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allcoding1

allcoding1

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allcoding1 (@allcoding1) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 22 590 obunachidan iborat bo'lib, Taʼlim toifasida 8 822-o'rinni va Hindiston mintaqasida 19 518-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 22 590 obunachiga ega bo‘ldi.

12 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -437 ga, so‘nggi 24 soatda esa -6 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 5.99% 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 1 353 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 13 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.

22 590
Obunachilar
-624 soatlar
-967 kunlar
-43730 kunlar
Postlar arxiv
// 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)); } }

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All codes are available Once check it 👇👇 https://www.instagram.com/allcoding1_official?igsh=ZHJpNXdpeWh1d2No

Short String
Short String

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Unique subarray sum solution
Unique subarray sum solution

Python
+3
Python

Tree lis Query
Tree lis Query

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

Python 3 Music melodies
+1
Python 3 Music melodies

Music melodies
Music melodies

This is the code Everyone write neatly All test cases passed Python 3 Infosys
This is the code Everyone write neatly All test cases passed Python 3 Infosys

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

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()

Python
Python

Sub set with LCM
Sub set with LCM

Minimum difference pairs in python
Minimum difference pairs in python

Equilibrium point Infosys all test cases passed
+1
Equilibrium point Infosys all test cases passed

Minimum unique sum
Minimum unique sum

def max_cost_split(s): n = len(s) max_cost = 0 for i in range(1, n): a = s[:i] b = s[i:] cost_a = len(set(a)) cost_b = len(set(b)) max_cost = max(max_cost, cost_a + cost_b) return n - max_cost Infosys max cost split code in python