allcoding1_official
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📈 Analytical overview of Telegram channel allcoding1_official
Channel allcoding1_official (@allcoding1_official) in the English language segment is an active participant. Currently, the community unites 78 714 subscribers, ranking 1 622 in the Technologies & Applications category and 3 757 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 78 714 subscribers.
According to the latest data from 30 October, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -1 600 over the last 30 days and by -27 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 1.38%. Within the first 24 hours after publication, content typically collects 0.62% reactions from the total number of subscribers.
- Post reach: On average, each post receives 1 091 views. Within the first day, a publication typically gains 486 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, dev.
📝 Description and content policy
Channel description not provided.
Thanks to the high frequency of updates (latest data received on 31 October, 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 Technologies & Applications category.
78 714
Subscribers
-2724 hours
-3217 days
-1 60030 days
Posts Archive
78 714
int solve(int n, vector& arr) {
while (arr.size() > 1) {
int len = INT_MAX;
int idx = -1;
for (int i = 0; i < arr.size() - 1; ++i) {
int d = arr[i + 1];
if (d == 0 || arr[i] == 0) {
continue;
}
int r = min(arr[i] % d, d % arr[i]);
if (r < len) {
len = r;
idx = i;
}
}
if (idx == -1) {
break;
}
int d = arr[idx + 1];
if (d != 0) {
arr[idx] = min(arr[idx] % d, d % arr[idx]);
}
arr.erase(arr.begin() + idx + 1);
}
return arr.size();
}
Women Day Mathematics Challenge
Telegram:-
78 714
#include <iostream>
#include <sql.h>
#include <sqlext.h>
#define MAX_QUERY_LEN 1000
int main() {
// Declare variables for ODBC connection
SQLHENV henv;
SQLHDBC hdbc;
SQLHSTMT hstmt;
SQLRETURN retcode;
// Allocate environment handle
retcode = SQLAllocHandle(SQL_HANDLE_ENV, SQL_NULL_HANDLE, &henv);
// Set the ODBC version to use
retcode = SQLSetEnvAttr(henv, SQL_ATTR_ODBC_VERSION, (SQLPOINTER)SQL_OV_ODBC3, 0);
// Allocate connection handle
retcode = SQLAllocHandle(SQL_HANDLE_DBC, henv, &hdbc);
// Connect to the data source
retcode = SQLConnect(hdbc, (SQLCHAR*)"your_datasource", SQL_NTS, (SQLCHAR*)"username", SQL_NTS, (SQLCHAR*)"password", SQL_NTS);
// Allocate statement handle
retcode = SQLAllocHandle(SQL_HANDLE_STMT, hdbc, &hstmt);
// Prepare the SQL query
char query[MAX_QUERY_LEN] = "SELECT DATE_FORMAT(date_column, '%Y-%m') AS month_year, source, COUNT(*) AS total_number FROM your_table WHERE YEAR(date_column) = '2022' AND (source = 'Jobs' OR source = 'Freelancers') GROUP BY month_year, source ORDER BY month_year ASC, source ASC";
// Execute the SQL query
retcode = SQLExecDirect(hstmt, (SQLCHAR*)query, SQL_NTS);
// Fetch and process the results
while (SQLFetch(hstmt) == SQL_SUCCESS) {
// Process the retrieved data here
}
// Free the handles
SQLFreeHandle(SQL_HANDLE_STMT, hstmt);
SQLDisconnect(hdbc);
SQLFreeHandle(SQL_HANDLE_DBC, hdbc);
SQLFreeHandle(SQL_HANDLE_ENV, henv);
return 0;
}
your_datasource, username, password, your_table, and date_column with your actual database connection details, table name, and column name.78 714
#include <iostream>
#include <string>
#include <iomanip>
#include <vector>
#include <algorithm>
struct Client {
std::string name;
double total_invested_in_bonds;
double total_invested_in_stocks;
};
bool compareByBonds(const Client &a, const Client &b) {
return a.total_invested_in_bonds > b.total_invested_in_bonds;
}
int main() {
std::vector<client> clients = {
{"Client1", 7500.0, 3000.0},
{"Client2", 6000.0, 5000.0},
{"Client3", 4000.0, 6000.0},
// Add more clients as needed
};
// Sort clients based on total_invested_in_bonds in descending order
std::sort(clients.begin(), clients.end(), compareByBonds);
// Display the result
std::cout << std::left << std::setw(15) << "Client Name"
<< std::setw(25) << "Total Invested in Bonds"
<< "Total Invested in Stocks" << std::endl;
for (const auto &client : clients) {
if (client.total_invested_in_bonds > 5000.00) {
std::cout << std::left << std::setw(15) << client.name
<< std::fixed << std::setprecision(2)
<< std::setw(25) << client.total_invested_in_bonds
<< client.total_invested_in_stocks << std::endl;
}
}
return 0;
}
You can modify the clients vector to include more clients with their respective investments in bonds and stocks. When you run this program, it will display the result according to the specified requirements.</client></algorithm></vector></iomanip></string></iostream>
Telegram:- @allcoding1_official
78 714
To find the ratio of A to C, we can use the given ratios for AB and BC.
