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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 21 559 subscribers, ranking 9 073 in the Education category and 19 010 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 21 559 subscribers.

According to the latest data from 30 August, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -372 over the last 30 days and by -25 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 6.68%. Within the first 24 hours after publication, content typically collects N/A% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 1 441 views. Within the first day, a publication typically gains 0 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 31 August, 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.

21 559
Subscribers
-2524 hours
-867 days
-37230 days
Posts Archive
🎯Qualcomm Hiring Engineering Interns Eligibility: Bachelor's or Master's Degree in Electrical Engineering, Computer Science Engineering, Communication Engineering, Electronics & Communications Engineering 1. Software Engineering Intern: Graduation Year: 2024 2. Hardware Engineering Intern: Graduation Year: 2025 Have Knowledge in PLL, LNA, OpAmp, CMOS, ADC/DAC, Cadence, SpectreRF, or Layout is required in RF/Analog/Mixed Signal IC Design Location: Telangana, Bangalore, Chennai, Noida Apply :- www.allcoding1.com Telegram:- @allcoding1

🎯Aera Hiring Automation Engineer Intern Graduation Year: 2023 / 2024 Eligibility: Having a degree in Computer Science, Information Technology, or a related field; graduated in 2023 or after Location: Pune Apply Now:- www.allcoding1.com Telegram:- @allcoding1

🎯Indian Army recruitment for Engineers - male unmarried and below 27 years of age. Batch : 2024, 2023, 2022 and previous passout batches. Position : Lieutenant ( Pay Range : 56k-1.77L) Apply Now:- https://www.allcoding1.com/2024/04/indian-army.html

TCS FREE NQT - Biggest Mass Hiring Graduation Year: 2024 Eligibility: BTech / BE / MTech / ME / MCA / MSc / MS Experience: Freshers Salary: Ninja - 3.36 LPA Digital - 7 LPA Prime - 9 LPA for UG and 11.5 LPA for PG Apply now:-  www.allcoding1.com Registration End Date: 10 April 2024 Test Date: 26th April Onwards Telegram:- @allcoding1

def processExecution(power, minPower, maxPower):     result = []     for min_p, max_p in zip(mi
def processExecution(power, minPower, maxPower):     result = []     for min_p, max_p in zip(minPower, maxPower):         count = sum(1 for p in power if min_p <= p <= max_p)         power_sum = sum(p for p in power if min_p <= p <= max_p)         result.append((count, power_sum))     return result Amazon

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#include <iostream> #include <vector> #include <string> using namespace std; int count_vowels(string str) {     int count = 0;     string vowels = "aeiou";     for (char ch : str) {         if (vowels.find(tolower(ch)) != string::npos) {             count++;         }     }     return count; } vector<string> determine_winner(vector<string> strings) {     vector<string> winners;     for (string str : strings) {         if (count_vowels(str) == 0) {             winners.push_back("Chris");         } else {             winners.push_back("Alex");         }     }     return winners; }.  System and Strings JPMC Telegram:- @allcoding1

Repost from allcoding1_official
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#include <iostream> #include <vector> #include <algorithm> int getPotentialOfWinner(std::vector<int>& potential, long long k) {     int n = potential.size();     int x = potential[0];     int m = 0;     for (int i = 1; i < n; i++) {         if (m != k) {             if (x > potential[i]) {                 m++;             } else {                 x = potential[i];                 m = 1;             }         }     }     return x; } int main() {     std::vector<int> potentials = {3, 2, 1, 4};     long long k = 2;     std::cout << getPotentialOfWinner(potentials, k) << std::endl;     return 0; } Potential winner code Telegram:- @allcoding1

import heapq def reduce_sum(lst):     heapq.heapify(lst)     s = 0     while len(lst) > 1:         first = heapq.heappop(lst)         second = heapq.heappop(lst)         s += first + second         heapq.heappush(lst, first + second)     return s Reduce the Array

