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لا تقع ضحية للمخادعين! تيليمتريو يكتشف ويُميّز هذه القنوات 👉 إذا كنت تريد رؤية العلامة، اشترك

Code With Virus

Code With Virus

الذهاب إلى القناة على Telegram

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

إظهار المزيد

📈 نظرة تحليلية على قناة تيليجرام Code With Virus

تُعد قناة Code With Virus (@codewithvirus) في القطاع اللغوي الإنكليزية لاعباً نشطاً. يضم المجتمع حالياً 10 559 مشتركاً، محتلاً المرتبة 10 501 في فئة التكنولوجيات والتطبيقات والمرتبة 43 593 في منطقة الهند.

📊 مؤشرات الجمهور والحراك

منذ تأسيسه في невідомо، حقق المشروع نمواً سريعاً وجمع 10 559 مشتركاً.

بحسب آخر البيانات بتاريخ 04 ديسمبر, 2025، تحافظ القناة على نشاط مستقر. خلال آخر 30 يوماً تغيّر عدد الأعضاء بمقدار -87، وفي آخر 24 ساعة بمقدار 0، مع بقاء الوصول العام مرتفعاً.

  • حالة التحقق: غير موثّقة
  • معدل التفاعل (ER): يبلغ متوسط تفاعل الجمهور 0‎%. وخلال أول 24 ساعة من النشر يحصد المحتوى عادةً N/A‎% من ردود الفعل نسبةً إلى إجمالي المشتركين.
  • وصول المنشورات: يحصل كل منشور على متوسط 0 مشاهدة. وخلال اليوم الأول يجمع عادةً 0 مشاهدة.
  • التفاعلات والاستجابة: يتفاعل الجمهور بانتظام؛ متوسط التفاعلات لكل منشور يبلغ 0.

📝 الوصف وسياسة المحتوى

يصف المؤلف القناة بأنها مساحة للتعبير عن الآراء الذاتية:
“Coding channel @codewithvirus. 👈 Main group @avirustech 👈 Accenture @Accenturavirustech 👈 IBM. @IBMavirustech 👈 Tech Mahindra. @TechMavirustech 👈”

بفضل وتيرة التحديث المرتفعة (أحدث البيانات بتاريخ 05 ديسمبر, 2025) تحافظ القناة على حداثتها ومستوى وصول مرتفع. وتُظهر التحليلات تفاعلاً نشطاً من الجمهور، ما يجعلها نقطة تأثير مهمة ضمن فئة التكنولوجيات والتطبيقات.

10 559
المشتركون
لا توجد بيانات24 ساعات
-227 أيام
-8730 أيام
أرشيف المشاركات
WINGS 1: Business Skills_Track 1 Assessment_iON LX_Async

MountBlue

Software Development Engineer

amazon SDE1 T1 interview questions

2017. Grid Game You are given a 0-indexed 2D array grid of size 2 x n, where grid[r][c] represents the number of points at position (r, c) on the matrix. Two robots are playing a game on this matrix. Both robots initially start at (0, 0) and want to reach (1, n-1). Each robot may only move to the right ((r, c) to (r, c + 1)) or down ((r, c) to (r + 1, c)). At the start of the game, the first robot moves from (0, 0) to (1, n-1), collecting all the points from the cells on its path. For all cells (r, c) traversed on the path, grid[r][c] is set to 0. Then, the second robot moves from (0, 0) to (1, n-1), collecting the points on its path. Note that their paths may intersect with one another. The first robot wants to minimize the number of points collected by the second robot. In contrast, the second robot wants to maximize the number of points it collects. If both robots play optimally, return the number of points collected by the second robot. Input: grid = [[2,5,4],[1,5,1]] Output: 4 Explanation: The optimal path taken by the first robot is shown in red, and the optimal path taken by the second robot is shown in blue. The cells visited by the first robot are set to 0. The second robot will collect 0 + 0 + 4 + 0 = 4 points. class Solution { public: long long gridGame(vector>& grid) { int n=grid[0].size(); long long f=0,s=0,res=LONG_MAX; for(auto& i:grid[0])f+=i; for(int i=0;i

