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 22 561 名订阅者,在 教育 类别中位列第 8 836,并在 印度 地区排名第 19 517 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 22 561 名订阅者。
根据 13 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 -442,过去 24 小时变化为 -20,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 6.17%。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 1 394 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 14 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。
22 561
订阅者
-2024 小时
-897 天
-44230 天
帖子存档
22 558
🎯Company Name Cognizant
Post Name Associate Software Engineer
Qualification Any Bachelor’s Degree
Branch Any Branches
Batch 2021, 2022, 2023 & 2024
Salary Up to ₹14 LPA
Experience Fresher
Job Location Chennai
Last DateASAP
Apply Now:- https://careers.cognizant.com/in/en/job/00046956561/Software-Engineer-Associate
Telegram:- @allcoding1_official
22 558
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22 558
🎯HCL is hiring
Role : Graduate Trainee
Batch : 2023, 2022 passouts
CTC : 4 LPA
Location : Lucknow, Nagpur
Apply now : https://forms.office.com/pages/responsepage.aspx?id=N-edGDrJWk-LaG9MqZQZEvdjEM5NiEhEl9yRzhUhvJpURE8yNUkxSDZYMDFWM0FXRU9OSUtMREEzMiQlQCN0PWcu
Telegram:- @allcoding1
22 558
🎯 Airtel is hiring
Role : Software Engineer l
CTC : 11+ LPA
Experience : 0 - 2 years
Apply Now: https://eeji.fa.em3.oraclecloud.com/hcmUI/CandidateExperience/en/sites/CX_1/job/125162/?utm_medium=jobshare
Telegram:- @allcoding1
22 558
🎯 Zog global is Hiring
Role: Software Developer Intern
Batch: 2023/2024/2025/2026
Duration: 3-6 months
Location: Remote
Apply now: https://zogglobal.com/jobs/software-developer-intern/
Telegram:- @allcoding1
22 558
🎯Company : Infosys
Post: 4000+
Qualification: UG, Graduation & PG
Selection Process: Interview & DV
Online Apply Date: 14th March 2024
Last Date of Online Apply: 27th April 2024
Apply now:- https://careers.infosys.com/instep/internship/instepregistrationpage.aspx
Telegram:- @allcoding1
22 558
🎯Company: TCS
Post: 6000+
Qualification: Graduate (All Stream)
Selection Process: Interview & DV
Online Apply Date: 08th March 2024
Last Date of Online Apply: 27th May 2024
Apply Now:- https://nextstep.tcs.com/campus/#/
Telegram:- @allcoding1_official
📢Not:- mobile users open in desktop mood
22 558
🎯 Dow Jones is hiring
Role : Web Developer
CTC : 12 - 15 LPA
Apply now : https://dowjones.jobs/mumbai-ind/web-developer/4AD2CBD937984D1DA86277BBBAC931DA/job/?vs=16061606
Telegram:- @allcoding1
22 558
🎯Matrimony is hiring
Role : software engineer intern
Batch : 2025,2024, 2023 passouts
Apply Now:- https://matrimonycareers.peoplestrong.com/job/detail/MCL_SE-I_983228
Telegram:- @allcoding1
22 558
🎯Locus is hiring
Role : DevOps Engineer
Experience : 1 - 2 years
CTC : 15-25 LPA
Apply Now:- https://locus.freshteam.com/jobs/mbURMTm9XO-R/devops?ft_source=4000030064&ft_medium=4000026063
Telegram:- @allcoding1
22 558
🎯Oracle is hiring
Role : Software Engineer l
Experience : 0 - 2+ years
Apply Now:- https://eeho.fa.us2.oraclecloud.com/hcmUI/CandidateExperience/en/sites/jobsearch/job/231334
Telegram:- @allcoding1
22 558
🎯 Global Logic is hiring
Role : Associate Analyst
Experience : 0 - 1 year
Apple Now:-
https://www.globallogic.com/careers/associate-analyst-dei-irc214321/
Telegram:- @allcoding1
22 558
def max_benefit(n,A, K):
n = len(A)
dp = [0] * (n + 1)
prefix_sum = [0] * (n + 1)
for i in range(1, n + 1):
prefix_sum[i] = prefix_sum[i - 1] + A[i - 1]
for i in range(1, n + 1):
min_val = float('inf')
max_val = float('-inf')
for j in range(1, K + 1):
if i - j >= 0:
min_val = min(min_val, prefix_sum[i] - prefix_sum[i - j])
max_val = max(max_val, prefix_sum[i] - prefix_sum[i - j])
dp[i] = max(dp[i], dp[i - j] + max(max_val, -min_val))
return dp[n] % MOD
Alternate Segment
Infosys
Telegram:- @allcoding1
22 558
#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
while (b != 0) {
int temp = b;
b = a % b;
a = temp;
}
return a;
}
vector<int> gcdArrays(vector<int>& arr) {
int n = arr.size();
int max_gcd = 0;
int max_length = 0;
for (int i = 0; i < n - 1; ++i) {
int current_gcd = arr[i];
for (int j = i + 1; j < n; ++j) {
current_gcd = gcd(current_gcd, arr[j]);
if (current_gcd > max_gcd) {
max_gcd = current_gcd;
max_length = j - i + 1;
} else if (current_gcd == max_gcd && (j - i + 1) > max_length) {
max_length = j - i + 1;
}
}
}
return {max_gcd, max_length};
}
DE Shaw
Telegram:- @allcoding1
22 558
Repost from allcoding1
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🇨🇵 , )
100 rupees
buy:- @meterials_available
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