Search in Rotated Sorted Array

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Example:

For [4, 5, 1, 2, 3] and target=1, return 2.

For [4, 5, 1, 2, 3] and target=0, return -1.

Challenge:

O(logN) time

Solution:

Since the time complexity is O(logN), we have to use binary search to reduce the SEARCH SPACE. Assume we have a sorted array [c, ..., b] and we rotate it at pivot a, then we have a rotated array [a, ..., b, c, ...., d], where c is the smallest value and b is the largest. A similar idea to Find Minimum in Rotated Sorted Array.

In this scenario, we could reduce the search space by following two cases:

1. a <= target <= nums[mid], where a <= nums[mid],
2. nums[mid] <= target <= d, where a > nums[mid].

Code:

public int search(int[] nums, int target) {
    if (nums == null || nums.length == 0) {
        return -1;
    }
    int start = 0;
    int end = nums.length - 1;
    while (start + 1 < end) {
        int mid = start + (end - start) / 2;
        if (nums[start] <= nums[mid]) {
            if (nums[start] <= target && target <= nums[mid]) {
                end = mid;
            } else {
                start = mid;
            }
        } else {
            if (nums[mid] <= target && target <= nums[end]) {
                start = mid;
            } else {
                end = mid;
            }
        }
    }
    if (nums[start] == target) {
        return start;
    }
    if (nums[end] == target) {
        return end;
    }
    return -1;
}