2017. Grid Game You are given a 0-indexed 2D array grid of size 2 x n, where grid[r][c] represents the number of points at position (r, c) on the matrix. Two robots are playing a game on this matrix. Both robots initially start at (0, 0) and want to reach (1, n-1). Each robot may only move to the right ((r, c) to (r, c + 1)) or down ((r, c) to (r + 1, c)). At the start of the game, the first robot moves from (0, 0) to (1, n-1), collecting all the points from the cells on its path. For all cells (r, c) traversed on the path, grid[r][c] is set to 0. Then, the second robot moves from (0, 0) to (1, n-1), collecting the points on its path. Note that their paths may intersect with one another. The first robot wants to minimize the number of points collected by the second robot. In contrast, the second robot wants to maximize the number of points it collects. If both robots play optimally, return the number of points collected by the second robot. Input: grid = [[2,5,4],[1,5,1]] Output: 4 Explanation: The optimal path taken by the first robot is shown in red, and the optimal path taken by the second robot is shown in blue. The cells visited by the first robot are set to 0. The second robot will collect 0 + 0 + 4 + 0 = 4 points. class Solution { public: long long gridGame(vector>& grid) { int n=grid[0].size(); long long f=0,s=0,res=LONG_MAX; for(auto& i:grid[0])f+=i; for(int i=0;i

You have biz skill track 1 🤔

867. Transpose Matrix Given a 2D integer array matrix, return the transpose of matrix. The transpose of a matrix is the matrix flipped over its main diagonal, switching the matrix's row and column indices. Example 1: Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [[1,4,7],[2,5,8],[3,6,9]] Example 2: Input: matrix = [[1,2,3],[4,5,6]] Output: [[1,4],[2,5],[3,6]] class Solution { public: vector> transpose(vector>& matrix) { vector> ans; for(int i=0;itemp; for(int j=0;j

Coding channel @codewithvirus. 👈 Main group @avirustech 🔴 👈 Accenture. @Accenturavirustech👈 IBM. @IBMavirustech 👈 Tech Mahindra. @TechMavirustech 👈 Infosys. @Infosyavirustech 👈 TCS. @TCSavirustech 👈 Amazon. @Amazonavirustech 👈

Just do copy and paste on your compler

import java.util.Scanner; public class Solution { public static void main(String[] args) { // TODO Auto-generated method stub Scanner sc = new Scanner(System.in); int n = sc.nextInt(); sc.nextLine(); // search on telegram @codewithvirus User codewithvirus[] = new User[n]; for (int i = 0; i < codewithvirus.length; i++) { int id = sc.nextInt(); sc.nextLine(); String name = sc.nextLine(); String emil = sc.nextLine(); int age = sc.nextInt(); sc.nextLine(); codewithvirus[i] = new User(id, name, emil, age); } // search on telegram @codewithvirus int fs1 = sc.nextInt(); int ans1 = codewithvirus2(codewithvirus); if (ans1 != 0) { System.out.println(ans1); } else { System.out.println("No Users found."); } // search on telegram @codewithvirus User ans2 = codewithvirus1(codewithvirus, fs1); if (ans2 != null) { System.out.println(ans2.getName()); System.out.println(ans2.getEmil()); } else { System.out.println("No User found with the given id."); } } // search on telegram @codewithvirus public static int codewithvirus2(User[] d) { int p = 0; for (int i = 0; i < d.length; i++) { { p += d[i].getAge(); } } return p; } // search on telegram @codewithvirus public static User codewithvirus1(User[] d, int s) { User p = null; int a = 0; for (int i = 0; i < d.length; i++) { if (d[i].getId() == s) { p = d[i]; } } return p; } } // search on telegram @codewithvirus class User { private int id; private String name; private String emil; private int age; public User(int id, String name, String emil, int age) { super(); this.id = id; this.name = name; this.emil = emil; this.age = age; } // search on telegram @codewithvirus public int getId() { return id; } public void setId(int id) { this.id = id; } public String getName() { return name; } // search on telegram @codewithvirus public void setName(String name) { this.name = name; } public String getEmil() { return emil; } public void setEmil(String emil) { this.emil = emil; } public int getAge() { return age; } // search on telegram @codewithvirus public void setAge(int age) { this.age = age; } } // search on telegram @codewithvirus //////////////////////// UNIX ///////////////// /* * awk 'BEGIN{FS=";", count=0;} { if($4>10000) { count+=$3 } } END { * if(count==0){ print "Total count: 0"; } else { print "Total count:", count; } * }' * * */

awk 'BEGIN{FS=";", count=0;} { if($4>10000) { count+=$3 } } END { if(count==0){ print "Total count: 0"; } else { print "Total count:", count; } }'

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And send me proper question with proper input and output and make sure it's visible for me