Chandan got bored playing with the arrays all the time. Therefore he has decided to buy a string \$S\$ consists of \$N\$ lower case alphabets. Once he purchased the string, He starts formulating his own terminologies over his string \$S\$. Chandan calls a string str A Balanced String if and only if the characters of the string str can be paritioned into two multisets \$M_1\$ and \$M_2\$ such that \$M_1=M_2\$ .

For example:

Strings like "abccba" , "abaccb" , "aabbcc" are all balanced strings as their characters can be partitioned in the two multisets \$M_1\$ and \$M2\$ such that \$M_1=M_2\$.

$$M1 = {a,b,c} \\ M2 = {c,b,a}$$

whereas strings like ababab , abcabb are not balanced at all.

Chandan wonders how many substrings of his string \$S\$ are Balanced Strings? Chandan is a little guy and do not know how to calculate the count of such substrings.

For input "abccba" Balanced substring are "cc" , "bccb" , "abccba" ie count=3 (Provided as per problem statement discussion) But I guess "aa", "bb", "cc", "abba", "acca", "cbbc" are also balanced sub string for the same input which makes count=6 Any wrong in my interpretation ?

Program

 import java.util.HashMap;
 import java.util.LinkedHashMap;
 import java.util.Scanner;
 import java.util.Set;

    public class BalancedStrings {
    static int  returnbalance(String input){
    HashMap<Character, Integer> characters=new LinkedHashMap<>();
    int count=0;
    Character c=null;
    for(int i=0;i<input.length();i++){
        c=input.charAt(i);
        if(characters.containsKey(c)==true){
            count=characters.get(c);
            characters.put(c, count+1);
        }
        else
            characters.put(c,1);        
    }
    int countunique=0;
    Set<Character> s=characters.keySet();       
    for(Character cc:s){            
        count=characters.get(cc);
        if(count%2==0)
            countunique++;
    }
    if(countunique!=s.size())
        return 0;
    Object[] sar= s.toArray();
    int len=0;
    for(int i=sar.length-1;i>=0;i--){
        len=len+i;
    }
        
    return len+countunique;
    
}
public static void main(String[] args) {
    Scanner in=new Scanner(System.in);
    int numberOfInputs=in.nextInt();
    in.nextLine();
    for(int i=0;i<numberOfInputs;i++){
        System.out.println(BalancedStrings.returnbalance(in.nextLine()));
    }

}

}