Day 19 Leet code series, today we will be picking the problem Search in Rotated Sorted Array (https://leetcode.com/problems/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 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
class Solution {
public:
int search(vector<int>& nums, int target) {
int l = 0, r = nums.size()-1;
while (l <= r) {
int mid = (l+r) / 2;
if (target == nums[mid])
return mid;
// there exists rotation; the middle element is in the left part of the array
if (nums[mid] > nums[r]) {
if (target < nums[mid] && target >= nums[l])
r = mid - 1;
else
l = mid + 1;
}
// there exists rotation; the middle element is in the right part of the array
else if (nums[mid] < nums[l]) {
if (target > nums[mid] && target <= nums[r])
l = mid + 1;
else
r = mid - 1;
}
// there is no rotation; just like normal binary search
else {
if (target < nums[mid])
r = mid - 1;
else
l = mid + 1;
}
}
return -1;
}
};
ADDITIONAL KNOWLEDGE 
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