[Solved] 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].

Question

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

Constraints:

• 1 <= nums.length <= 5000
• -104 <= nums[i] <= 104
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• -104 <= target <= 104

Python Solution

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        if nums==None or len(nums)==0:
            return -1
        
        l = 0
        r = len(nums)-1

        while l<r:
            m = l +(r-l)//2

            if nums[m]>nums[r]:
                l = m+1
            else:
                r=m

        start = l
        l = 0
        r = len(nums)-1

        if target>=nums[start] and target<=nums[r]:
            l = start
        else:
            r=start-1

        while(l<=r):
            m = l+(r-l)//2
            if nums[m]==target:
                return m
            elif target<nums[m]:
                r=m-1
            else :
                l=m+1
        return -1