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allcoding1

allcoding1

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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 21 578 名订阅者,在 教育 类别中位列第 9 062,并在 印度 地区排名第 19 025

📊 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 21 578 名订阅者。

根据 29 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 -360,过去 24 小时变化为 -8,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 5.78%。内容发布后 24 小时内通常能获得 1.10% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 1 247 次浏览,首日通常累积 238 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 30 八月, 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。

21 578
订阅者
-824 小时
-787
-36030
帖子存档
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Willey edge
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Willey edge

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Telegram:- @allcoding1 Juspay =int(input()) dic={} def addnode(u,v): dic[u].append(v) for i in range(N): n1=int(input()) dic[n1]=[] edge=int(input()) for i in range(edge): u,v=map(int,input().split()) addnode(u,v) start=int(input()) end=int(input()) tr=[] tr.append(start) def traverse(dic,start): for i in dic[start]: if(i==end): return if(len(dic[start])==0): return else: tr.append(i) traverse(dic,i) traverse(dic,start) tr=sorted(tr) for i in tr: print(i,end=" ")

Nagging react newbie Telegram:- @allcoding1
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Nagging react newbie Telegram:- @allcoding1

import java.util.Scanner; public class Main {     public static void main(String[] args) {         Scanner scanner = new Scanner(System.in);                 int n = scanner.nextInt();         int[] left = new int[n];         int[] right = new int[n];                 for (int i = 0; i < n; i++) {             left[i] = scanner.nextInt();             right[i] = scanner.nextInt();         }                 int sumLeft = 0;         int sumRight = 0;                 for (int i = 0; i < n; i++) {             sumLeft += left[i];             sumRight += right[i];         }                 if (sumLeft % 2 == 0 && sumRight % 2 == 0) {             System.out.println(0);             return;         }                 boolean possible = false;         for (int i = 0; i < n; i++) {             int newSumLeft = sumLeft - left[i] + right[i];             int newSumRight = sumRight - right[i] + left[i];             if (newSumLeft % 2 == 0 && newSumRight % 2 == 0) {                 possible = true;                 break;             }         }                 if (possible) {             System.out.println(1);         } else {             System.out.println(-1);         }     } }

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