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allcoding1
前往频道在 Telegram
📈 Telegram 频道 allcoding1 的分析概览
频道 allcoding1 (@allcoding1) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 22 561 名订阅者，在 教育 类别中位列第 8 836，并在 印度 地区排名第 19 517 位。
📊 受众指标与增长动态
自 невідомо 创建以来，项目保持高速增长，吸引了 22 561 名订阅者。
根据 13 六月, 2026 的最新数据，频道保持稳定运转。过去 30 天订阅人数变化为 -442，过去 24 小时变化为 -20，整体触达仍然可观。
- 认证状态： 未认证
- 互动率 (ER)： 平均受众互动率为 6.17%。内容发布后 24 小时内通常能获得 1.25% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 1 394 次浏览，首日通常累积 283 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 2。
- 主题关注点： 内容集中在 dsa, stack, namaste, javascript, learning 等核心主题上。
📝 描述与内容策略
尚未提供频道描述。
凭借高频更新（最新数据采集于 14 六月, 2026），频道始终保持新鲜度与高覆盖。分析显示受众积极互动，使其成为 教育 类别中的关键影响点。
22 561
订阅者
-2024 小时
-897 天
-44230 天
帖子存档
22 561
#include <iostream>
#include <vector>
#include <algorithm>
int getPotentialOfWinner(std::vector<int>& potential, long long k) {
int n = potential.size();
int x = potential[0];
int m = 0;
for (int i = 1; i < n; i++) {
if (m != k) {
if (x > potential[i]) {
m++;
} else {
x = potential[i];
m = 1;
}
}
}
return x;
}
int main() {
std::vector<int> potentials = {3, 2, 1, 4};
long long k = 2;
std::cout << getPotentialOfWinner(potentials, k) << std::endl;
return 0;
}
Potential winner code
Telegram:- @allcoding1
22 561
import heapq
def reduce_sum(lst):
heapq.heapify(lst)
s = 0
while len(lst) > 1:
first = heapq.heappop(lst)
second = heapq.heappop(lst)
s += first + second
heapq.heappush(lst, first + second)
return s
Reduce the Array
22 561
#include <iostream>
#include <vector>
#include <set>
using namespace std;
int getSmallestArea(vector<vector<int>>& grid) {
int rows = grid.size();
if (rows == 0) return 0;
int cols = grid[0].size();
if (cols == 0) return 0;
set<int> rowsSet, colsSet;
for (int i = 0; i < rows; ++i) {
for (int j = 0; j < cols; ++j) {
if (grid[i][j] == 1) {
rowsSet.insert(i);
colsSet.insert(j);
}
}
}
int width = colsSet.empty() ? 0 : *colsSet.rbegin() - *colsSet.begin() + 1;
int height = rowsSet.empty() ? 0 : *rowsSet.rbegin() - *rowsSet.begin() + 1;
return width * height;
}
shipping space
Salesforce
22 561
Repost from allcoding1
+9📌IT learning courses
📌All programing courses
📌Abdul bari courses
📌Ashok IT
100 rupees
Contact:- @meterials_available
22 561
MOD = 10**9 + 7
def solve(arrival_departure):
arrival_departure.sort(key=lambda x: x[1])
prev_departure = -1
total_stations = 0
for arrival, departure in arrival_departure:
if arrival > prev_departure:
total_stations += 1
prev_departure = departure
return total_stations % MOD
def main():
N = int(input())
arrival_departure = []
for _ in range(N):
arrival, departure = map(int, input().split())
arrival_departure.append((arrival, departure))
result = solve(arrival_departure)
print(result)
if name == "main":
main()
Trains Code
Python
HackWithInfy
Telegram:- @allcoding1
22 561
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int MOD = 1e9 + 7;
int main() {
int N, Q;
cin >> N;
vector<int> A(N);
for (int i = 0; i < N; ++i) {
cin >> A[i];
}
cin >> Q;
long long sum = 0;
for (int q = 0; q < Q; ++q) {
int type, L, R, X, i , zero1 , zero2;
cin >> type;
if (type == 1) {
cin >> L >> R >> X;
for (int j = L-1; j < R; ++j) {
A[j] = min(A[j], X);
}
} else if (type == 2) {
cin >> i >> zero1 >>zero2;
sum = (sum + A[i - 1]) % MOD;
}
}
cout << sum << endl;
return 0;
}
Replace by Minimum Code
C++
HackWithInfy
Telegram:- @allcoding1
22 561
Repost from allcoding1
+9📌IT learning courses
📌All programing courses
📌Abdul bari courses
📌Ashok IT
100 rupees
Contact:- @meterials_available
22 561
https://www.allcoding1.com/2023/12/allcoding1-answers.html?m=1
🆓 Copy ur question and past you get Answer
22 561
#include <iostream>
#include <vector>
#include <unordered_map>
const int MOD = 1000000007;
int n, m, startRow, startColumn, moves, q;
std::vector<std::vector<int>> obstacles;
bool isObstacle(int x, int y) {
for (auto& obstacle : obstacles) {
if (obstacle[0] == x && obstacle[1] == y) {
return true;
}
}
return false;
}
int countPaths(int x, int y, int movesLeft, std::unordered_map<std::string, int="">& dp) {
if (x < 0 y < 0 x >= n || y >= m) {
return 1;
}
if (movesLeft == 0 || isObstacle(x, y)) {
return 0;
}
std::string key = std::to_string(x) + ":" + std::to_string(y) + ":" + std::to_string(movesLeft);
if (dp.find(key) != dp.end()) {
return dp[key];
}
int paths = countPaths(x + 1, y, movesLeft - 1, dp) % MOD;
paths = (paths + countPaths(x - 1, y, movesLeft - 1, dp)) % MOD;
paths = (paths + countPaths(x, y + 1, movesLeft - 1, dp)) % MOD;
paths = (paths + countPaths(x, y - 1, movesLeft - 1, dp)) % MOD;
dp[key] = paths;
return paths;
}
int main() {
std::cin >> n >> m >> startRow >> startColumn >> moves >> q;
obstacles.resize(q, std::vector<int>(2));
for (int i = 0; i < q; ++i) {
std::cin >> obstacles[i][0] >> obstacles[i][1];
}
std::unordered_map<std::string, int=""> dp;
int totalPaths = countPaths(startRow, startColumn, moves, dp);
std::cout << totalPaths << std::endl;
return 0;
}
C++
In a town called Gridland, there lived a man named Alex who had a special skill. He could move around a grid like a big square map. The grid had obstacles, things he couldn't go through.
The grid is described by its size n x m, and Alex starts at a specífic position [startRow, startColumn). There are also obstacles in fixed positions.
HackWithInfy
@allcoding1
22 561
#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
22 561
#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;
}
22 561
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()22 561
#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;
}
22 561
#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;
}
22 561
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.
22 561
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;
}22 561
#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
