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### 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

` 5 1 4 2 8 `

`here 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 exist`
```PASS1
case0 1 //check 0 and first element
1 5 4 2 8
case1 2 //check 1 and 2 element
1 4 5 2 8
case2 3
1 4 2 5 8
case3 4
1 4 2 5 8
swap istrue```
```PASS2 //first pass completed now do second pass
case0 1
1 4 2 5 8
case1 2
1 2 4 5 8
case2 3
1 2 4 5 8
case3 4
1 2 4 5 8
swap istrue
PASS3 // third pass

case0 1
1 2 4 5 8
case1 2
1 2 4 5 8
case2 3
1 2 4 5 8
case3 4
1 2 4 5 8
swap isfalse
```

`since swap is false we break from the loop and do not go for fourth and fifth pass`

Program:-

```package sorting;

public class BubbleSort {

//function to print array
public static void print(int[] arr)
{
for(int i=0;i<=arr.length-1;i++)
System.out.print(" "+arr[i]);
System.out.println(" ");
}

//function to swap elements
public static int[] swap(int[] arr,int i, int j)
{
int temp;
temp=arr[i];
arr[i]=arr[j];
arr[j]=temp;

return arr;
}

public static void main(String[] args)
{
int[] arr={5,1,4,2,8};
print(arr);
//declared to check the number of passes
int pass=1;

while(pass!=arr.length)
{
/*declared to check if there is no swap then we are
working on already sorted array and can break the loop */
boolean swap=false;

System.out.println("PASS"+pass);

for(int i=0;i<arr.length-1;i++)
{
if(arr[i]>arr[i+1])
{
arr=swap(arr,i,i+1);
swap=true;
}

System.out.println("case"+i+" "+(i+1));

print(arr);
}

System.out.println("swap is"+swap);
if(swap==false)
{
break;
}
pass++;
}

}

}
```

### Basics of System Design

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 = 10^31 MB= 10^6 = 1024 KB 1 GB= 10^9 = 1024 MB1 TB = 10^12 = 1024 GB
Memory we need to see in BytesRequests 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*3600) = 4.8 requests per second
memory:-
if we take 20:80 principal where 20 percent is write and 80 percent is …

### 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 …