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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 22 601 subscribers, ranking 8 831 in the Education category and 19 534 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 22 601 subscribers.

According to the latest data from 10 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -430 over the last 30 days and by -7 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 5.78%. Within the first 24 hours after publication, content typically collects 1.40% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 1 307 views. Within the first day, a publication typically gains 317 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
  • 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 11 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 Education category.

22 601
Subscribers
-724 hours
-907 days
-43030 days
Posts Archive
using System; using System.Collections.Generic; class Program { static void Main() { int N = int.Parse(Console.ReadLine()); var graph = new Dictionary>(); var indegrees = new Dictionary(); for (int i = 0; i < N; i++) { var edge = Console.ReadLine().Split(); string from = edge[0]; string to = edge[1]; if (!graph.ContainsKey(from)) graph[from] = new List(); graph[from].Add(to); if (!indegrees.ContainsKey(from)) indegrees[from] = 0; if (!indegrees.ContainsKey(to)) indegrees[to] = 0; indegrees[to]++; } var words = Console.ReadLine().Split(); var levels = new Dictionary(); var queue = new Queue(); foreach (var node in indegrees.Keys) { if (indegrees[node] == 0) { levels[node] = 1; // Root level is 1 queue.Enqueue(node); } } while (queue.Count > 0) { var current = queue.Dequeue(); int currentLevel = levels[current]; if (graph.ContainsKey(current)) { foreach (var neighbor in graph[current]) { if (!levels.ContainsKey(neighbor)) { levels[neighbor] = currentLevel + 1; queue.Enqueue(neighbor); } indegrees[neighbor]--; if (indegrees[neighbor] == 0 && !levels.ContainsKey(neighbor)) { queue.Enqueue(neighbor); } } } } int totalValue = 0; foreach (var word in words) { if (levels.ContainsKey(word)) { totalValue += levels[word]; } else { totalValue += -1; } } Console.Write(totalValue); } } String Puzzle C+

int main() { string s; cin >> s; int n = s.length(), res = 0; vector v(n); for (int i = 0; i < n; ++i) cin >> v[i]; int lw = s[0] - '0', lwv = v[0]; for (int i = 1; i < n; ++i) { if (s[i] - '0' == lw) { res += min(lwv, v[i]); lwv = max(lwv, v[i]); } else { lw = s[i] - '0'; lwv = v[i]; } } cout << res; return 0; } Form alternating string Change accordingly before submitting to avoid plagiarism

#include<stdio.h> int main() {     int n;     scanf("%d",&n);     int arr[n];     for(int i=0;i<n;i++)     {         scanf("%d",&arr[i]);     }     int count=0;     int num=arr[0];     for(int i=1;i<n;i++)     {        if(num!=arr[i])             count++;     }     printf("%d",count); }

Tcs exam
Tcs exam

#include <iostream> #include <vector> #include <map> #include <set> #include <cmath> #include <algorithm> using namespace std; struct Line { int x1, y1, x2, y2; }; int countCells(Line line, pair<int, int> star, bool split) { if (line.x1 == line.x2) { if (split) { return min(abs(star.second - line.y1), abs(star.second - line.y2)) + 1; } else { return abs(line.y1 - line.y2) + 1; } } else { if (split) { return min(abs(star.first - line.x1), abs(star.first - line.x2)) + 1; } else { return abs(line.x1 - line.x2) + 1; } } } bool intersects(Line a, Line b, pair<int, int>& intersection) { if (a.x1 == a.x2 && b.y1 == b.y2) { if (b.x1 <= a.x1 && a.x1 <= b.x2 && a.y1 <= b.y1 && b.y1 <= a.y2) { intersection = {a.x1, b.y1}; return true; } } if (a.y1 == a.y2 && b.x1 == b.x2) { if (a.x1 <= b.x1 && b.x1 <= a.x2 && b.y1 <= a.y1 && a.y1 <= b.y2) { intersection = {b.x1, a.y1}; return true; } } return false; } int main() { int N, K; cin >> N; vector<Line> lines(N); for (int i = 0; i < N; ++i) { cin >> lines[i].x1 >> lines[i].y1 >> lines[i].x2 >> lines[i].y2; if (lines[i].x1 > lines[i].x2 || (lines[i].x1 == lines[i].x2 && lines[i].y1 > lines[i].y2)) { swap(lines[i].x1, lines[i].x2); swap(lines[i].y1, lines[i].y2); } } cin >> K; map<pair<int, int>, vector<Line>> stars; for (int i = 0; i < N; ++i) { for (int j = i + 1; j < N; ++j) { pair<int, int> intersection; if (intersects(lines[i], lines[j], intersection)) { stars[intersection].push_back(lines[i]); stars[intersection].push_back(lines[j]); } } } int asiylam = 0; for (auto& star : stars) { if (star.second.size() / 2 == K) { vector<int> intensities; for (auto& line : star.second) { intensities.push_back(countCells(line, star.first, true)); } asiylam += *min_element(intensities.begin(), intensities.end()); } } cout << asiylam << endl; return 0; } Magic Star Intensity Code C++ TCS Exam

int main() { int n, m, k, d = 1, rv = 0; cin >> n >> m; vector> f(n); for (int i = 0, u, v; i < m; ++i) { cin >> u >> v; f[u].insert(v); f[v].insert(u); } cin >> k; vector w(n, true); rv = n; while (rv < k) { vector nw(n, false); for (int i = 0; i < n; ++i) { int cnt = 0; for (int fr : f[i]) cnt += w[fr]; if (w[i] && cnt == 3) nw[i] = true; else if (!w[i] && cnt < 3) nw[i] = true; } w = nw; rv += count(w.begin(), w.end(), true); ++d; } cout << d; return 0; } Office Rostering C+ TCS exam

Industrial code
Industrial code

#include<stdio.h> int main() {     int n;     scanf("%d",&n);     int arr[n];     for(int i=0;i<n;i++)     {         scanf("%d",&arr[i]);     }     int count=0;     int num=arr[0];     for(int i=1;i<n;i++)     {        if(num!=arr[i])             count++;     }     printf("%d",count); } C Language TCS 1st Qsn --------------------------------------------------------- N=int(input()) K=int(input()) price=list(map(int,input().split())) vol=list(map(int,input().split())) maxvol=0 volu=0 maxvol=max(vol) for i in range(0,N):     if (maxvol==vol[i] and price[i]<=K):         K=K-price[i]         volu=maxvol for i in range(0,N):     for j in range(i+1,N+1):         if (price[i]<=K and price[i]==price[j]):             if (vol[i]>vol[j]):                 volu=volu+vol[i]                 K=K-price[i]             else:                 volu=volu+vol[j]                 K=K-price[j]         elif (price[i]<=K and price[i]!=price[j]):             K=K-price[i]             ------- include<stdio.h> int main() {     int n;     scanf("%d",&n);     int arr[n];     for(int i=0;i<n;i++)     {         scanf("%d",&arr[i]);     }     int count=0;     int num=arr[0];     for(int i=1;i<n;i++)     {        if(num!=arr[i])             count++;     }     printf("%d",count); } Array Code in C language

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allcoding1 - Statistics & analytics of Telegram channel @allcoding1