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📈 Análisis del canal de Telegram allcoding1
El canal allcoding1 (@allcoding1) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 21 559 suscriptores, ocupando la posición 9 073 en la categoría Educación y el puesto 19 010 en la región India.
📊 Métricas de audiencia y dinámica
Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 21 559 suscriptores.
Según los últimos datos del 30 agosto, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -372, y en las últimas 24 horas de -25, conservando un alto alcance.
- Estado de verificación: No verificado
- Tasa de interacción (ER): El promedio de interacción de la audiencia es 6.68%. Durante las primeras 24 horas tras publicar, el contenido suele obtener N/A% de reacciones respecto al total de suscriptores.
- Alcance de las publicaciones: Cada publicación recibe en promedio 1 441 visualizaciones. En el primer día suele acumular 0 visualizaciones.
- Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 0.
- Intereses temáticos: El contenido se centra en temas clave como dsa, stack, namaste, javascript, learning.
📝 Descripción y política de contenido
No se ha proporcionado la descripción del canal.
Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 31 agosto, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Educación.
21 559
Suscriptores
-2524 horas
-867 días
-37230 días
Archivo de publicaciones
21 554
#include <iostream>
#include <vector>
#include <unordered_map>
const int MOD = 1000000007;
int countUniqueArrangements(std::vector<std::pair<int, int="">>& dominoes) {
std::unordered_map<int, int=""> dp;
dp[0] = 1; // Base case
int n = dominoes.size();
for(int i = 1; i <= n; ++i) {
std::unordered_map<int, int=""> next_dp;
for(auto& domino : dominoes) {
int key = dp.count(domino.first) ? domino.first : domino.second;
next_dp[key] = (next_dp[key] + dp[key ^ domino.first ^ domino.second]) % MOD;
}
dp = std::move(next_dp);
}
int total_arrangements = 0;
for(auto& p : dp) {
total_arrangements = (total_arrangements + p.second) % MOD;
}
return total_arrangements;
}
int main() {
int N;
std::cin >> N;
std::vector<std::pair<int, int="">> dominoes(N);
for(int i = 0; i < N; ++i) {
std::cin >> dominoes[i].first >> dominoes[i].second;
}
int total_arrangements = countUniqueArrangements(dominoes);
std::cout << total_arrangements << std::endl;
return 0;
}
C++
HackWithInfy
@allcoding1
21 554
#include <iostream>
#include <string>
#include <algorithm>
int findLargestX(std::string& str) {
int max_x = 0;
int sum = 0;
for (char c : str) {
if (isdigit(c)) {
sum += c - '0';
} else if (c == '-') {
max_x = std::max(max_x, sum);
sum = 0;
}
}
return max_x;
}
int main() {
std::string input = "63-1+2-1+3+4-9-1+2-3-3+4+9";
int largest_x = findLargestX(input);
std::cout << "Largest value of x: " << largest_x << std::endl;
return 0;
}
21 554
import sys
def gcd(a, b):
while b:
a, b = b, a % b
return a
def Get_ans(N, A):
mod = 1000000007
prime_factors = [2, 3, 5, 7, 11, 13]
def count_good_partitions_gcd(gcd_val):
res = 0
for i in range(1, gcd_val + 1):
if all(i % pf != 0 for pf in prime_factors):
res += pow(2, gcd_val // i - 1, mod)
res %= mod
return res
total_gcd = A[0]
for i in range(1, N):
total_gcd = gcd(total_gcd, A[i])
result = count_good_partitions_gcd(total_gcd)
return result
def main():
N = int(sys.stdin.readline().strip())
A = []
for _ in range(N):
A.append(int(sys.stdin.readline().strip()))
result = Get_ans(N, A)
print(result)
if __name__ == "__main":
main()21 554
#include <iostream>
#include <vector>
#include <string>
using namespace std;
const int MOD = 1e9 + 7;
int countGoodStrings(int N, int M, string S) {
long long totalGoodStrings = 1;
for (int i = 0; i < N; i++) {
totalGoodStrings = (totalGoodStrings * M) % MOD;
}
return totalGoodStrings;
}
int main() {
int N, M;
cin >> N >> M;
string S;
cin >> S;
int result = countGoodStrings(N, M, S);
cout << result << endl;
return 0;
}
21 554
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int settleAccounts(vector<int> A) {
int sum = 0;
for (int num : A) {
sum += num;
}
return -sum;
}
int main() {
int N;
cin >> N;
vector<int> A(N);
for (int i = 0; i < N; i++) {
cin >> A[i];
}
int result = settleAccounts(A);
cout << result << endl;
return 0;
}
21 554
Bob is a trader who gives loans to his friends when they need them and takes loans when he needs them. He has decided to leave the trade and work as a Software Engineer, so he wants to settle his accounts.
