### Calculating size of User and Cache storage

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

2^10=10^3
then 60*(10^3)^2 = 2^20 *2^6

which means we almost need 26 bits for storing ids. If we have any separate system to generate these ids then 26*60*10^6=1560*10^6=1.5 GB approx.

Cache storage :-

calcluting an idea for cache storage is not tough once we have an idea of calculating total storage . Cache normally sits between end user and application servers and caters read requests for the website. Nearly 60-80 percent of users do not apply any modification to the data of website and just read data. To avoid load on application servers which are modifying data for write requests and involve in lot of performance on data we use cache servers which just caters data as a replica of original one. So if we have 500 GB of original data , 100 GB of data can be served through cache.

There can be multiple techniques we can use for cache such as LRU , LFU which i will explain in coming articles.

### Bubble Sort in JAVA

What is bubble sort?

It is a sorting technique that is based on the comparison.Here we compare adjacent element, if the first element is larger than the second we swap each other. We do the same procedure again and again until array do not sort completely.

Example:-
5 1 4 2 8

51428here pass is nothing but iterating the loops equal to number of elements in the array but if it already sorted before then we can break the loop anddo existPASS1 case0 1 //check 0 and first element15428 case1 2 //check 1 and 2 element14528 case2 314258 case3 414258 swap istruePASS2 //first pass completed now do second pass case0 114258 case1 212458 case2 312458 case3 412458 swap istrue PASS3 // third pass case0 112458 case1 212458 case2 312458 case3 412458 swap isfalse since swap is false we break from the loop and do not go for fourth and fifth pass
Program:-

package sorting;publicclassBubbleSort{//function to print arraypublicstaticvoidprint(int[] arr){for(int i=0;i<=arr.length-1;i++) Syst…

### Sorting Implementation in JAVA

Sorting is a technique to order the data in a way such that data will remain in lowest to highest form. There are various techniques involved in sorting data.I have added the implementation of few important sorting techniques. Please click on link to read in detail:-
Bubble Sort
Selection Sort
Insertion Sort
Merge Sort