public static String isValid(String s) {
            // Write your code here
            HashMap<Character , Integer> charFreq = new HashMap<>() ;
            for(char c:s.toCharArray())
            {
                int count = charFreq.getOrDefault(c, 0) +1 ;
                charFreq.put(c, count);
            }
            
            HashMap<Integer,Integer> freqCount = new HashMap<>();
            //freq => count 
            // 2   => 2   : aabbccc (a freq = 2 and b freq =2 so freq 2 => have count 2)  
            for(int freq : charFreq.values())
            {
                int count = freqCount.getOrDefault(freq , 0)+1 ;
                freqCount.put(freq , count);
            }
            
            if(freqCount.size() == 1)
            {
                return "YES" ;
            }
            
            if(freqCount.size() == 2)
            {
                int i =0 , f1 = 0 , f2 = 0 , c1 = 0 , c2 = 0 ;
                for(int freq : freqCount.keySet())
                {
                    if(i == 0)
                    {
                        f1 = freq ;
                        c1 = freqCount.get(freq);
                    }else{
                        f2 = freq ;
                        c2 = freqCount.get(freq);
                    }
                    i++ ;
                }
                
                if((c1 == 1 && f1 == 1) || (c2 == 1 && f2 == 1))
                { 
                    //int this condition size == 2 (only 2 frequecy)
                    // so if any freq == 1 and it's count == 1
                    // then if we remove that so it become valid 
                    // so it's a "YES" 
                    return "YES";
                }
                
                if(Math.abs(f1 -f2) == 1)
                {
                    //frequecy diffrence should be 1 
                    // example f1 = 2 and f2 = 3 => so if we remove 1 from f2 
                    // f2 = (3 -1) = 2 
                    // it can be  become valid string 
                    // can be not deffinatly 
                    /*
                        The two frequencies differ by exactly 1
                        The higher frequency (3) appears only once
                        We can remove 1 character from higher frequency
                         to make its frequency = 2
                         All characters would then have frequency 2
                    */
                                       
                    if((c1 == 1 && f1 > f2) || (c2 == 1 && f2 > f1))
                    {
                        /*
                        //eg => "aabbb"(true)
                            f1 = 2 , c1 = 1
                            f2 = 3 ,c2 = 1
                        
                         or "aabbbccc" (false)
                         f1 = 2 , c1 = 1 
                         f2 = 3 , c2 = 2
                         */
                         // if count = 1 
                         //then it freqncy shoubld be greter (+1)
                         //becouse remove there by one make the string valid 
                         //if count = 1 and freqncy is smaller ex => 2
                         // then even 2-1 = 1 
                         // one charecter still left 
                         //so not a valid string 
                        return "YES";
                    }
                }
                
            }
            
           
            return "NO"; // if freqCount greter than 2 then it's not valid 
                        // eg. => 2 , 1 , 3 => not valid

        }