Even or Odd String Interview Question - Hacker Rank

Even or Odd String Interview Question - Hacker Rank
Goal: To determine if it is a even or odd string.
Input: Number and Array of String containing lower case
Output:
For each string compute:
[ Char[0]^Number  X  Char[1]^Number   X   …… (All character of String[0])  +
Char[0]^Number   X Char[1]^Number    X   …… (All character of String[1])  +
…………… ] is Even Number or Odd Number ?
Here Char[i] is the ASCII value of character.
For example:
For input: abc bd    and 2
=  (97^2  X   98^2   X    99^2)  + (98^2   X   100^2)
=  ( 9409 X 9604 X 9801 )  + ( 9604 X 10000)
=  885657916836 + 96040000
=  885753956836
So it should return Even.
The catch here is not to compute complex mathematical formulas. There are very basic mathematical checks that needs to be followed to compute if this whole sum is even or odd.
1. Even^AnyNumber = Even & Odd^AnyNumber = Odd
So you do not need to compute the powers.
2. Even X Any Number = Even
Odd X Odd = Odd
So you just need to identify whether there is any even number present in the multiplication.
3. For sum: if Odd numbers occur odd times, final sum will be Odd.
So you need to compute the odd count

    public static void main(String args[]){

         

        String data[] = {"aceace","ceceaa","abdbdbdbakjkljhkjh"};

        System.out.println(checkComputation(data));

         

        String data2[] = {"azbde","abcher","acegk"};

        System.out.println(checkComputation(data2));

                 

    }

     

    public static  boolean checkComputation(String data[] ){

         

        boolean isStringEvenArray[] = new boolean[data.length] ;

        int tempVal;

        boolean isEvenPresent = false;

         

        for(int i=0; i<data.length;i++){

            isEvenPresent= false;

             

            for(int j=0; j<data[i].length();j++ ){

                tempVal = (int) data[i].charAt(j);

                if(tempVal%2==0){

                    isEvenPresent = true;

                }

            }          

            isStringEvenArray[i] = isEvenPresent;          

        }

         

        int oddCount=0;    

        for(int i=0; i<data.length; i++){           

            if(!isStringEvenArray[i])

                oddCount++;        

        }

         

        if(oddCount%2!=0)return false;

        else return true;

         

    }

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