Arrays: Find Numbers with Even Number of Digits

public class Nodigeven {
    static int even=0;
    static int count=0;
    static int findnodigeven(int[] arr) {
        for(int i=0;i<arr.length;i++) {
            while(arr[i]!=0) {
                arr[i]= arr[i]/10;
                count++;
            }
            if(count%2==0) {
                even++;
            }
        }
return even;
        }
    public static void main(String[] args) {
        int[] arr = {122,255,133,14,15,16};
        System.out.println(findnodigeven(arr));
    }   
}

am getting wrong output i.e : 1 .. can you tell me where i am going wrong here in the code

>Solution :

You need to change where you initialize count in your code – it never resets back to 0 and therefore the logic is incorrect for detecting if a number has an even number of digits.

public class Nodigeven {
    static int even = 0;
    static int findnodigeven(int[] arr) {
        for(int i = 0; i < arr.length; i++) {
            int count = 0; // This variable was in the wrong spot!
            while(arr[i] != 0) {
                arr[i] = arr[i] / 10;
                count++;
            }
            if(count % 2 == 0) {
                even++;
            }
        }
        return even;
    }
    public static void main(String[] args) {
        int[] arr = {122,255,133,14,15,16};
        System.out.println(findnodigeven(arr));
    }   
}