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 20 708 名订阅者，在 教育 类别中位列第 9 717，并在 印度 地区排名第 19 457 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 20 708 名订阅者。
根据 31 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 -415，过去 24 小时变化为 -18，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 2.72%。内容发布后 24 小时内通常能获得 1.04% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 563 次浏览，首日通常累积 216 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 01 十一月, 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。
20 708
订阅者
-1824 小时
-997 天
-41530 天
帖子存档
20 707
Pepsico is hiring for Finance Analyst
Qualification : BBA, B.com, MBA, M.Com
Passedout year : 2019,2020,2021,2022
Experience: Freshers
Location: Hyderabad
Salary : 5 to 7 LPA
Apply Now :- http://www.allcoding1.com
Telegram:- @allcoding1
20 707
🎯Cognizant Hiring as Programmer Analyst Trainee With 4.5 LPA
Job Role : Programmer Analyst Trainee(PAT)
Qualification : B.E/B.Tech/M.E/M.Tech/MCA/M.Sc
Batch : 2020/2021/2022
CTC : Rs 4.5 LPA
APPLY Now:- http://www.allcoding1.com
Deadline -16 oct 11:59pm
Telegram:- @allcoding1
20 707
🎯Amazon Off Campus Recruitment 2022 : Hiring for Freshers Technical Associate
Designation : Technical Associate, Books
Eligibility : B.E/B.Tech
Batch : 2021/2022
Salary : 5 LPA
Apply Now:- http://www.allcoding1.com
Telegram:- @allcoding1
20 707
🎯KPMG Recruitment 2022 Hiring Graduate Freshers As Analyst Across India
Job Title : Analyst
Qualification : Any Degree
Batch : 2022/2021 or Any
Salary : 4 LPA*
Apply Now:- http://www.allcoding1.com
Telegram:- @allcoding1
20 707
🎯Cognizant Hiring as Programmer Analyst Trainee With 4.5 LPA
Job Role : Programmer Analyst Trainee(PAT)
Qualification : B.E/B.Tech/M.E/M.Tech/MCA/M.Sc
Batch : 2020/2021/2022
CTC : Rs 4.5 LPA
APPLY Now:- http://www.allcoding1.com
Telegram:- @allcoding1
20 707
🎯Altera Digital Health Off Campus Drive 2022 | Associate Software Engineer | 6 LPA
Job Title : MS Associate Software Engineer – SQL Query, Fresher
Qualification : B.E/B.Tech/BSC/M.E/M.Tech/MCA/M.Sc
Batch : 2020/2021/2022 Batch or Any
Salary : 6 LPA
Apply Now:- http://www.allcoding1.com
Telegram:- @allcoding1
20 707
🎯Zoho Recruitment 2022 – Hiring Web Developers | 4 LPA+
Job Title : Web Developer
Qualification : Any Degree
Batch : Any
Salary : 4 LPA+
Apply Now:- http://www.allcoding1.com
Telegram:- @allcoding1
20 707
A = int()
B = int()
N = A*10+B
D = A+B
If D%2==0:
Print (Not Googly)
Elif D%2!=0:
Print(Googly)
Else:
Print(error)
Googly Number
Code in python
Telegram:- @allcoding1
20 707
🎯Altera Digital Health Off Campus Drive 2022 | Associate Software Engineer | 6 LPA
Job Title : MS Associate Software Engineer – SQL Query, Fresher
Qualification : B.E/B.Tech/BSC/M.E/M.Tech/MCA/M.Sc
Batch : 2020/2021/2022 Batch or Any
Salary : 6 LPA
Apply Now:- http://www.allcoding1.com
Telegram:- @allcoding1
20 707
🎯Zoho Recruitment 2022 – Hiring Web Developers | 4 LPA+
Job Title : Web Developer
Qualification : Any Degree
Batch : Any
Salary : 4 LPA+
Apply Now:- http://www.allcoding1.com
Telegram:- @allcoding1
20 707
🎯Infosys Off Campus Recruitment Drive 2022 For Engineering Freshers
Job Title : Specialist Programmer and Digital Specialist Engineer
Qualification : B.E/B.Tech/M.E/M.Tech/MCA/M.Sc
Batch : 2020/2021/2022 Batch
Salary : INR 6.25-9.5 LPA
Apply Now:- http://www.allcoding1.com
Telegram:- @allcoding1
20 707
def variantsCount(n, s0, k, b, m, a):
if n <=0:
return 0
sum_array = [s0]
for i in range(1,n):
s_i = ((k * sum_array[i - 1] + b) % m) + 1 + sum_array[i - 1]
sum_array.append(s_i)
# two pointer to calculate the count
left = 0
right = len(sum_array) -1
count = 0
while left <= right:
area = sum_array[left] * sum_array[right]
if area <= a:
count += 2*(right - left) + 1
left +=1
elif area > a:
right -=1
return count
Paint the ceiling code
Python3
Telegram:- @allcoding1
20 707
def variantsCount(n, s0, k, b, m, a):
if n <=0:
return 0
sum_array = [s0]
for i in range(1,n):
s_i = ((k * sum_array[i - 1] + b) % m) + 1 + sum_array[i - 1]
sum_array.append(s_i)
# two pointer to calculate the count
left = 0
right = len(sum_array) -1
count = 0
while left <= right:
area = sum_array[left] * sum_array[right]
if area <= a:
count += 2*(right - left) + 1
left +=1
elif area > a:
right -=1
return count
Paint the ceiling code
Python3
Telegram:- @@allcoding1
