Problem 1295.java

class Solution {
    public int findNumbers(int[] nums) {
        int temp=0;

        for(int i=0;i<nums.length;i++){
            int count=0;
            while(nums[i]>0){
                count++;
                nums[i]=nums[i]/10;
            }
            if(count % 2 ==0){
                temp++;
            }
            
        }
        return temp;
    }
}
/* Given an array nums of integers, return how many of them contain an even number of digits. 

 //Method 1 - count by conventional while loop method
//Method 2 - use Math.log to count number of digits
//Method 3 - converrt to string and find length of string

class Solution {
    public int findNumbers(int[] nums) {
        int c=0;
        for(int i=0;i<nums.length;i++){
            int d = digits(nums[i]);
            if(even(d))
            c++;
        }
        return c;
    }

    static int digits(int n){  //M-2 instead of counting like this use 'return (int)(Math.log10(n)) + 1; '  in place of whole function
        int d=0;
        while(n!=0){
            d++;
            n/=10;
        }
        return d;
    }

    static boolean even(int n){
        if(n%2==0) return true;
        else return false;
    }
}


//Method - 3
 class Solution {
    public int findNumbers(int[] nums) {
        int c=0;
        for(int i=0;i<nums.length;i++){
            String a = Integer.toString(nums[i]);
            int l =  a.length();
            if(even(l))
            c++;
        }
        return c;
    }
    static boolean even(int n){
        if(n%2==0) return true;
        else return false;
    }
} */