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  1. Prepare
  2. Algorithms
  3. Strings
  4. Sherlock and the Valid String
  5. Discussions

Sherlock and the Valid String

  • akshatbhawsar111
     + 0 comments

    I am using this java solution but still getting 2 test case failed anyone hlep me , thanks in advance

    Map<Character,Integer> ma=new HashMap<>();

        for(int i=0;i<s.length();i++)
        {
            if(!ma.containsKey(s.charAt(i)))
            {
                ma.put(s.charAt(i),1);
            }
            else {
                ma.computeIfPresent(s.charAt(i),(k,v)->++v);
            }
        }

         Integer[] op= ma.values().toArray(Integer[]::new);

        Arrays.sort(op);

        if(op[0]==op[op.length-1])
        {
            return "YES";
        }
  if(op[0]==1 && op[1]==op[op.length-1])
        {
           return "YES";
        }
       
            if(op[0]==op[1] && op[1]==op[op.length-2] && (op[1]==op[op.length-1]-1))
            {
                return "YES";
            }
        
    

return "NO";
    }