allcoding1
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📈 Analytical overview of Telegram channel allcoding1
Channel allcoding1 (@allcoding1) in the English language segment is an active participant. Currently, the community unites 20 819 subscribers, ranking 9 647 in the Education category and 19 582 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 20 819 subscribers.
According to the latest data from 23 October, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -422 over the last 30 days and by -15 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 1.26%. Within the first 24 hours after publication, content typically collects 1.18% reactions from the total number of subscribers.
- Post reach: On average, each post receives 263 views. Within the first day, a publication typically gains 246 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 1.
- Thematic interests: Content is focused on key topics such as dsa, stack, namaste, javascript, learning.
📝 Description and content policy
Channel description not provided.
Thanks to the high frequency of updates (latest data received on 24 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 Education category.
20 819
Subscribers
-1524 hours
-987 days
-42230 days
Posts Archive
20 819
+1https://youtube.com/@TechHelp-kn9fb?si=ZrRasvSTTjfAn-o8
Report this channel 👆👆👆
#Scammer
9929302889
Be safe
20 819
🎯Wipro Recruitment 2024 For Banking Operations
Location: Chennai
Qualification: Any Graduate
Work Experience: Freshers / Experience
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
20 819
🎯Deloitte National Level Hiring 2024 As Analyst Trainee | 4-8 LPA
Job Role Analyst Trainee
Qualification BE, B Tech, ME, M Tech, MCA
Location PAN India
Salary 4-8 LPA
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
20 819
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
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
20 819
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
20 819
#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;
}
Telegram:- @allcoding1
20 819
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 ✅
Telegram:- @allcoding1
20 819
#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 ✅
Telegram:- @allcoding1
20 819
🎯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
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
20 819
SQL, IBM, TESTING and SAP support anybody who wants contact on Instagram.
https://www.instagram.com/allcoding1_official?igshid=OGQ5ZDc2ODk2ZA==
20 819
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
Telegram:- @allcoding1
20 819
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
20 819
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
Telegram:- @allcoding1_official
