Search through sorted list (ascending) By given number, find the position where the number should go. Example given array: { 1, 3, 5, 7 } Number 6 will go after index 2. (after 5 and before 7) NOTE: Array must be sorted. (Not checking) NOTE: This is my algorithm. It might not be efficient but I think its doing the job. Complexity Big-O : O(lg n)

My Bin Search

abstract class MySearch {

    static int lastPivot = -1;

    /**
     * THIS IS MY ALGORITHM it might not be efficient but I didn't use external sources! :D
     * @param arr <b>MUST BE SORTED ARRAY!</b>
     * @return Output is POSITION - which the value is less than the position+1. Returns -1 if the number is
     * minimal from all the values of the array.
     */
    public static int binarySearch(int[] arr, int number) {

        int H = arr.length - 1;
        int L = 0,M;

        while(true) {
            M = (L + H) / 2;

            //if headers L and H are so close, compare both to number.
            if(H-L == 1) {
                if(number > arr[H])
                    return H;
                else if(number < arr[L])
                    return L-1;
                else
                    return L;
            }

            //best case senario
            if(number == arr[M])
                return M;

            if (number < arr[M])
                //H = M - 1;
                H = M;
            else //number is less than arr[M]
                //L = M + 1;
                L = M;
        }
    }

    public static void main(String[] args) {
        int[] arr = new int[]{5,9,11,20,100,103};
        int number = 14;
        System.out.println("Number " + number + " will go after index " + binarySearch(arr, number));
    }