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### How to design a Node in Tree?

How to design a Node in Tree?

There are three main components of a tree in a node.

1) Integer holding data.
2) Left pointer holding node in a left subtree.
3) Right pointer holding node in a right subtree.

The following design is having data of int type and left,right pointers of a node to the subtrees.

```package com.BST;

public class Node {

int data;
Node left;
Node right;
/**
* @return the data
*/
public Node(int data)
{
this.left=null;
this.right=null;
this.data=data;
}

public int getData() {
return data;
}
/**
* @param data the data to set
*/
public void setData(int data) {
this.data = data;
}
/**
* @return the left
*/
public Node getLeft() {
return left;
}
/**
* @param left the left to set
*/
public void setLeft(Node left) {
this.left = left;
}
/**
* @return the right
*/
public Node getRight() {
return right;
}
/**
* @param right the right to set
*/
public void setRight(Node right) {
this.right = right;
}
/* (non-Javadoc)
* @see java.lang.Object#toString()
*/
@Override
public String toString() {
return "Node [data=" + data + ", left=" + left + ", right=" + right
+ "]";
}

}
```

### 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 …