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 20 637 subscribers, ranking 9 754 in the Education category and 19 472 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 20 637 subscribers.
According to the latest data from 06 November, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -405 over the last 30 days and by -19 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 3.59%. Within the first 24 hours after publication, content typically collects 1.08% reactions from the total number of subscribers.
- Post reach: On average, each post receives 742 views. Within the first day, a publication typically gains 223 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 07 November, 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.
20 637
Subscribers
-1924 hours
-957 days
-40530 days
Posts Archive
20 637
🎯Global Edge Hiring | 2022 graduates
CTC - 4LPA
Direct Link :- https://www.allcoding1.com/2022/08/global-edge-hiring-2022-graduatesctc.html
Telegram :- @allcoding1
20 637
#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
20 637
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
20 637
#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
20 637
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
20 637
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
20 637
# 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
