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allcoding1
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allcoding1 (@allcoding1) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 22 561 obunachidan iborat bo'lib, Taʼlim toifasida 8 836-o'rinni va Hindiston mintaqasida 19 517-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 22 561 obunachiga ega bo‘ldi.
13 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -442 ga, so‘nggi 24 soatda esa -20 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 6.17% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.25% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 1 394 marta ko‘riladi; birinchi sutkada odatda 283 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, learning kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Kanal uchun tavsif kiritilmagan.
Yuqori yangilanish chastotasi (oxirgi ma’lumot 14 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
22 561
Obunachilar
-2024 soatlar
-897 kunlar
-44230 kunlar
Postlar arxiv
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
