Data Structures and Algorithms :

Basic Programs :
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if-else
Functions & How to call functions
loops
Standard programs
Patterns
Arrays
Strings
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Write a program to print wheather a number is Even or Odd

Even numbers : 0, 2 , 4, 6, 8
Odd numbers : 1,3, 5, 7,9...

if a number is divisible by 2 then the number said to be Even else odd

% Modulus opearator it returns remainder

21%2==1 (odd if remainder and even if remainder 0)
22%2==0
23%2==1
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Instead hard coding I want to take input from the end user and map them to our local variables
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package test;

import java.util.*;
public class Checker {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		
		if(n%2==0) {
			System.out.println("n is an even number");
		}else {
			System.out.println("n is an Odd number");
		}
		
	}

}
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2. Write a program to print n natural number

for(initialization ; conditioncheck;incr/decr){

}
1.initialization
2.condition check
3.statements
4.increment/decrement

2,3,4 steps will repet till your condition got failed
=================================================
import java.util.*;
public class Checker {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scan=new Scanner(System.in);
		System.out.println("Please enter the value of N");
		int n=scan.nextInt();
		
		System.out.println("The Values are");
		for(int i=1;i<=n;i++) {
			System.out.println(i);
		}
		//console : 1,2,3
		//i=1 ; 1<=50
		//i=2  2<=50
		//i=3  3<=50
		
	}

}
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3. Wap to print all the even numbers till n
  
we need to apply filter 

if(i%2==0)
sop("print even number")
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import java.util.*;
public class Checker {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scan=new Scanner(System.in);
		System.out.println("Please enter the value of N");
		int n=scan.nextInt();
		
		System.out.println("The Values are");
		for(int i=1;i<=n;i++) {
			if(i%2==0) {
				System.out.println(i);
			}
			
		}
		//console : 1,2,3
		//i=1 ; 1<=50
		//i=2  2<=50
		//i=3  3<=50
		
	}

}
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3 Programs 
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Prime number I will explain in 3 ways by optimizing the iterations
4th Approach based on binary search

The number having 2 factors is called prime number 
Those 2 factors are 1 and itself

will run a loop from 1 to <=n
will calculate count of all factors

if x is factor of n what is the basic requirement :
n%x==0


code : 
               for(int i=1;i<=n;i++) {
			if(n%i==0) {
				count=count+1;
			}
		}
		if(count==2) {
			System.out.println("It is a prime number");
		}else {
			System.out.println("It is not a prime number");
		}
============================
n=5
i= 1 to 5

5%1==0 (crct) count=1
5%2!=0      count=1
5%3!=0      count=1
5%4!=0     count=1
5%5==0   count=2
=================================
n=1 lakh  how many iterations loop is going
1 lakh 
50k iterations

response time of Amazon has down by few milli seconds 

number 2 to n/2

23/2 = 2 to 11

if you find a factor between 2 to n/2
count+=1
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		if(n<=1) {
			System.out.println("It is not a prime 				number");
		}else {
			for(int i=2;i<=n/2;i++) {
				if(n%i==0) {
					count=count+1;
					break;
				}
			}
			if(count==0) {
				System.out.println("It is a prime number");
			}else {
				System.out.println("It is not a prime number");
			}
		}
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if n=23
 we will check from 2 to 23/2 = 11

if(count==0)
sop("It is a prime number")
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1 lakh to 50k 

1 crore to 50 lakhs

 2 to sqrt(100)
  1 to 100
  2 to 50
  2 to 10
===========================
17 ==> 4

2 to 4

17%2!=0
17%3!=0
17%4!=0

if(count==0)
=======================
10000000
10000=10^7

Square root(some big number)= 1 lakh
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Binary Search class
mid=1 + some big/2
mid*mid==number
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Arrays , Strings , Map ,Stack , Queue , LL , Trees , GrapSorting algos , searching , Greedy , Dynamic programming , Recursion
2 pointer algos , sliding window algos
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