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 21 554 subscribers, ranking 9 078 in the Education category and 18 983 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 21 554 subscribers.
According to the latest data from 31 August, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -369 over the last 30 days and by -5 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 6.77%. Within the first 24 hours after publication, content typically collects N/A% reactions from the total number of subscribers.
- Post reach: On average, each post receives 1 460 views. Within the first day, a publication typically gains 0 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 0.
- 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 01 September, 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.
21 554
Subscribers
-524 hours
-817 days
-36930 days
Posts Archive
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
