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📈 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

📊 受众指标与增长动态

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
int solve(vector& arr) { int n = arr.size(); if (n == 0) return 0; int maxLen = 1; int currLen = 1; bool f = true; for (i
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int solve(vector<int>& 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++ Telegram:- @allcoding1 L&T

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 L&t Q) Ever - increasing & ever decreasing array segments

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 L&t Q) Encryption by digits

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def check_first_and_last_char(Str): if Str[0].isdigit() and not Str[-1].isdigit(): print("Yes") elif Str[-1].isdigit() and no
def check_first_and_last_char(Str):     if Str[0].isdigit() and not Str[-1].isdigit():         print("Yes")     elif Str[-1].isdigit() and not Str[0].isdigit():         print("No")     else:         print("Invalid input") Str = input() check_first_and_last_char(Str) @allcoding1

def check_digit_position(s): if s[0].isdigit() and not s[-1].isdigit(): return "Yes" elif not s[0].isdigit() and s[-1].isdigi
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)) @allcoding1

Deloitte exam send Questions 👇

def check_digit_position(s): if s[0].isdigit() and not s[-1].isdigit(): return "Yes" elif not s[0].isdigit() and s[-1].isdigi
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)) Telegram:- @allcoding1

🎯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

Deloitte exam Ans

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Python 2nd code Deloitte exam Telegram:- @allcoding1
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Python 2nd code Deloitte exam Telegram:- @allcoding1

1st code Python Deloitte exam Telegram:- @allcoding1
1st code Python Deloitte exam Telegram:- @allcoding1

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