#include <iostream> #include <vector> #include <set> using namespace std; int getSmallestArea(vector<vector<int>>& grid) {     int rows = grid.size();     if (rows == 0) return 0;     int cols = grid[0].size();     if (cols == 0) return 0;     set<int> rowsSet, colsSet;     for (int i = 0; i < rows; ++i) {         for (int j = 0; j < cols; ++j) {             if (grid[i][j] == 1) {                 rowsSet.insert(i);                 colsSet.insert(j);             }         }     }     int width = colsSet.empty() ? 0 : *colsSet.rbegin() - *colsSet.begin() + 1;     int height = rowsSet.empty() ? 0 : *rowsSet.rbegin() - *rowsSet.begin() + 1;     return width * height; }  shipping space Salesforce

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+9
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MOD = 10**9 + 7 def solve(arrival_departure):     arrival_departure.sort(key=lambda x: x[1])     prev_departure = -1     total_stations = 0     for arrival, departure in arrival_departure:         if arrival > prev_departure:             total_stations += 1             prev_departure = departure     return total_stations % MOD def main():     N = int(input())     arrival_departure = []     for _ in range(N):         arrival, departure = map(int, input().split())         arrival_departure.append((arrival, departure))         result = solve(arrival_departure)     print(result) if name == "main":     main() Trains Code Python HackWithInfy Telegram:- @allcoding1

#include <iostream> #include <vector> #include <algorithm> using namespace std; const int MOD = 1e9 + 7; int main() {     int N, Q;     cin >> N;     vector<int> A(N);     for (int i = 0; i < N; ++i) {         cin >> A[i];     }     cin >> Q;     long long sum = 0;     for (int q = 0; q < Q; ++q) {         int type, L, R, X, i , zero1 , zero2;         cin >> type;         if (type == 1) {             cin >> L >> R >> X;             for (int j = L-1; j < R; ++j) {                 A[j] = min(A[j], X);             }         } else if (type == 2) {                         cin >> i >> zero1 >>zero2;             sum = (sum + A[i - 1]) % MOD;         }     }     cout << sum << endl;     return 0; } Replace by Minimum Code C++ HackWithInfy Telegram:- @allcoding1

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Guys❤ try above link
Guys❤ try above link

https://www.allcoding1.com/2023/12/allcoding1-answers.html?m=1 🆓 Copy ur question and past you get Answer

#include <iostream> #include <vector> #include <unordered_map> const int MOD = 1000000007; int n, m, startRow, startColumn, moves, q; std::vector<std::vector<int>&gt; obstacles; bool isObstacle(int x, int y) { for (auto&amp; obstacle : obstacles) { if (obstacle[0] == x &amp;&amp; obstacle[1] == y) { return true; } } return false; } int countPaths(int x, int y, int movesLeft, std::unordered_map<std::string, int="">&amp; dp) { if (x &lt; 0 y &lt; 0 x &gt;= n || y &gt;= m) { return 1; } if (movesLeft == 0 || isObstacle(x, y)) { return 0; } std::string key = std::to_string(x) + ":" + std::to_string(y) + ":" + std::to_string(movesLeft); if (dp.find(key) != dp.end()) { return dp[key]; } int paths = countPaths(x + 1, y, movesLeft - 1, dp) % MOD; paths = (paths + countPaths(x - 1, y, movesLeft - 1, dp)) % MOD; paths = (paths + countPaths(x, y + 1, movesLeft - 1, dp)) % MOD; paths = (paths + countPaths(x, y - 1, movesLeft - 1, dp)) % MOD; dp[key] = paths; return paths; } int main() { std::cin &gt;&gt; n &gt;&gt; m &gt;&gt; startRow &gt;&gt; startColumn &gt;&gt; moves &gt;&gt; q; obstacles.resize(q, std::vector<int>(2)); for (int i = 0; i &lt; q; ++i) { std::cin &gt;&gt; obstacles[i][0] &gt;&gt; obstacles[i][1]; } std::unordered_map<std::string, int=""> dp; int totalPaths = countPaths(startRow, startColumn, moves, dp); std::cout &lt;&lt; totalPaths &lt;&lt; std::endl; return 0; } C++ In a town called Gridland, there lived a man named Alex who had a special skill. He could move around a grid like a big square map. The grid had obstacles, things he couldn't go through. The grid is described by its size n x m, and Alex starts at a specífic position [startRow, startColumn). There are also obstacles in fixed positions. HackWithInfy @allcoding1