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
🎯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
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_official
85 483
int solve(vector<int>& A) { int s = 0; for (const auto&n: A) { string sn = to_string(n); char md = *max_element(sn.begin().
sn.end());
s += md-'0';
}
return s;
}
C++
Telegram:- @allcoding1_official
L&t
Encryption by digits
85 483
int solve(vector<int>& arr) { int narr.size(); if (n == 0) return 0;
int maxLen = 1;
int currLen = 1;
bool f = true;
for (int i = 1; i < n; ++i) { if ((arr[i] > arr[i-1] && f) || (arr[i] < arr[i-1] && !f)) { ++currLen; } else { maxLen = max(maxLen, currLen); currLen = 2;
f = arr[i] > arr[i-1]
}
}
maxLen = max(maxLen, currLen); return maxLen;
}
C++
Telegram:- @allcoding1_official
L&t
Q) Ever - increasing & ever decreasing array segments
85 483
int solve(vector& arr) { int n = arr.size(); if (n == 0) return 0;
int maxLen = 1; int currLen = 1; bool f = true;
for (int i = 1; i < n; ++i) { if ((arr[i] > arr[i-1] && f) II (arr[i] < arr[i-1] && !f)) { ++currLen; } else {
maxLen = max(maxLen, currLen); currLen = 2; f = arr[i] > arr[i-1]; }
maxLen = max(maxLen, currLen); return maxLen; }
C++
85 483
Repost from allcoding1_official
Cognizant is hiring❤️
Apply link:
https://t.me/itjoblinks
Join our group for daily job posts
#paid promotion
85 483
Repost from allcoding1_official
Guys❤
👉 I am doing promotions (real Or fake) I don't know
👉I'm not forcing you. Your decision
👉I am not a responsible to you and money
👉 I need money that why I'm doing promotions try understand
85 483
🎯Position: Accountant Trainee
Location: Coimbatore
Experience: Fresher
Qualification: B.Com. 2023 passed out
https://careers.banfico.com/jobs/Careers/31483000015926001/Accountant-Trainee?source=CareerSite&t=807
85 483
Cognizant is hiring❤️
Apply link:
https://t.me/itjoblinks
Join our group for daily job posts
#paid promotion
85 483
Guys❤
👉 I am doing promotions (real Or fake) I don't know
👉I'm not forcing you. Your decision
👉I am not a responsible to you and money
👉 I need money that why I'm doing promotions try understand
85 483
Cognizant is hiring❤️
Apply link:
https://t.me/itjoblinks
Join our group for daily job posts
#paid promotion
85 483
Cognizant is hiring❤️
Apply link:
https://t.me/itjoblinks
Join our group for daily job posts
#paid promotion
85 483
🎯Qualcomm Off Campus Recruitment
Job Position:- IT Software Developer (UX/UI/Fullstack)
Job ID: 3058496
Qualification:- B.Tech
Batch:- 2019/2020/2021
Job Location:- Hyderabad, India
Department: Software Engineering
Salary Package:- As per Company Standards
Job Type:- Full-time
Last Date:- ASAP
Apply Now:- https://www.allcoding1.com/2024/02/qualcomm-off-campus-recruitment-2024.html?m=1
Telegram:- @allcoding1_official
85 483
def check_digit_position(s):
if s[0].isdigit() and not s[-1].isdigit():
return "Yes"
elif not s[0].isdigit() and s[-1].isdigit():
return "No"
else:
return "Invalid input"
input_str = input().strip()
print(check_digit_position(input_str))
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