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Code With Virus

Code With Virus

Kanalga Telegram’da o‘tish

Coding channel @codewithvirus. 👈 Main group @avirustech 👈 Accenture @Accenturavirustech 👈 IBM. @IBMavirustech 👈 Tech Mahindra. @TechMavirustech 👈

Ko'proq ko'rsatish

📈 Telegram kanali Code With Virus analitikasi

Code With Virus (@codewithvirus) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 10 559 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 10 501-o'rinni va Hindiston mintaqasida 43 593-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 10 559 obunachiga ega bo‘ldi.

04 Dekabr, 2025 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -87 ga, so‘nggi 24 soatda esa 0 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 0% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining N/A% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 0 marta ko‘riladi; birinchi sutkada odatda 0 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 0 ta reaksiya keladi.

📝 Tavsif va kontent siyosati

Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
Coding channel @codewithvirus. 👈 Main group @avirustech 👈 Accenture @Accenturavirustech 👈 IBM. @IBMavirustech 👈 Tech Mahindra. @TechMavirustech 👈

Yuqori yangilanish chastotasi (oxirgi ma’lumot 05 Dekabr, 2025 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

10 559
Obunachilar
Ma'lumot yo'q24 soatlar
-227 kunlar
-8730 kunlar
Postlar arxiv

#include <bits/stdc++.h> using namespace std; vector<string> rows; char alphabet; string findPrecedingAlphabets() { int r = -1, c = -1; for (int i = 0; i < rows.size(); i++) { int idx = rows[i].find(alphabet); if (idx != string::npos) { r = i; c = idx; break; } } if (r == -1) return "alphabet not found"; string a = ""; if (r > 0) a += rows[r - 1][c]; if (c > 0) a += rows[r][c - 1]; return a; } int main() { int n; cin >> n; for (int i = 0; i < n; i++) { string row; cin >> row; rows.push_back(row); } cin >> alphabet; cout << findPrecedingAlphabets() << endl; return 0; }

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