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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 22 442 名订阅者,在 教育 类别中位列第 8 828,并在 印度 地区排名第 19 108

📊 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 22 442 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 -364,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 4.97%。内容发布后 24 小时内通常能获得 1.18% 的反应,占订阅者总量。
  • 帖子覆盖: 每篇帖子平均可获得 1 116 次浏览,首日通常累积 264 次浏览。
  • 互动与反馈: 受众积极参与,单帖平均反应数为 1
  • 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 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 教育 类别中的关键影响点。

22 442
订阅者
-1424 小时
-787
-36430
帖子存档
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Rube Goldberg machine
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Rube Goldberg machine

photo content
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C++ Ant Drop Code Telegram - @allcoding1
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C++ Ant Drop Code Telegram - @allcoding1

Java Find your topic Code Telegram - @allcoding1
Java Find your topic Code Telegram - @allcoding1

Ant Drop Code Telegram - @allcoding1
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Ant Drop Code Telegram - @allcoding1

#include<bits/stdc++.h> int max(int a, int b); /* Returns length of LCS for X[0..m-1], Y[0..n-1] */ int lcs( char *X, char *Y, int m, int n ) { if (m == 0 || n == 0) return 0; if (X[m-1] == Y[n-1]) return 1 + lcs(X, Y, m-1, n-1); else return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n)); } /* Utility function to get max of 2 integers */ int max(int a, int b) { return (a > b)? a : b; } /* Driver program to test above function */ int main() { char X[] = "AGGTAB"; char Y[] = "GXTXAYB"; int m = strlen(X); int n = strlen(Y); printf("Length of LCS is %d", lcs( X, Y, m, n ) ); return 0; } C++ Longest Special Subsequences Telegram - @allcoding1

Python 3.8 language Telegram:- @allcoding1
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Python 3.8 language Telegram:- @allcoding1

Jerwin JD: n = int(input ()) flag=0 While(flag==0): n = n+1 IfLen(set(list(str(n))))== Len(str(n)): flag=1 break Print(n) Two frnds Accenture code Telegram:- @allcoding1

C++ TERRAFORM Code Telegram - @allcoding1
C++ TERRAFORM Code Telegram - @allcoding1

#include void alphabet_Z_Pattern(int N) { int index, side_index, size; int Top = 1, Bottom = 1, Diagonal = N - 1; for (index
#include <stdio.h> void alphabet_Z_Pattern(int N) { int index, side_index, size; int Top = 1, Bottom = 1, Diagonal = N - 1; for (index = 0; index < N; index++) printf("%d ", Top++); printf("\n"); for (index = 1; index < N - 1; index++) { for (side_index = 0; side_index < 2 * (N - index - 1); side_index++) printf(" "); printf("%d", Diagonal--); printf("\n"); } for (index = 0; index < N; index++) printf("%d ", Bottom++); } int main() { int N = 5; alphabet_Z_Pattern(N); return 0; } C language Telegram:- @allcoding1

C++17 Bingo Code Telegram - @allcoding1
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C++17 Bingo Code Telegram - @allcoding1

C++ Build road code Telegram:- @allcoding1
C++ Build road code Telegram:- @allcoding1

Java Binary Sort Code Telegram:- @allcoding1
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Java Binary Sort Code Telegram:- @allcoding1

Python Minimum swap Code Telegram:-@allcoding1
Python Minimum swap Code Telegram:-@allcoding1

Pangram code Python3
Pangram code Python3

🎯Global Edge (Capgemini Engineering) is Hiring Role: Software Analyst Qualifications: BE/B.Tech graduates Branch: CS, IT, ECE or related streams Percentage Criteria: Minimum 65% throughout academics Experience: Freshers CTC: 4 LPA Job Location: Bangalore Apply Now:- http://www.allcoding1.com Telegram:- @allcoding1

Virtusa 2nd Round Scooby Code Telegram:- @allcoding1
Virtusa 2nd Round Scooby Code Telegram:- @allcoding1

Virtusa exam Rajesh Code Python Telegram:-
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Virtusa exam Rajesh Code Python Telegram:-

Virtusa exam Telegram:- @allcoding1
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Virtusa exam Telegram:- @allcoding1