Given:
AB = 2:3
BC = 4:5
We want to find the ratio of A to C.
One way to solve this is to express both AB and BC in terms of a common variable and then use that to find the ratio of A to C.
Let's express AB and BC in terms of a common multiple of B.
Let's assume the common multiple is x. Then:
AB = 2x : 3x
BC = 4x : 5x
Now, to find the ratio of A to C, we can multiply the two ratios together:
A:B * B:C = A:C
So:
(2x : 3x) * (4x : 5x)
= (2x * 4x : 3x * 5x)
= (8x^2 : 15x^2)
Therefore, the ratio of (A to C is 8:15)
78 714
#include <iostream>
#include <vector>
using namespace std;
bool isValidPermutation(const vector<int>& permutation) {
for (int i = 0; i < permutation.size(); i++) {
if (permutation[i] % (i + 1) != 0 || (i + 1) % permutation[i] != 0) {
return false;
}
}
return true;
}
void generatePermutations(vector<int>& nums, int start, vector<vector<int>>& result) {
if (start == nums.size()) {
result.push_back(nums);
return;
}
for (int i = start; i < nums.size(); i++) {
swap(nums[start], nums[i]);
generatePermutations(nums, start + 1, result);
swap(nums[start], nums[i]);
}
}
int countValidPermutations(int N) {
vector<int> nums(N);
for (int i = 0; i < N; i++) {
nums[i] = i + 1;
}
vector<vector<int>> permutations;
generatePermutations(nums, 0, permutations);
int count = 0;
for (const auto& perm : permutations) {
if (isValidPermutation(perm)) {
count++;
}
}
return count;
}
int main() {
int N = 2;
cout << "Number of valid permutations: " << countValidPermutations(N) << endl;
return 0;
}78 714
int min(string str){
unordered_map<char,int>mp;
for(char :str){
mp[ch]++;
}
unodered_set<int>d;
for(auto it:mp){
d.insert(it.second);
}
return d.size();
}
Music Teacher
Only 4 test cases pass
Telegram:- @allcoding1_official
78 714
#include <iostream>
#include <cmath>
struct Circle {
double x; // x-coordinate of the center
double y; // y-coordinate of the center
double r; // radius
};
double distanceBetweenCenters(Circle A, Circle B) {
return sqrt(pow((B.x - A.x), 2) + pow((B.y - A.y), 2));
}
int main() {
Circle A, B;
// Example values
A.x = 0;
A.y = 0;
A.r = 5;
B.x = 10;
B.y = 0;
B.r = 7;
double distance = distanceBetweenCenters(A, B);
double sumOfRadii = A.r + B.r;
double differenceOfRadii = abs(A.r - B.r);
if (distance == sumOfRadii) {
std::cout << "The circles are touching at a single point." << std::endl;
} else if (distance < sumOfRadii) {
std::cout << "The circles are intersecting." << std::endl;
} else if (distance == differenceOfRadii) {
std::cout << "The circles are touching from within or without." << std::endl;
} else {
std::cout << "The circles are not intersecting." << std::endl;
}
if ((A.x == B.x) && (A.y == B.y) && (A.r == B.r)) {
std::cout << "The circles are concentric." << std::endl;
}
return 0;
}
C++
Telegram:- @allcoding1_official
