allcoding1
前往频道在 Telegram
📈 Telegram 频道 allcoding1 的分析概览
频道 allcoding1 (@allcoding1) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 21 554 名订阅者,在 教育 类别中位列第 9 078,并在 印度 地区排名第 18 983 位。
📊 受众指标与增长动态
自 невідомо 创建以来,项目保持高速增长,吸引了 21 554 名订阅者。
根据 31 八月, 2026 的最新数据,频道保持稳定运转。过去 30 天订阅人数变化为 -369,过去 24 小时变化为 -5,整体触达仍然可观。
- 认证状态: 未认证
- 互动率 (ER): 平均受众互动率为 6.77%。内容发布后 24 小时内通常能获得 N/A% 的反应,占订阅者总量。
- 帖子覆盖: 每篇帖子平均可获得 1 460 次浏览,首日通常累积 0 次浏览。
- 互动与反馈: 受众积极参与,单帖平均反应数为 0。
- 主题关注点: 内容集中在 dsa, stack, namaste, javascript, learning 等核心主题上。
📝 描述与内容策略
尚未提供频道描述。
凭借高频更新(最新数据采集于 01 九月, 2026),频道始终保持新鲜度与高覆盖。分析显示受众积极互动,使其成为 教育 类别中的关键影响点。
21 554
订阅者
-524 小时
-817 天
-36930 天
帖子存档
21 554
🎯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
21 554
🎯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
21 554
🎯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
21 554
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
21 554
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
21 554
#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
21 554
Repost from allcoding1_official
📌IT learning courses
📌All programing courses
📌Abdul bari courses
📌Ashok IT
100 rupees
Contact:- @meterials_available
21 554
#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
21 554
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
21 554
#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
21 554
Repost from allcoding1
📌IT learning courses
📌All programing courses
📌Abdul bari courses
📌Ashok IT
100 rupees
Contact:- @meterials_available
21 554
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
21 554
#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
21 554
Repost from allcoding1
📌IT learning courses
📌All programing courses
📌Abdul bari courses
📌Ashok IT
100 rupees
Contact:- @meterials_available
21 554
https://www.allcoding1.com/2023/12/allcoding1-answers.html?m=1
🆓 Copy ur question and past you get Answer
21 554
#include <iostream>
#include <vector>
#include <unordered_map>
const int MOD = 1000000007;
int n, m, startRow, startColumn, moves, q;
std::vector<std::vector<int>> obstacles;
bool isObstacle(int x, int y) {
for (auto& obstacle : obstacles) {
if (obstacle[0] == x && obstacle[1] == y) {
return true;
}
}
return false;
}
int countPaths(int x, int y, int movesLeft, std::unordered_map<std::string, int="">& dp) {
if (x < 0 y < 0 x >= n || y >= 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 >> n >> m >> startRow >> startColumn >> moves >> q;
obstacles.resize(q, std::vector<int>(2));
for (int i = 0; i < q; ++i) {
std::cin >> obstacles[i][0] >> obstacles[i][1];
}
std::unordered_map<std::string, int=""> dp;
int totalPaths = countPaths(startRow, startColumn, moves, dp);
std::cout << totalPaths << 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
