rotatedBinarySearch.java

/*

There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].
Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
You must write an algorithm with O(log n) runtime complexity.

Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Example 3:
Input: nums = [1], target = 0
Output: -1

*/


public class rotatedBinarySearch {
    public static void main(String[] args) {
        int[] arr = {4, 5, 6, 7, 0, 1, 2};
        System.out.println(findPivot(arr));
    }

    static int binarySearch(int[] arr, int target, int start, int end) {
        while (start <= end) {
            int mid = start + (end - start)/2;
            if (target < arr[mid]) end = mid - 1;
            else if (target > arr[mid]) start = mid + 1;
            else return mid;
        }
        return -1;
    }

    static int search(int[] nums, int target) {

        int pivot = findPivot(nums);

        if (pivot == -1) {
            return binarySearch(nums, target, 0, nums.length - 1);
        }

        // if pivot is found, you have found 2 asc sorted arrays

        // if pivot element = target
        if (nums[pivot] ==  target) return pivot;

        // if the target is greater than start element than search in the left half because it contains numbers greater than start
        if (target >= nums[0]) return binarySearch(nums, target, 0, pivot - 1);

        // else search in the right half where the elements are lesser than start
        return binarySearch(nums, target, pivot + 1, nums.length - 1);

    }

    static int findPivot(int[] arr) {
        int start = 0;
        int end = arr.length - 1;
        while (start <= end) {
            int mid = start + (end - start)/2;
            //  4 cases over here
            if (mid < end && arr[mid] > arr[mid + 1]) return mid;
            if (mid < start && arr[mid] < arr[mid - 1]) return mid - 1;
            if (arr[mid] <= arr[start]) end = mid - 1;
            else start = mid + 1;
        }
        return -1;
    }
}