Question

Below code is to find how many times a number is shown in an array.
For Example:

1,2,2,2,3,4,5,5,5,5
number 2 = 3 times
number 5 = 4 times.

What is the time complexity in Java for the below code?
What is the best way to solve this problem in respect to Time complexity?

public static void main(String[]args)
    {
        int[] data = {1,1,2,3,4,4,4,5,6,7,8,8,8,8};
        System.out.println(count(data,8));


    }


public static int count(int[] a, int x)
{
    int count=0;
    int index=0;
    while(index<a.length)
    {
        if(a[index]==x)
        {
            count++;
        }
        index++;

    }
    return count;
}

Answer 1

is it o(n) ,logN or else?

You look at every element once so it is O(n)

if it is o(n), can I can i do this task with LogN complicity?

Do a binary search with a low and high value searching for the value just less than the one you search for and just above. The difference between these two result will tell you how many there are. This is an O(Log N) search.