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allcoding1

allcoding1

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📈 Analytical overview of Telegram channel allcoding1

Channel allcoding1 (@allcoding1) in the English language segment is an active participant. Currently, the community unites 21 531 subscribers, ranking 9 159 in the Education category and 19 101 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 21 531 subscribers.

According to the latest data from 01 September, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -367 over the last 30 days and by -14 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 7.16%. Within the first 24 hours after publication, content typically collects 1.25% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 1 543 views. Within the first day, a publication typically gains 270 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 0.
  • Thematic interests: Content is focused on key topics such as dsa, stack, namaste, javascript, learning.

📝 Description and content policy

Channel description not provided.

Thanks to the high frequency of updates (latest data received on 02 September, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.

21 531
Subscribers
-1424 hours
-797 days
-36730 days
Posts Archive
class Solution { public:      void bfs(vector<vector<int>>&vis,vector<vector<char>>&grid,int i,int j,int n,int m)     {         vis[i][j]=1;         queue<pair<int,int>>q;         q.push({i,j});         while(!q.empty())         {             int row=q.front().first;             int col=q.front().second;             q.pop();             int delrow[4]={1,0,-1,0};             int delcol[4]={0,1,0,-1};                    for(int k=0;k<=3;k++){                     int nrow=row+delrow[k];                     int ncol=col+delcol[k];                     if(nrow>=0 and nrow<n and ncol>=0 and ncol<m and grid[nrow][ncol]=='1' and !vis[nrow][ncol])                     {                         vis[nrow][ncol]=1;                         q.push({nrow,ncol});                     }                 }             }         }     int numIslands(vector<vector<char>>& grid) {     int n=grid.size();         int m=grid[0].size();         vector<vector<int>>vis(n,vector<int>(m,0));         int cnt=0;         for(int i=0;i<n;i++)         {             for(int j=0;j<m;j++)             {                 if(!vis[i][j] and grid[i][j]=='1')                 {                     cnt++;                     bfs(vis,grid,i,j,n,m);                 }             }         }         return cnt;     } }; Zeta ✅ Telegram:- @allcoding1

class Solution { public:      void bfs(vector<vector<int>>&vis,vector<vector<char>>&grid,int i,int j,int n,int m)     {         vis[i][j]=1;         queue<pair<int,int>>q;         q.push({i,j});         while(!q.empty())         {             int row=q.front().first;             int col=q.front().second;             q.pop();             int delrow[4]={1,0,-1,0};             int delcol[4]={0,1,0,-1};                    for(int k=0;k<=3;k++){                     int nrow=row+delrow[k];                     int ncol=col+delcol[k];                     if(nrow>=0 and nrow<n and ncol>=0 and ncol<m and grid[nrow][ncol]=='1' and !vis[nrow][ncol])                     {                         vis[nrow][ncol]=1;                         q.push({nrow,ncol});                     }                 }             }         }     int numIslands(vector<vector<char>>& grid) {     int n=grid.size();         int m=grid[0].size();         vector<vector<int>>vis(n,vector<int>(m,0));         int cnt=0;         for(int i=0;i<n;i++)         {             for(int j=0;j<m;j++)             {                 if(!vis[i][j] and grid[i][j]=='1')                 {                     cnt++;                     bfs(vis,grid,i,j,n,m);                 }             }         }         return cnt;     } }; Zeta ✅ Telegram:- @allcoding1

any paid promotion DM :- @Priya_i

Fake #scammer This not my ID @allcoding1_officiaI

KPMG Off Campus Drive Hiring 2024 | Front End Developer | 4 LPA+ Job Title : Front End Developer Qualification : B.E / B.Tech Batch : Any Batch Package : 4 LPA+ Apply Now:- www.allcoding1.com Telegram:- @allcoding1

SELECT category, title, total_stock FROM ( SELECT p.category, p.title, SUM(w.quantity) AS total_stock FROM products p JOIN wa
SELECT   category,   title,   total_stock FROM (   SELECT     p.category,     p.title,     SUM(w.quantity) AS total_stock   FROM     products p   JOIN     warehouse w ON p.product_id = w.product_id   GROUP BY     p.category, p.title   HAVING     total_stock > 10 ) AS filtered_data ORDER BY   category ASC, title ASC, total_stock DESC; IBM✅ Telegram:- @allcoding1

IBM✅ SQL Telegram:- @allcoding1
+1
IBM✅ SQL Telegram:- @allcoding1

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return min(e-b+1, numChapters) Python 3✅ IBM Telegram:- @allcoding1
+1
return min(e-b+1, numChapters) Python 3✅ IBM Telegram:- @allcoding1

🎯Phone pe hiring Job role:- Advisor,ONDC Qualification:- Any Experience:- 0-2years Location:- Bangalore Apply Now:- www.allcoding1.com Telegram:- @allcoding1_official

#scammer Please report 👇 @offical1allcoding , Vds , @Allcodingoffical The changing names Guys ♥️ be careful don't pay money
+1
#scammer Please report 👇 @offical1allcoding , Vds , @Allcodingoffical The changing names Guys ♥️ be careful don't pay money

Guys ♥️ @allcoding, @Allcodingoffical1, @offical1allcoding @Allcodingoffical, @codingoffical and @Allcodingofficalmain it's n
Guys ♥️ @allcoding, @Allcodingoffical1, @offical1allcoding @Allcodingoffical, @codingoffical and @Allcodingofficalmain it's not me don't lose your money #scammer Please report Please share with your friends and telegram Group's

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int solve(vector&gt; arr) { &nbsp;&nbsp;&nbsp; int n = arr.size(); &nbsp;&nbsp;&nbsp; vector&gt; dp(n, vector(n)); &nbsp;&nbs
int solve(vector<vector<int>> arr) {     int n = arr.size();     vector<vector<int>> dp(n, vector<int>(n));     int maxi = 0;     for(int i = 0; i < n; i++) {         for(int j = 0; j < n; j++) {             if(i == 0 || j == 0) {                 dp[i][j] = arr[i][j];             } else if(arr[i][j] == 1) {                 dp[i][j] = min({dp[i-1][j], dp[i][j-1], dp[i-1][j-1]}) + 1;             }             maxi = max(maxi, dp[i][j]);         }     }     return maxi; } IBM C++ Telegram:- @allcoding1