allcoding1
Kanalga Telegram’da o‘tish
Ko'proq ko'rsatish
📈 Telegram kanali allcoding1 analitikasi
allcoding1 (@allcoding1) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 21 545 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, 21 545 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.
21 545
Obunachilar
-524 soatlar
-817 kun
-36930 kun
Postlar arxiv
21 545
Repost from allcoding1
🎯Qualcomm Off Campus Recruitment
Job Position:- IT Software Developer (UX/UI/Fullstack)
Job ID: 3058496
Qualification:- B.Tech
Batch:- 2019/2020/2021
Job Location:- Hyderabad, India
Department: Software Engineering
Salary Package:- As per Company Standards
Job Type:- Full-time
Last Date:- ASAP
Apply Now:- https://www.allcoding1.com/2024/02/qualcomm-off-campus-recruitment-2024.html?m=1
Telegram:- @allcoding1
21 545
500 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🇩🇪 )
₹50
Contact:- @meterials_available
21 545
![int main() { ll N; cin >> N; vector w(N); for(ll i = 0; i < N; ++i) { cin >> w[i]; } vector s(N, 15); for(ll i](https://cdn.tlmtr.io/_next/static/media/empty-channel.35ed232f.svg)
int main() {
ll N;
cin >> N;
vector<ll> w(N);
for(ll i = 0; i < N; ++i) {
cin >> w[i];
}
vector<ll> s(N, 15);
for(ll i = N-2; i >= 0; --i) {
for(ll j = i+1; j < N; ++j) {
if(w[i] < w[j]) {
s[i] = 10;
for(ll k = j+1; k < N; ++k) {
if(w[j] > w[k]) {
s[i] = 5;
break;
}
}
break;
}
}
}
ll g = 0;
for(ll sc : s) {
g += sc;
}
cout << g << endl;
return 0;
}
chair manufacture
@allcoding1
21 545
int main() {
ll N;
cin >> N;
vector<ll> u(N);
for(ll i = 0; i < N; ++i) {
cin >> u[i];
}
ll M, X;
cin >> M >> X;
vector<vector<ll>> a(N);
for(ll i = 0; i < M; ++i) {
ll x, y;
cin >> x >> y;
a[x].push_back(y);
a[y].push_back(x);
}
vector<bool> v(N, false);
queue<ll> q;
q.push(0);
v[0] = true;
ll m = 0;
while(!q.empty()) {
ll s = q.size();
ll us = 0;
for(ll i = 0; i < s; ++i) {
ll n = q.front();
q.pop();
us += u[n];
for(ll nb : a[n]) {
if(!v[nb]) {
q.push(nb);
v[nb] = true;
}
}
}
m = max(m, us);
}
cout << m << endl;
return 0;
}
apartment
@allcoding1
21 545
class TreeNode:
def init(self, val):
self.val = val
self.children = []
def build_tree(units, relationships):
nodes = {i: TreeNode(units[i]) for i in range(len(units))}
for rel in relationships:
parent, child = rel
nodes[parent].children.append(nodes[child])
return nodes[0]
def max_electricity_per_floor(root):
max_electricity = 0
floor_electricity = {}
def dfs(node, level):
nonlocal max_electricity
if level not in floor_electricity:
floor_electricity[level] = 0
floor_electricity[level] += node.val
max_electricity = max(max_electricity, floor_electricity[level])
for child in node.children:
dfs(child, level + 1)
dfs(root, 0)
return max_electricity
num = int(input())
units = list(map(int, input().split()))
numRel, memConnect = map(int, input().split())
relationships = [tuple(map(int, input().split())) for _ in range(numRel)]
root = build_tree(units, relationships)
result = max_electricity_per_floor(root)
print(result)
Apartment one
@allcoding1
21 545
Repost from allcoding1_official
X(1), (2)
int count;
// Line 1
Antint i) { count 1)
int getcount () {return count)
// Line 5
public class Book
public static void main(String[] args)
System.out.println(AB.X.count + AB.Y.count); // Line 10
Ans)Compile time error at "// Line 10"
@allcoding1_official
21 545
Repost from allcoding1_official
Which of the statement(s) given below is/are correct?
1)Routers can be static or dynamic
2)Routers connect two or more separate networks,
Ans) both 1 and 2
@allcoding1_official
21 545
Repost from allcoding1_official
In MS Word, how can a document be split into multiple parts such that each part can have different formats and layouts
Ans)By creating sections
@allcoding1_official
21 545
Repost from allcoding1_official
How many pairs of letters are there in the word "DYNASTY" which has as many letters between them (from both the sides) as in the alphabets from A to Z?
A) 2
@allcoding1_official
21 545
Repost from allcoding1_official
Mostly, it is students from English mediun schools who appear in this competition
@allcoding1_official
