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allcoding1_official

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

📊 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 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
-1 36930
帖子存档
🎯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

sticker.webp0.13 KB

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

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

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

int solve(vector&amp; arr) { int n = arr.size(); if (n == 0) return 0; int maxLen = 1; int currLen = 1; bool f = true; for (i
+2
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++

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Cognizant is hiring❤️ Apply link: https://t.me/itjoblinks Join our group for daily job posts #paid promotion
Cognizant is hiring❤️ Apply link: https://t.me/itjoblinks Join our group for daily job posts #paid promotion

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

Cognizant is hiring❤️ Apply link: https://t.me/itjoblinks Join our group for daily job posts #paid promotion

Cognizant is hiring❤️ Apply link: https://t.me/itjoblinks Join our group for daily job posts #paid promotion

any paid promotion DM :- @Priya_i

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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))