33. Search in Rotated Sorted Array

1. Description

There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is possibly left rotated at an unknown 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 left rotated by 3 indices 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.

2. Example

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

3. Constraints

• 1 <= nums.length <= 5000
• -10$^{4}$ <= nums[i] <= 10$^{4}$
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• -10$^{4}$ <= target <= 10$^{4}$

4. Solutions

Binary Search

n = nums.size()
Time complexity: O(logn)
Space complexity: O(1)

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class Solution {
public:
    int search(const vector<int> &nums, int target) {
        int left = 0, right = nums.size();
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                return mid;
            } else if (
                nums.front() <= nums[mid] && nums[mid] < target || // all left
                nums[mid] < target && target < nums.front() || // all right
                nums[mid] >= nums.front() && target < nums.front()) { // nums[mid] left, target right
                left = mid + 1;
            } else {
                right = mid;
            }
        }

        return -1;
    }
};