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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 22 397 subscribers, ranking 8 807 in the Education category and 18 942 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 22 397 subscribers.
According to the latest data from 27 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -383 over the last 30 days and by -21 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 4.45%. Within the first 24 hours after publication, content typically collects 1.42% reactions from the total number of subscribers.
- Post reach: On average, each post receives 997 views. Within the first day, a publication typically gains 318 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
- 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 28 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 Education category.
22 397
Subscribers
-2124 hours
-927 days
-38330 days
Posts Archive
22 397
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
ll solve(ll n)
{
ll position = 1;
ll m = 1;
while (!(n & m)) {
m = m << 1;
position++;
}
return position;
}
int main() {
// your code goes here
ll n,p;
cin>>n;
vector<int> ans(n);
for(int i=0;i<n;i++)
{
cin>>p;
ans[i] = solve(p+1);
}
for(int i=0;i<n;i++)
cout<<ans[i]<<"\n";
return 0;
}
C++
Telegram:- @allcoding1
22 397
class Solution
{
public:
vector<int> frequncey(string s)
{
vector<int> freq(26, 0);
for (int i = 0; i < s.length(); i++)
freq[s[i] - 'a']++;
return freq;
}
vector<string> wordSubsets(vector<string> &words1, vector<string> &words2)
{
vector<string> ans;
vector<int> Freq(26, 0);
for (auto &x : words2)
{
vector<int> freq1 = frequncey(x);
for (int i = 0; i < 26; i++)
Freq[i] = max(freq1[i], Freq[i]);
}
for (auto &x : words1)
{
vector<int> freq1 = frequncey(x);
bool flag = true;
for (int i = 0; i < 26; i++)
{
if (freq1[i] < Freq[i])
{
flag = false;
break;
}
}
if (flag)
ans.push_back(x);
}
return ans;
}
};
C++
Word Subsets
Telegram:- @allcoding1
22 397
#include<bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
string str;
cin >> str;
int n = str.length(), ans = 0;
for(int i = 0; i < n; i++){
if(str[i] == '5' || str[i] == '6')
continue;
else
ans += 1;
}
cout << ans << endl;
}
return 0;
}
Beautiful Number
Telegram:- @allcoding1
22 397
array AR of size N
perfect_sum(arr, s, result) :
x = [0]*len(arr)
j = len(arr) - 1
while (s > 0) :
x[j] = s % 2
s = s // 2
j -= 1
sum = 0
for i in range(len(arr)) :
if (x[i] == 1) :
sum += arr[i]
if (sum == result) :
print("{",end="");
for i in range(len(arr)) :
if (x[i] == 1) :
print(arr[i],end= ", ");
print("}, ",end="")
def print_subset(arr, K) :
x = pow(2, len(arr))
for i in range(1, x):
perfect_sum(arr, i, K)
# Driver code
arr = [ ]
n = int(input("Enter length of array : "))
s=int(input("Enter sum : "))
for i in range(n):
ele=int(input("Enter element : "))
arr.append(ele)
print_subset(arr, s)
Python
Telegram:- @allcoding1
22 397
def minOperations(arr1, arr2, i, j):
# Base Case
if arr1 == arr2:
return 0
if i >= len(arr1) or j >= len(arr2):
return 0
# If arr[i] < arr[j]
if arr1[i] < arr2[j]:
# Include the current element
return 1 \
+ minOperations(arr1, arr2, i + 1, j + 1)
# Otherwise, excluding the current element
return max(minOperations(arr1, arr2, i, j + 1),
minOperations(arr1, arr2, i + 1, j))
# Function that counts the minimum
# moves required to sort the array
def minOperationsUtil(arr):
brr = sorted(arr);
# If both the arrays are equal
if(arr == brr):
# No moves required
print("0")
# Otherwise
else:
# Print minimum operations required
print(minOperations(arr, brr,)
Python
Q) minimum operations
Telegram - @allcoding1
22 397
# A Naive recursive python program to find minimum of coins
# to make a given change V
import sys
# m is size of coins array (number of different coins)
def minCoins(coins, m, V):
# base case
if (V == 0):
return 0
# Initialize result
res = sys.maxsize
# Try every coin that has smaller value than V
for i in range(0, m):
if (coins[i] <= V):
sub_res = minCoins(coins, m, V-coins[i])
# Check for INT_MAX to avoid overflow and see if
# result can minimized
if (sub_res != sys.maxsize and sub_res + 1 < res):
res = sub_res + 1
//@allcoding1_official
return res
# Driver program to test above function
coins = [9, 6, 5, 1]
m = len(coins)
V = 11
print("Minimum coins required is",minCoins(coins, m, V))
Python 3
Telegram:- @allcoding1
22 397
Repost from allcoding1_official
array AR of size N
perfect_sum(arr, s, result) :
x = [0]*len(arr)
j = len(arr) - 1
while (s > 0) :
x[j] = s % 2
s = s // 2
j -= 1
sum = 0
for i in range(len(arr)) :
if (x[i] == 1) :
sum += arr[i]
if (sum == result) :
print("{",end="");
for i in range(len(arr)) :
if (x[i] == 1) :
print(arr[i],end= ", ");
print("}, ",end="")
def print_subset(arr, K) :
x = pow(2, len(arr))
for i in range(1, x):
perfect_sum(arr, i, K)
# Driver code
arr = [ ]
n = int(input("Enter length of array : "))
s=int(input("Enter sum : "))
for i in range(n):
ele=int(input("Enter element : "))
arr.append(ele)
print_subset(arr, s)
Python
Telegram :- @allcoding1_official
22 397
Repost from allcoding1_official
class CountOccurrences {
public static void main(String[] args) {
String str = "aabbccd";
char a[]=str.toCharArray();
int cnt = 0;
for (int i = 0; i < a.length; i++) {
for (int j = i + 1; j < a.length; j++)
{
if ((a[i] == a[j])) {
cnt++;
}
}
}
System.out.println(cnt);
}
}
Java
Telegram:- @allcoding1_official
22 397
Number Of Subsequences Code For Infosys
int A[] = {10,13,7,8,14,11};
int n = 6;
int memo[6];//initialized with -1s;
int count(int currIndex){
if (currIndex == n-1) return 1;
if (memo[currIndex] != -1) return memo[currIndex];
int res = 0;
for (int i=currIndex+1 ; i<n ; i++){
if (abss(A[currIndex] - A[i]) <= 3){
res += count(i);
}
}
memo[currIndex] = res;
return res;
}
c++
Telegram:- @allcoding1
