Your window size should not exceed k

if it is exceeding k you need to release the characters
======================================================
//Earlier we have checked for unique strings 

int release=0;

for(int i=0;i<str.length();i++)
{
   Character ch=str.charAt(i);
   if(hm.containsKey(ch))
   {
      int x=hm.get(ch);
      x++;
      hm.put(ch,x);
   }
   else{
     hm.put(ch,1)
  }

  //
  while(hm.size()>k)
  {
   	  Character ch=str.charAt(release);
     int x=hm.get(ch);
     x--;
     hm.put(ch,x)
     if(hm.get(ch)==0)
     {
         hm.remove(ch);
     }
      release++;
  }

   count=count+(i-release+1);
  //
  max=Math.max(max,(i-release+1));
}
return max or count
====================================
k=3
str="geeksforgeeks"

g  1
e  2
k  1
s  1

https://www.naukri.com/code360/problems/distinct-characters_2221410?leftPanelTabValue=SUBMISSION

import java.util.*;
public class Solution {

	public static int kDistinctChars(int k, String str) {
		// Write your code here
		int count=0;
		int release=0;
		int max=-1;
		HashMap<Character,Integer> hm=new HashMap<>();
		for(int i=0;i<str.length();i++){
			Character ch=str.charAt(i);
			if(hm.containsKey(ch)){
				int x=hm.get(ch);
				x++;
				hm.put(ch,x);
			}else{
				hm.put(ch,1);
			}
			while(hm.size()>k){
				Character relchar=str.charAt(release);
				int y=hm.get(relchar);
				y--;
				hm.put(relchar,y);
				if(hm.get(relchar)==0){
					hm.remove(relchar);
				}
				release++;
			}
			max=Math.max(max,(i-release+1));
		}
		return max;
	}

}
===================================================================
Total Unique elements in each window
==================================================================