Concepts of Computer Science

user storage:- it can be solved using same scenario as mentioned in first article . https://tech.nazarmubeenworks.com/2019/08/basics-of-system-design-chapter-0.html Suppose there are 12 Million of users are adding every year meaning 1 Million per month. So if we consider 5 year there will be around 60 million of users. Now in terms of data we can look to our DB table and get information about it. A user will be generally having name , id , address , some forien keys . Lets assume 10 columns with each column on an average storing 4 byte of data. 10*4 = 40 bytes of data for one user. 60 * 10^6 * 40 = 2400 * 10^6 = 2.4 * 10^9 = 2.4 GB of data we need to store only user values. Also while calculating storage for the user there is also one important point we need to remember is of ids. If we are going to have 60 Million of users which means 60*10^6 users so unique id’s will be as we know below figures are almost equal ...

This article is first one from the series of articles dedicated to system design interviews. Here i am going to present the base scenario to consider before starting to solve system design problems. Questions to ask? 1) what is the number of requests a website will recieve in a day/month/second? 2) what is the amount of memory a website will deal in a day/month/second? 3) what is the number of servers that can accomodate these requests? To answer this , first we need to remember the below numbers:- 1 million = 10 lakh = 1000000 = 10^6 1 billion = 1000 million = 10^9 1 KB = 1024 B = 10^3 1 MB= 10^6 = 1024 KB 1 GB= 10^9 = 1024 MB 1 TB = 10^12 = 1024 GB Memory we need to see in Bytes Requests we need to see in numbers example :- suppose a website recieves 100M requests every month then:- requests per day = request per month /24 = 416700 requests requests per second = requests per day / (24*360...

Learn in a different way :- Strings stacks and queues heaps searching hash table sorting recursion dynamic programming greedy algorithms graphs tree Binary Search Tree Linked List Array Parallel programming and concurrency design problems system design availability and scalability  Below topics are good to have object oriented programming language details ( java , python) object oriented design tools ( bash , git , maven , jira , jenkins , docker , kubernetes) database

There is no secret to doing effective programming but following a sequence of steps and procedures can help in building great programs. Today I will take you on a tour to some of the best practices that are integrants of perfect programs. Algorithms are important: - You can’t deny the fact that algorithms are important and before start to write any program, Algorithms are must to write. For example, if there is a program about finding the sum of two numbers? What will you write ?The steps can be :- 1)   Get the first number. 2)   Get the second number. 3)   Add both numbers. This is what algorithm is writing about ,a list of statements .From the above three statements you can conclude boundary cases (at least two number should be there in input), mathematical function(Sum is needed here) , storage capacity(amount of memory to be assign to the variables), number of methods by analyzing repeat steps (reduce replete codes) and many other things. During alg...

Left Rotation in an Array using Java There are two methods implemented by me here :- 1) Using extra Array   Get new position of  a current element using formula.  Add the element in the new position in a new array. 2) Using same Array:- This is little bit complex hold variable at index i. find new index for variable. hold variable present at new index. place variable at index i to this new index. assign i to new index variable. repeat steps until count is not equal to lenght of an array which simply means all variable got covered. package Arrays ; /* * 12345 to 34512 */ /* * sample input * 5 4(d=4 that is 4 rotations) 1 2 3 4 5 sample output 51234 */ public class LeftRotation { public static void printArray ( int [] arr ) { for ( int i =0; i < arr . length ; i ++) { System . out . print ( " " + arr [ i ]); } System . out . println ( " " ); } //method ...