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allcoding1 (@allcoding1) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 22 543 obunachidan iborat bo'lib, Taʼlim toifasida 8 854-o'rinni va Hindiston mintaqasida 19 507-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 22 543 obunachiga ega bo‘ldi.
14 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -445 ga, so‘nggi 24 soatda esa -14 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 6.31% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.25% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 1 423 marta ko‘riladi; birinchi sutkada odatda 282 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 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 16 Iyun, 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.
22 543
Obunachilar
-1424 soatlar
-947 kunlar
-44530 kunlar
Postlar arxiv
22 543
+1https://youtube.com/@TechHelp-kn9fb?si=ZrRasvSTTjfAn-o8
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22 543
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Qualification: Any Graduate
Work Experience: Freshers / Experience
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22 543
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Job Role Analyst Trainee
Qualification BE, B Tech, ME, M Tech, MCA
Location PAN India
Salary 4-8 LPA
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22 543
Cognizant Off Campus Hiring | Data Analyst – Fresher | Upto 7.5 LPA
Job RoleData Analyst
QualificationB.E/B.Tech
BatchAny Batch
CTC/SalaryUpto 7.5 LPA
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22 543
def count(a, X):
a.sort()
ans = 0
n = len(a)
for i in range(n):
if (i < n // 2):
ans += max(0, a[i] - X)
elif (i == n // 2):
ans += abs(X - a[i])
else:
ans += max(0, X - a[i]);
return ans
IBM
Telegram:- @allcoding1
22 543
#include<bits/stdc++.h>
using namespace std;
void dfs(int node, vector<int>& vis, vector<vector<int>>& adj) {
vis[node] = 1;
for(auto it : adj[node]) {
if(!vis[it]) {
dfs(it, vis, adj);
}
}
}
int main() {
int n, m;
cin >> n >> m;
vector<vector<int>> adj(n);
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
int x;
cin >> x;
if(x == 1) {
adj[i].push_back(j);
adj[j].push_back(i);
}
}
}
vector<int> vis(n, 0);
int cc = 0;
for(int i = 0; i < n; i++) {
if(!vis[i]) {
dfs(i, vis, adj);
cc++;
}
}
cout << cc << endl;
return 0;
}
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22 543
int solve(vector<int>& nums) {
vector<int> s(nums);
sort(s.begin(), s.end());
int i = 0, j = nums.size() - 1;
while (i < nums.size() && nums[i] == s[i]) {
i++;
}
while (j > i && nums[j] == s[j]) {
j--;
}
return j - i + 1;
}
Arrange the heights
Apple ✅
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22 543
#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll solve(ll k, ll s) {
auto comb = [](ll n) {
return n * (n - 1) / 2;
};
if (k > 3 * s) return 0;
ll ans = comb(k + 2);
if (k > s) ans -= 3 * comb(k - s + 1);
if (k - 2 >= 2 * s) ans += 3 * comb(k - 2 * s);
return ans;
}
int main() {
ll n, s;
cin >> n >> s;
cout << solve(s,n) << endl;
return 0;
}
Distribute Car Toy
Service Now ✅
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22 543
🎯Amazon off Campus Drive 2024 Hiring Freshers As Associate Quality Services | 2.52-4 LPA
Job Role : Associate Quality Services
Qualification : BE/BTech/MCA/BSc/BCA/MSc
Experience : Freshers
Salary CTC : Rs 2.52-4 LPA
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22 543
SQL, IBM, TESTING and SAP support anybody who wants contact on Instagram.
https://www.instagram.com/allcoding1_official?igshid=OGQ5ZDc2ODk2ZA==
22 543
string solve(string bs) {
map<string, string> nb = {
{"001", "C"},
{"010", "G"},
{"011", "A"},
{"101", "T"},
{"110", "U"},
{"000", "DNA"},
{"111", "RNA"}
};
string ds = "";
string t = nb[bs.substr(0, 3)];
for(int i = 3; i < bs.length(); i += 3) {
string b = bs.substr(i, 3);
if(nb.find(b) != nb.end()) {
string x = nb[b];
if(t == "DNA" && x == "U") {
x = "T";
}
ds += x;
} else {
ds += "Error";
}
}
return ds;
}
DNA✅
IBM
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22 543
bool isPal(int n) {
int r, s = 0, t;
t = n;
while (n > 0) {
r = n % 10;
s = (s * 10) + r;
n = n / 10;
}
return (t == s);
}
int firstPal(int n) {
int i = 1;
while (true) {
if (isPal(i)) {
int d = 1 + log10(i);
if (d == n)
return i;
}
i++;
}
}
void login(int d, string u, string p) {
map<string, string> users = {
{"user1", "pass1"},
{"user2", "pass2"},
{"user3", "pass3"},
{"user4", "pass4"},
{"user5", "pass5"}
};
if (users.find(u) != users.end() && users[u] == p) {
int t = firstPal(d);
cout << "Welcome " << u << " and the generated token is: token-" << t << endl;
} else {
cout << "UserId or password is not valid, please try again." << endl;
}
}
IBM✅
Telegram:- @allcoding1
22 543
Here's a Python program to simulate the given problem:
`python
def print_terrain(terrain):
for row in terrain:
print(''.join(row))
def flow_water(terrain, n):
water_level = int(terrain[n // 2][n // 2])
terrain[n // 2][n // 2] = 'W'
def can_flow(x, y, direction):
if direction == 'N':
return x > 0 and terrain[x-1][y] != 'W' and int(terrain[x-1][y]) <= water_level
elif direction == 'S':
return x < n - 1 and terrain[x+1][y] != 'W' and int(terrain[x+1][y]) <= water_level
elif direction == 'E':
return y < n - 1 and terrain[x][y+1] != 'W' and int(terrain[x][y+1]) <= water_level
elif direction == 'W':
return y > 0 and terrain[x][y-1] != 'W' and int(terrain[x][y-1]) <= water_level
def flow(x, y):
if can_flow(x, y, 'N'):
terrain[x-1][y] = 'W'
return True
if can_flow(x, y, 'S'):
terrain[x+1][y] = 'W'
return True
if can_flow(x, y, 'E'):
terrain[x][y+1] = 'W'
return True
if can_flow(x, y, 'W'):
terrain[x][y-1] = 'W'
return True
return False
while True:
print_terrain(terrain)
has_flown = False
for i in range(n):
for j in range(n):
if terrain[i][j] == 'W':
if flow(i, j):
has_flown = True
if not has_flown:
water_level += 1
print(f"Cannot flow, increasing water level to {water_level}")
break
if any(cell == 'W' and (i == 0 or j == 0 or i == n - 1 or j == n - 1) for i, row in enumerate(terrain) for j, cell in enumerate(row)):
print("Reached edge, exiting.")
break
n = 7
terrain = [
[494, 88, 89, 778, 984, 726, 587],
[340, 959, 220, 301, 639, 280, 290],
[666, 906, 632, 824, 127, 505, 787],
[673, 499, 843, 172, 193, 613, 154],
[544, 211, 124, 60, 575, 572, 389],
[635, 170, 174, 946, 593, 314, 300],
[620, 167, 931, 780, 416, 954, 275]
]
flow_water(terrain, n)
Python
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