allcoding1
前往频道在 Telegram
📈 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 21 545 名订阅者，在 教育 类别中位列第 9 078，并在 印度 地区排名第 18 983 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 21 545 名订阅者。
根据 31 八月, 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 -369，过去 24 小时变化为 -5，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 6.77%。内容发布后 24 小时内通常能获得 N/A% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 1 460 次浏览，首日通常累积 0 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 01 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。
21 545
订阅者
-524 小时
-817 天
-36930 天
帖子存档
21 533
Repost from allcoding1_official
+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
#paid promotion
21 533
Repost from allcoding1
🎯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
21 533
Repost from allcoding1
🎯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
21 533
Repost from allcoding1
+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
#paid promotion
21 533
🎯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
21 533
🎯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
21 533
+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
#paid promotion
21 533
🎯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
21 533
def helper(reward):
nums = []
for i in reward:
heappush(nums, -i)
ans = -heappop(nums)
i = 1
while nums:
temp = -heappop(nums)
if temp - i <= 0: break
ans += temp - i
i += 1
return ans
@allcoding1
21 533
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
Q) compressing array
21 533
🎯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
21 533
public class Solution {
public static int longestArraySegment(int array_length, List<Integer> arr) {
int maxLength = 1;
int currentLength = 1;
for (int i = 1; i < array_length; i++) {
if (arr.get(i) > arr.get(i - 1)) {
currentLength++;
} else {
currentLength = 1;
}
maxLength = Math.max(maxLength, currentLength);
}
currentLength = 1;
for (int i = 1; i < array_length; i++) {
if (arr.get(i) < arr.get(i - 1)) {
currentLength++;
} else {
currentLength = 1;
}
maxLength = Math.max(maxLength, currentLength);
}
return maxLength;
}
Ever increasing and ever decreasing - L&T
Java
Telegram:- @allcoding1
