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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 85 467 subscribers, ranking 1 502 in the Technologies & Applications category and 3 471 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 85 467 subscribers.
According to the latest data from 25 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -1 422 over the last 30 days and by -71 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 2.42%. Within the first 24 hours after publication, content typically collects 0.88% reactions from the total number of subscribers.
- Post reach: On average, each post receives 2 071 views. Within the first day, a publication typically gains 749 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 4.
- 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 26 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.
85 467
Subscribers
-7124 hours
-3077 days
-1 42230 days
Posts Archive
85 467
Cognizant Off Campus Hiring | Data Analyst – Fresher | Upto 7.5 LPA
Job Role Data Analyst
Qualification B.E/B.Tech
Batch Any Batch
CTC/Salary Upto 7.5 LPA
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1_official
85 467
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:-
85 467
int solve(vector& nums) {
vector 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:-
85 467
void solve(string x){
ll n = x.size();
vector> dp(n,vector(n,0));
for(ll i=0;i
85 467
#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;
}
85 467
int getMinOperations(vector arr) {
int n=arr.size();
int operations = 0;
for (int i = 1; i < n - 1; i++)
{
if (arr[i] < 0 && arr[i - 1] < 0)
{
arr[i] = 1e9;
operations++;
}
else if (arr[i] < 0 && arr[i - 1] > 0 && arr[i + 1] > 0)
{
if ((arr[i] + arr[i - 1] < 0) || (arr[i] + arr[i + 1] < 0))
{
arr[i] = 1e9;
operations++;
}
}
}
return operations;
}
Make The Array Postive ✅
85 467
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<vector<int>> freq(n, vector<int> (26, 0));
string s;
for(int i = 0; i < n; i++) {
string s;
cin >> s;
for(auto &ch : s)
freq[i][ch - 'a']++;
}
long long ans = 0;
for(int i = 0; i < 26; i++) {
int minValue = INT_MAX;
for(int j = 0; j < n; j++)
minValue = min(minValue, freq[j][i]);
ans += minValue;
}
cout << ans << endl;
}
Service Now ✅
85 467
#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 ✅
85 467
🎯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
85 467
def decodeSequence(binarySequence):
nucleobases = {'001': 'C', '011': 'A', '101': 'T','010': 'G', '110': 'U'}
if len(binarySequence) % 3 != 0:
print("Error: The length of the input string should be a multiple of 3.")
return
decodedSequence = ""
for i in range(0, len(binarySequence), 3):
chunk = binarySequence[i:i+3]
if i == 0 and chunk == '000':
identifier = 'DNA'
elif i == 0 and chunk == '111':
identifier = 'RNA'
else:
nucleobase = nucleobases.get(chunk, 'X')
if nucleobase == 'U' and identifier == 'DNA':
nucleobase = 'T'
elif nucleobase == 'T' and identifier == 'RNA':
nucleobase = 'U'
decodedSequence += nucleobase
print(decodedSequence)
binaryInput = input("Enter the binary sequence: ")
decodeSequence(binaryInput)
DNA
IBM
Telegram:- @allcoding1_official
85 467
string solve(string bs) {
map 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
85 467
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 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✅
85 467
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)