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allcoding1_official

allcoding1_official

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📈 Analytical overview of Telegram channel allcoding1_official

Channel allcoding1_official (@allcoding1_official) in the English language segment is an active participant. Currently, the community unites 85 467 subscribers, ranking 1 502 in the Technologies & Applications category and 3 471 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 85 467 subscribers.

According to the latest data from 25 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -1 422 over the last 30 days and by -71 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 2.42%. Within the first 24 hours after publication, content typically collects 0.88% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 2 071 views. Within the first day, a publication typically gains 749 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 4.
  • Thematic interests: Content is focused on key topics such as dsa, stack, namaste, javascript, dev.

📝 Description and content policy

Channel description not provided.

Thanks to the high frequency of updates (latest data received on 26 June, 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 Technologies & Applications category.

85 467
Subscribers
-7124 hours
-3077 days
-1 42230 days
Posts Archive
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Copy and paste question

for_in range(int(input())): n, x = map(int, input().split()) if n == 1 and x == 0: print(1) continue alice = 2 remaining = n -1 if x != 0 and (x == n or x == n - 1): print(-1) elif x == 0: for j in range(1, n + 1): print(j, end=" ") print() else: a = list(range(1, n + 1)) a.pop() a.insert(max(0, n-x-2), n) for j in a: print(j, end="") print() Codechef Greedy Lis

string make_string_S_to_T(string S) {     string T=“programming”;     bool possible = false;     int M = T.length();     int N = S.length();     for (int i = 0; i <= M; i++) {         int prefix_length = i;         int suffix_length = M - i;         string prefix = S.substr(0, prefix_length);         string suffix = S.substr(N - suffix_length, suffix_length);         if (prefix + suffix == T) {             possible = true;             break;         }     }     if (possible)         return "YES";     else         return "NO"; } Deleting substring ✅ Zeta Telegram:- @allcoding1_official

#include using namespace std; vector solution(vector a, int n, int k) { &nbsp;&nbsp;&nbsp; vector v; &nbsp;&nbsp;&nbsp; deque
#include <bits/stdc++.h> using namespace std; vector<int> solution(vector<int> a, int n, int k) {     vector<int> v;     deque<int> dq;     for (int i = 0; i < n; i++) {         while (!dq.empty() && dq.front() <= i - k)             dq.pop_front();         while (!dq.empty() && a[dq.back()] <= a[i])             dq.pop_back();         dq.push_back(i);         if (i >= k - 1)             v.push_back(a[dq.front()]);     }     return v; } int main() {     int n, k;     cin >> n >> k;     vector<int> a(n);     for (int i = 0; i < n; i++)         cin >> a[i];     vector<int> result = solution(a, n, k);     for (int i = 0; i < result.size(); i++)         cout << result[i] << " ";     return 0; }.  //cricket match ✅ Zeta

class Solution { public:      void bfs(vector>&vis,vector>&grid,int i,int j,int n,int m)     {         vis[i][j]=1;         queue>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=0 and ncol>& grid) {     int n=grid.size();         int m=grid[0].size();         vector>vis(n,vector(m,0));         int cnt=0;         for(int i=0;i

#include <bits/stdc++.h> using namespace std; vector<int> solution(vector<int> a, int n, int k) {     vector<int> v;     deque<int> dq;     for (int i = 0; i < n; i++) {         while (!dq.empty() && dq.front() <= i - k)             dq.pop_front();         while (!dq.empty() && a[dq.back()] <= a[i])             dq.pop_back();         dq.push_back(i);         if (i >= k - 1)             v.push_back(a[dq.front()]);     }     return v; } int main() {     int n, k;     cin >> n >> k;     vector<int> a(n);     for (int i = 0; i < n; i++)         cin >> a[i];     vector<int> result = solution(a, n, k);     for (int i = 0; i < result.size(); i++)         cout << result[i] << " ";     return 0; }.  //cricket match ✅ Zeta

string make_string_S_to_T(string S) {     string T=“programming”;     bool possible = false;     int M = T.length();     int N = S.length();     for (int i = 0; i <= M; i++) {         int prefix_length = i;         int suffix_length = M - i;         string prefix = S.substr(0, prefix_length);         string suffix = S.substr(N - suffix_length, suffix_length);         if (prefix + suffix == T) {             possible = true;             break;         }     }     if (possible)         return "YES";     else         return "NO"; } Deleting substring ✅ Zeta

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Fake #scammer @allcoding1_officiaI

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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✅

IBM✅ SQL
+1
IBM✅ SQL

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

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🎯Google Internship 2024 | Software Student Training in Engineering Program | Apply Now Job Role : Software Student Training in Engineering Program Qualification : B.E/B.Tech/B.Sc Experience : Freshers Last Date : 19 January 2024 Apply Now:- www.allcoding1.com Telegram:- @allcoding1_official

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