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📈 Аналитический обзор Telegram-канала allcoding1

Канал allcoding1 (@allcoding1) языкового сегмента Английский является активным участником. Сейчас сообщество объединяет 21 554 подписчиков, занимая 9 078 место в категории Образование и 18 983 место в регионе Индия.

📊 Показатели аудитории и динамика

С момента создания невідомо проект демонстрирует стремительный рост, собрав аудиторию из 21 554 подписчиков.

Согласно последним данным от 31 августа, 2026, канал показывает стабильную активность. За последние 30 дней изменение числа участников составило -369, а за последние 24 часа — -5, при этом общий охват остаётся высоким.

  • Статус верификации: Не верифицирован
  • Уровень вовлечённости (ER): Средний показатель вовлечённости аудитории составляет 6.77%. В первые 24 часа после публикации контент обычно набирает N/A% реакций от общего числа подписчиков.
  • Охват публикаций: В среднем каждый пост получает 1 460 просмотров. В течение первых суток публикация набирает 0 просмотров.
  • Реакции и взаимодействия: Аудитория активно поддерживает контент: среднее количество реакций на один пост — 0.
  • Тематические интересы: Контент сосредоточен на ключевых темах, таких как dsa, stack, namaste, javascript, learning.

📝 Описание и контентная политика

Описание канала не предоставлено.

Благодаря высокой частоте обновлений (последние данные получены 01 сентября, 2026) канал поддерживает актуальность и высокий уровень охвата публикаций. Аналитика показывает, что аудитория активно взаимодействует с контентом, что делает его важной точкой влияния в категории Образование.

21 554
Подписчики
-524 часа
-817 дней
-36930 день
Архив постов
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#include<bits/stdc++.h> using namespace std; const int MOD = 1e9 + 7; vector<vector<vector<string>>> dp; vector<string> solve(string& alice, string& bob, int i, int j) {     if(i == alice.size() || j == bob.size()) {         return {""};     }     if(dp[i][j].size() != 0) {         return dp[i][j];     }     if(alice[i] == bob[j]) {         vector<string> tmp = solve(alice, bob, i+1, j+1);         for(string& s : tmp) {             s = alice[i] + s;         }         return dp[i][j] = tmp;     }     vector<string> left = solve(alice, bob, i+1, j);     vector<string> right = solve(alice, bob, i, j+1);     if(left[0].size() > right[0].size()) {         return dp[i][j] = left;     }     if(left[0].size() < right[0].size()) {         return dp[i][j] = right;     }     left.insert(left.end(), right.begin(), right.end());     sort(left.begin(), left.end());     left.erase(unique(left.begin(), left.end()), left.end());     return dp[i][j] = left; } int main() {     int T;     cin >> T;     while(T--) {         string alice, bob;         cin >> alice >> bob;         dp = vector<vector<vector<string>>>(alice.size(), vector<vector<string>>(bob.size()));         vector<string> trips = solve(alice, bob, 0, 0);         for(string& trip : trips) {             cout << trip << endl;         }     }     return 0; }

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#include <iostream> #include <vector> #include <algorithm> using namespace std; class Solution { public:     int maxSumOptimalArrangement(vector<int>& coins) {         vector<int> positive;         vector<int> negative;         for (int coin : coins) {             if (coin >= 0)                 positive.push_back(coin);             else                 negative.push_back(coin);         }         sort(positive.rbegin(), positive.rend());                 sort(negative.begin(), negative.end());         int totalSum = 0;         for (size_t i = 0; i < max(positive.size(), negative.size()); ++i) {             if (i < positive.size())                 totalSum += positive[i];             if (i < negative.size())                 totalSum -= negative[i];         }         return totalSum;     } }; ZS campus beat code Circuit Board Telegram:- @allcoding1

📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Work
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#include #include #define MOD 1000000007 int compare(const void *a, const void *b) { &nbsp;&nbsp;&nbsp; return (*(int*)a - *(
#include<stdio.h> #include<stdlib.h> #define MOD 1000000007 int compare(const void *a, const void *b) {     return (*(int*)a - *(int*)b); } int main() {     int N;     scanf("%d", &N);     if (N <= 1) {         printf("NO HOURS\n");         return 0;     }     int *A = (int*)malloc(N * sizeof(int));     for (int i = 0; i < N; i++) {         scanf("%d", &A[i]);     }     qsort(A, N, sizeof(int), compare);     int count = 0;     for (int i = 0; i < N - 1; i++) {         for (int j = i + 1; j < N; j++) {             if ((A[i] + A[j]) % 60 == 0) {                 count = (count + 1) % MOD;             }         }     }     if (count > 0) {         printf("%d\n", count);     } else {         printf("NO HOURS\n");     }     free(A);     return 0; } Hours Count Telegram:-

Repost from allcoding1_official
📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Work
+8
📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 All courses (100 rupees) Contact:- @meterials_available Proofs:- https://t.me/+mP2YTsMihZozNWQ1

Repost from allcoding1_official
📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Work
+8
📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 All courses (100 rupees) Contact:- @meterials_available Proofs:- https://t.me/+mP2YTsMihZozNWQ1

Repost from allcoding1_official
📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Work
+8
📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 All courses (100 rupees) Contact:- @meterials_available

Repost from allcoding1_official
📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Work
+8
📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 100 rupees Contact:- @meterials_available

Repost from allcoding1_official
📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Work
+8
📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 100 rupees Contact:- @meterials_available

Exam Ans:- @allcoding1 Jobs post:- @allcoding1_official

Salesperson ID's - IBM
Salesperson ID's - IBM

Lexicographical smallest substring - IBM
Lexicographical smallest substring - IBM

def max_call_executives(n, start_times, end_times): timeline = [0] * (24 * 60 + 1) for i in range(n): start = int(start_times
def max_call_executives(n, start_times, end_times):     timeline = [0] * (24 * 60 + 1)         for i in range(n):         start = int(start_times[i][:2]) * 60 + int(start_times[i][2:])         end = int(end_times[i][:2]) * 60 + int(end_times[i][2:])         timeline[start] += 1         timeline[end] -= 1         max_executives = 0     current_executives = 0     for i in range(len(timeline)):         current_executives += timeline[i]         max_executives = max(max_executives, current_executives)         return max_executives Call Centre

import heapq def distance(x, y): return x*x + y*y def nearest_houses(P, T, queries): distances = [] heapq.heapify(distances)
import heapq def distance(x, y):     return x*x + y*y def nearest_houses(P, T, queries):     distances = []     heapq.heapify(distances)     for query in queries:         if query[0] == 1:             x, y = query[1], query[2]             heapq.heappush(distances, distance(x, y))         else:             nearest = heapq.nsmallest(T, distances)[-1]             print(nearest) Nearest House

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📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Work
+8
📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 100 rupees Contact:- @meterials_available

#include #include using namespace std; const int MOD = 1e9 + 7; int F(int i, int k, int n, vector &amp;level, vector &amp;dp)
#include <iostream> #include <vector> using namespace std; const int MOD = 1e9 + 7; int F(int i, int k, int n, vector<int> &level, vector<int> &dp) {     if (i == n) return 1;     if (dp[i] != -1) return dp[i];     int ans = 0, odds = 0;     vector<int> hash(n + 10, 0);     for (int j = i; j < n; j++) {         if (++hash[level[j]] % 2 == 0) odds -= 1;         else odds += 1;         if (odds <= k) {             ans = (ans + F(j + 1, k, n, level, dp)) % MOD;         }     }     return dp[i] = ans; } int countValidPartitions(vector<int> level, int k) {     int n = level.size();     vector<int> dp(n, -1);     return F(0, k, n, level, dp); } DE Shaw