
for (int~j=1;j(pr[i]));
}
for (int~k=j+1;k p[j]) r = max(r, static_cast(pr[k]));
}
if(l>0~88~r>0) m = max(m, I + pr[j] + r);
}
return m;
}
85 483
int
getMinlength(vector<int>a,int k)
{
int ans=1;
long long mul=1;
for(int i=0;i<a.size();i++){
if(mul*a[i]>k){
ans++;
mul=a[i];
}
else{
mul*=a[i];
}
}
return ans;
}
L&T
@allcoding1_official
Q) compressing array
