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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 85 483 subscribers, ranking 1 507 in the Technologies & Applications category and 3 480 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 85 483 subscribers.
According to the latest data from 24 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -1 369 over the last 30 days and by -35 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 2.60%. Within the first 24 hours after publication, content typically collects 0.55% reactions from the total number of subscribers.
- Post reach: On average, each post receives 2 222 views. Within the first day, a publication typically gains 474 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 3.
- 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 25 June, 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.
85 483
Subscribers
-3524 hours
-2687 days
-1 36930 days
Posts Archive
85 467
#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
85 467
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)
85 467
#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;
}85 467
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
85 467
#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
85 467
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,m;
cin>>n>>m;
vector<int>ans(n);
if(n<=m){
for(int i=0;i<n;i++){
ans[i]=i+1;
}
}
else{
int k=n/m;
// int r=n%m;
int j=0;
for(int i=0;i<n;i++){
ans[i]=j+1;
j++;
j%=m;
}
}
for(int i=0;i<n;i++){
cout<<ans[i]<<" ";
}
cout<<endl;
}
King Dreams ✅
Intuit
Telegram:- @meterials_available
85 467
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85 467
class Solution {
public:
int minOperations(vector& nums, int x, int y) {
int l = 0;
int r = ranges::max(nums);
while (l < r) {
const int m = (l + r) / 2;
if (isPossible(nums, x, y, m))
r = m;
else
l = m + 1;
}
return l;
}
private:
bool isPossible(const vector& nums, int x, int y, int m) {
long long additionalOps = 0;
for (const int num : nums)
additionalOps += max(0LL, (num - 1LL * y * m + x - y - 1) / (x - y));
return additionalOps <= m;
}
};
Job Execution ✅
