SearchInRotatedSortedArray.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
 */

public class SearchInRotatedSortedArray {
    public static void main(String[] args) {
        int[] nums={4,5,6,7,0,1,2};
        int ans=search(nums,0);
        System.out.println(ans);
    }
    public static int search(int[] nums, int target) {
        int pivot = findPivot(nums);

        // if you did not find a pivot, it means the array is not rotated
        if (pivot == -1) {
            // just do normal binary search
            return binarySearch(nums, target, 0 , nums.length - 1);
        }

        // if pivot is found, you have found 2 asc sorted arrays
        if (nums[pivot] == target) {
            return pivot;
        }

        if (target >= nums[0]) {
            return binarySearch(nums, target, 0, pivot - 1);
        }

        return binarySearch(nums, target, pivot + 1, nums.length - 1);
    }
    public static int binarySearch(int[] arr, int target, int start, int end) {
        while(start <= end) {
            // find the middle element
//            int mid = (start + end) / 2; // might be possible that (start + end) exceeds the range of int in java
            int mid = start + (end - start) / 2;

            if (target < arr[mid]) {
                end = mid - 1;
            } else if (target > arr[mid]) {
                start = mid + 1;
            } else {
                // ans found
                return mid;
            }
        }
        return -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;
    }
}