allcoding1_official
Open in Telegram
📈 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
![function ArithGeoll(arr) { let arith = true; let d = Math.abs(arr[0]- arr[1]); for (let i = 2; i < arr.length; i++) { if (](https://cdn.tlmtr.io/_next/static/media/empty-channel.35ed232f.svg)
function ArithGeoll(arr) { let arith = true; let d = Math.abs(arr[0]-
arr[1]);
for (let i = 2; i < arr.length; i++)
{ if (Math.abs(arr[i] - arr[i -
1]) !== d) {
arith = false;
}
}
if (arith) {
return "Arithmetic"; }
let geo = true; d = arr[1]/ arr[0];
for (let i = 2; i < arr.length; i++)
if (Math.abs(arr[i] / arr[i-
{
1]) !== d) {
geo = false;
}
}
if (geo) { return "Geometric";
}
if (larith && !geo) { return "-1";
}
}
78 714
function ArithGeoll(arr) { let arith = true; let d = Math.abs(arr[0]-
arr[1]);
for (let i = 2; i < arr.length; i++)
{ if (Math.abs(arr[i] - arr[i -
1]) !== d) {
arith = false;
}
}
if (arith) {
return "Arithmetic"; }
let geo = true; d = arr[1]/ arr[0];
for (let i = 2; i < arr.length; i++)
if (Math.abs(arr[i] / arr[i-
{
1]) !== d) {
geo = false;
}
}
78 714
Repost from allcoding1_official
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78 714
struct Request {
string id;
string start_time;
string end_time;
};
bool compare(Request a, Request b) {
return a.start_time < b.start_time;
}
int solve(vector requests) {
sort(requests.begin(), requests.end(), compare);
int max_concurrent = 0;
int current_concurrent = 0;
string current_end_time = "00:00:00";
for (Request request : requests) {
if (request.start_time >= current_end_time) {
current_concurrent--;
} else {
current_concurrent++;
if (current_concurrent > max_concurrent) {
max_concurrent = current_concurrent;
}
}
if (request.end_time > current_end_time) {
current_end_time = request.end_time;
}
}
return max_concurrent;
}
78 714
Repost from allcoding1_official
+2500 TB Tutorials + Books + Courses + Trainings + Workshops + Educational Resources
🔹Data science
🔹Python
🔹Artificial Intelligence
🔹AWS Certified
🔹Cloud
🔹BIG DATA
🔹Data Analytics
🔹BI
🔹Google Cloud Platform
🔹IT Training
🔹MBA
🔹Machine Learning
🔹Deep Learning
🔹Ethical Hacking
🔹SPSS
🔹Statistics
🔹Data Base
🔹Learning language resources ( English🏴 , French🇨🇵 , German🇩🇪 )
₹300
Contact:- @meterials_available
78 714
#define vv(a) vector<vector<int>>(a)
int solve(int n, int m, vector<vector<int>> &a)
{
vv(right)(n, vector<int>(m, 0));
vv(down)(n, vector<int>(m, 0));
vv(left)(n, vector<int>(m, 0));
for (int i = 0; i < n; i++)
{
int cnt = 0;
for (int j = m - 1; j >= 0; j--)
{
right[i][j] = cnt;
if (a[i][j] == 1)
cnt++;
}
}
for (int i = 0; i < n; i++)
{
int cnt = 0;
for (int j = 0; j < m; j++)
{
left[i][j] = cnt;
if (a[i][j] == 1)
cnt++;
}
}
for (int j = 0; j < m; j++)
{
int cnt = 0;
for (int i = n - 1; i >= 0; i--)
{
down[i][j] = cnt;
if (a[i][j] == 1)
cnt++;
}
}
int ans = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
if (a[i][j] == 1)
{
ans += (right[i][j] * down[i][j]);
ans += (left[i][j] * down[i][j]);
}
}
}
return ans;
}
C++
Telegram:- @allcoding1_official
