RotatedBinarySearch.java

package search;


public class RotatedBinarySearch {


    /*
     * Binary Search using recursion
     * Time complexity - Olog(n)
     * Memory Complexity - O(logn) - as it will consume memory on the stack
     * */
    public static int rotatedBinarySearch(int[] arr, int n, int low, int high) {
        if(low > high) return -1;

        int middle = low + ((high-low)/2);
        if(arr[middle] == n) return middle;

        /*
         * Step 1: check if the n value is less than the middle value
         * Step 2: Check if the first value is less than the middle value
         * Step 3: Check if the first value is less than the n value
         * */
        if(n < arr[middle] && arr[low] < arr[middle] && n>= arr[low]) {
            return rotatedBinarySearch(arr, n, low, middle-1);

        /*
         * Step 1: check if the n value is greater than the middle value
         * Step 2: Check if the middle value is less than the highest value
         * Step 3: Check if the highest value is <= the n value
         * */

        } else if(n > arr[middle] && arr[middle] < arr[high] && n <= arr[high]) {
            return rotatedBinarySearch(arr, n, middle+1, high);

        /*
         * Step 1: check if the first value is greater than the middle value
         * */

        } else if(arr[low] > arr[middle]) {
            return rotatedBinarySearch(arr, n, low, middle-1);

        /*
         * Step 1: check if the last value is less than the middle value
         * */
        } else if(arr[high] < arr[middle]) {
            return rotatedBinarySearch(arr, n, middle+1, high);
        }

        return -1;
    }


    public static void main(String[] args) {
        int[] arr = {5, 6, 7, 8, 9, 10, 1, 2, 3};
        int n = 3;
        System.out.println(rotatedBinarySearch(arr, n, 0, arr.length-1));
    }
}