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allcoding1

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 22 558 名订阅者,在 教育 类别中位列第 8 854,并在 印度 地区排名第 19 507

📊 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 22 558 名订阅者。

根据 14 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 -445,过去 24 小时变化为 -14,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 6.31%。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 1 423 次浏览,首日通常累积 282 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 15 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。

22 558
订阅者
-1424 小时
-947
-44530
帖子存档
Minimum moves Wissen
Minimum moves Wissen

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def helper(reward): nums = [] for i in reward: heappush(nums, -i) ans = -heappop(nums) i = 1 while nums: temp = -heappop(nums
def helper(reward): nums = [] for i in reward: heappush(nums, -i) ans = -heappop(nums) i = 1 while nums: temp = -heappop(nums) if temp - i <= 0: break ans += temp - i i += 1 return ans @allcoding1

int getMinlength(vector<int>a,int k) { int ans=1; long long mul=1; for(int i=0;i<a.size();i++){ if(mul*a[i]>k){ ans++; mul=a[i]; } else{ mul*=a[i]; } } return ans; } L&T @allcoding1 Q) compressing array

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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

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
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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