Sliding Window :
==================
we need to make sure the window is Valid according to the 
problem statement

2 types of window 
window refers to Substring or SubArray

1. constant window

  in a window of size we need no of Unique elements in each window

1, 2, 2 , 3,4,5

k=3

2. Variable size window

longest substring with out having duplicate characters
  str =geeksforgeeks
  eksforg
The size of the window will varying

we need to maintain 2 pointers 
start pointer and end pointer

start=0;
end=0
getyueksforgeeks

tyue
max=end-start+1
HashMap is empty
end=2
start=0

t   1
y   1
u  1

will check e is there in the map or not



you need to maintain one hashmap

for(end=0;end<str.length();end++)
{
  str.charAt(start)
}
we need to make sure content in the HashMap should be valid 
between start and end pointer

first I will check wheather g is there in Map or not
Once you find your window is invalid
you need to start releasing characters from the map

m  1
n= 1

naman =
n
na
nam
nama X
naman X
----------
a
am
ama X
aman X
------------
m
ma
man
------------
a
an
-----
n
==============================

import java.util.*;

public class Solution 
{

	public static int countDistinctSubstrings(String str) 
	{
		//	  Write your code here.
        str=str.trim();
		int start=0;
        int end=0;
        int sum=0;
        HashMap<Character,Integer> hm=new HashMap<>();
        for(end=0;end<str.length();end++){
            Character ch=str.charAt(end);
            while(hm.containsKey(ch)){
                //you need to get every pointed by strat
                //you need to remove same from the map
                //geeksfor
                Character c=str.charAt(start);
                int x=hm.get(c);
                x--;
                hm.put(c,x);
                if(hm.get(c)==0)
                {
                    hm.remove(c);
                }
                start++;
                
            }
            hm.put(ch,1);
            sum=sum+(end-start+1);
        }
        return sum+1;
	}
}