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allcoding1 (@allcoding1) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 21 554 obunachidan iborat bo'lib, Taʼlim toifasida 9 078-o'rinni va Hindiston mintaqasida 18 983-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 21 554 obunachiga ega bo‘ldi.

31 Avgust, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -369 ga, so‘nggi 24 soatda esa -5 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 6.77% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining N/A% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 1 460 marta ko‘riladi; birinchi sutkada odatda 0 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 0 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, learning kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Kanal uchun tavsif kiritilmagan.

Yuqori yangilanish chastotasi (oxirgi ma’lumot 01 Sentabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

21 554
Obunachilar
-524 soatlar
-817 kunlar
-36930 kunlar
Postlar arxiv
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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

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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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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