allcoding1
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š Analytical overview of Telegram channel allcoding1
Channel allcoding1 (@allcoding1) in the English language segment is an active participant. Currently, the community unites 20 877 subscribers, ranking 9 630 in the Education category and 19 588 in the India region.
š Audience metrics and dynamics
Since its creation on Š½ŠµŠ²ŃŠ“Š¾Š¼Š¾, the project has demonstrated rapid growth, gathering an audience of 20 877 subscribers.
According to the latest data from 20 October, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -410 over the last 30 days and by -14 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 1.03% reactions from the total number of subscribers.
- Post reach: On average, each post receives 263 views. Within the first day, a publication typically gains 215 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.
š Description and content policy
Channel description not provided.
Thanks to the high frequency of updates (latest data received on 21 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.
20 877
Subscribers
-1424 hours
-767 days
-41030 days
Posts Archive
20 865
//Equilibrium path
#include <bits/stdc++.h>
using namespace std;
string trim(string str) {
str.erase(0, str.find_first_not_of(' '));
str.erase(str.find_last_not_of(' ') + 1);
return str;
}
int solve(int N, vector<int> A) {
int total_sum = accumulate(A.begin(), A.end(), 0);
int left_sum = 0;
int equilibrium_count = 0;
for (int i = 0; i < N; ++i) {
int right_sum = total_sum - left_sum - A[i];
if (left_sum == right_sum) {
equilibrium_count++;
}
left_sum += A[i];
}
return equilibrium_count;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
string inputline;
getline(cin, inputline);
int N = stoi(trim(inputline));
vector<int> A(N);
for (int j = 0; j < N; j++) {
getline(cin, inputline);
A[j] = stoi(trim(inputline));
}
int result = solve(N, A);
cout << result << endl;
return 0;
}
20 865
// coins game
import java.util.*;
class HelloWorld {
static int helper(int a, int b, int i ,int[]x,int n){
if(i==x.length) return 0;
int left=(x[i]>n)?0:Math.abs(x[i]-a)+helper(x[i],b,i+1,x,n);
int right=(x[i]>n)?0:Math.abs(x[i]-b)+helper(a,x[i],i+1,x,n);
return Math.min(left,right);
}
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int N=sc.nextInt();
int Q=sc.nextInt();
int A=sc.nextInt();
int B=sc.nextInt();
int[]x=new int[Q];
for(int i=0;i<Q;i++){
x[i]=sc.nextInt();
}
System.out.println(helper(A,B,0,x,N));
}
}
20 865
All codes are available
Once check it šš
https://www.instagram.com/allcoding1_official?igsh=ZHJpNXdpeWh1d2No
20 865
import sys
def get_ans(N, K, A):
def main():
N = int(sys.stdin.readline().strip())
K = int(sys.stdin.readline().strip())
A = []
for _ in range(N):
A.append(int(sys.stdin.readline().strip()))
result = get_ans(N, K, A)
print(result)
if name == "main":
main()
Subset with LCM code in python
20 865
def longest_equal_subarray():
n = int(input())
A = [int(input()) for _ in range(n)]
A = [-1 if x == 0 else 1 for x in A]
prefix_sum_map = {}
prefix_sum = 0
max_length = 0
for i in range(n):
prefix_sum += A[i]
if prefix_sum == 0:
max_length = i + 1
if prefix_sum in prefix_sum_map:
max_length = max(max_length, i - prefix_sum_map[prefix_sum])
else:
prefix_sum_map[prefix_sum] = i
return max_length
print(longest_equal_subarray())
Infosys Longest Subarray code
20 865
class TreeNode:
def init(self, value=0, left=None, right=None):
self.value = value
self.left = left
self.right = right
def count_nodes(node, counts):
if node is None:
return
if node.value in counts:
counts[node.value] += 1
else:
counts[node.value] = 1
count_nodes(node.left, counts)
count_nodes(node.right, counts)
def find_double_roots(root):
counts = {}
count_nodes(root, counts)
double_roots = [value for value, count in counts.items() if count > 1]
return double_roots
def main():
# Example tree:
# 1
# / \
# 2 3
# / \
# 2 4
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(2)
root.left.right = TreeNode(4)
result = find_double_roots(root)
print("Nodes with double roots:", result)
if name == "main":
main()
