Search in Rotated Sorted Array.cpp

/*
Search in Rotated Sorted Array
==============================

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is rotated at an unknown pivot index k (0 <= 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 rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

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

Constraints:
1 <= nums.length <= 5000
-104 <= nums[i] <= 104
All values of nums are unique.
nums is guaranteed to be rotated at some pivot.
-104 <= target <= 104

Follow up: Can you achieve this in O(log n) time complexity?
*/

class Solution
{
public:
  int search(vector<int> &nums, int target)
  {
    int i = 0, j = nums.size() - 1, n = nums.size();

    while (i <= j)
    {
      int mid = (i + j) / 2;
      if (nums[mid] == target)
        return mid;
      if (nums[i] <= nums[mid])
      {
        if (nums[i] <= target && nums[mid] >= target)
          j = mid - 1;
        else
          i = mid + 1;
      }
      else
      {
        if (nums[mid] <= target && nums[j] >= target)
          i = mid + 1;
        else
          j = mid - 1;
      }
    }
    return -1;
  }
};