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allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 79 039 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 607-o'rinni va Hindiston mintaqasida 3 824-o'rinni egallagan.
π Auditoriya koβrsatkichlari va dinamika
Π½Π΅Π²ΡΠ΄ΠΎΠΌΠΎ sanasidan buyon loyiha tez oβsib, 79 039 obunachiga ega boβldi.
24 Oktabr, 2026 dagi oxirgi maβlumotlarga koβra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 636 ga, soβnggi 24 soatda esa -24 ga oβzgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya oβrtacha 1.75% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.67% ini tashkil etuvchi reaksiyalarni toβplaydi.
- Post qamrovi: Har bir post oβrtacha 1 383 marta koβriladi; birinchi sutkada odatda 527 ta koβrish yigβiladi.
- Reaksiyalar va oβzaro taβsir: Auditoriya faol: har bir postga oβrtacha 0 ta reaksiya keladi.
- Tematik yoβnalishlar: Kontent dsa, stack, namaste, javascript, dev kabi asosiy mavzularga jamlangan.
π Tavsif va kontent siyosati
Kanal uchun tavsif kiritilmagan.
Yuqori yangilanish chastotasi (oxirgi maβlumot 25 Oktabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli boβlib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim taβsir nuqtasiga aylantirishini koβrsatadi.
79 039
Obunachilar
-2424 soatlar
-3787 kun
-1 63630 kun
Postlar arxiv
79 039
//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;
}
79 039
// 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));
}
}
79 039
All codes are available
Once check it ππ
https://www.instagram.com/allcoding1_official?igsh=ZHJpNXdpeWh1d2No
79 039
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
79 039
This is the code
Everyone write neatly
All test cases passed
Python 3 Infosys
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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
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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()