You are given an array A of size N representing Bob's dealings. If the value of A[i] is less than zero it means that Bob has taken a loan of absolute value of A[i]. Otherwise, he has given a loan of value A[i].
In order to settle accounts, Bob wants to divide all his dealing into some months.
Find the maximum value of minimum money that Bob will take every month to settle his accounts.
Note:
• Bob can distribute any number of dealings in any number of months.
Input Format
The first line contains an integer, N, denoting the number of elements in A.
Each line i of the N subsequent lines (where 0 < i < N) contains an integer describing A[i].
Medium 1: Bob's Dealings
Constraints
1 <= N <= 1000
-10^9 <= A <= 10^9
Sample Test Cases
Case 1
Input:
6
10
10
-40
-40
10
10
Output:
-20
Explanation:
Here, N=6
A=[10, 10, 40, 40, 10, 10]
Bob can divide his dealings into two months the first three dealings in the first month and the second three dealings in the second month, and then the answer will be -20. (a negative number of units means that he needs to pay 20 units).
Hence, 20 is the minimum amount of money that Bob has to take per month to settle his dealings.
Case 2
Input
-7
7
Output:
4
Explanation:
Here, N=5 A=[2,-3,5,-7,7]
Bob can split his dealings into only one month which will cost him 2+ (-3)+5+(-7)+7=4.
Hence, 4 is the maximum amount of money that Bob has to take in one month to settle his dealings.
Case 3
Input:
3
-30
-20
-10
Output:
-30
Explanation:
Here, N=3
A=[-30, -20, -10]
Bob can split his dealings into two months as he cansplit the first dealing in one month for which he
has to take a value of 30 as a loan.
Then he can split the second and third dealings inthe second month for which he has to take a loan on 20 + 10 = 30.
Hence, the maximum amount of money that Coach Yasserhas to take in one month to settle his dealings isequal to 30.
21 554
Here's the C++ code to solve the problem:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int settleAccounts(vector<int> A) {
int sum = 0;
for (int num : A) {
sum += num;
}
return -sum;
}
int main() {
int N;
cin >> N;
vector<int> A(N);
for (int i = 0; i < N; i++) {
cin >> A[i];
}
int result = settleAccounts(A);
cout << result << endl;
return 0;
}21 554
#include <iostream>
#include <string>
using namespace std;
bool isGoodString(string s) {
int count = 0;
for (char c : s) {
if (c == 'a') {
count++;
} else {
count--;
}
if (count < 0) {
return false;
}
}
return count == 0;
}
int main() {
string s = "aaabbbaaa";
cout << "Is the string good? " << (isGoodString(s) ? "Yes" : "No") << endl;
return 0;
}
Count Good Triples
C++
HackWithInfy
21 554
#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
const int MOD = 1e9 + 7;
vector<vector<int>> tree;
vector<int> a;
unordered_map<int, int> countMap;
bool checkPalindrome(unordered_map<int, int>& countMap) {
int oddCount = 0;
for (auto& it : countMap) {
if (it.second % 2 != 0) oddCount++;
if (oddCount > 1) return false;
}
return true;
}
int dfs(int node) {
int ans = 0;
countMap[a[node]]++;
if (checkPalindrome(countMap)) {
ans = 1;
}
for (auto& child : tree[node]) {
ans += dfs(child);
ans %= MOD;
}
countMap[a[node]]--; // Backtrack to remove the current node's count
return ans;
}
int main() {
int n;
cin >> n;
tree.resize(n + 1);
a.resize(n + 1);
vector<int> par(n + 1);
for (int i = 2; i <= n; i++) {
cin >> par[i];
tree[par[i]].push_back(i);
}
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
int ans = dfs(1);
cout << ans << endl;
return 0;
}
c++
Palindromic subtrees
21 554
Repost from allcoding1
+9📌IT learning courses
📌All programing courses
📌Abdul bari courses
📌Ashok IT
100 rupees
Contact:- @meterials_available
21 554
Repost from allcoding1
+9📌IT learning courses
📌All programing courses
📌Abdul bari courses
📌Ashok IT
100 rupees
Contact:- @meterials_available
