# idea : binary search
# Time Complexity : O(log n)

# Sorting is not allowed because it costs O(n log n), so we must handle it using binary search without sorting.
# One way is to check where the sorted order is disrupted.

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        left, right = 0, len(nums)-1
        
        while left <= right:
            mid = (left + right) // 2
            
            if nums[mid] == target:
                return mid
            
            if nums[left] <= nums[mid]:
                if nums[left] <= target < nums[mid]:
                    right = mid - 1
                else:
                    left = mid + 1
            
            else:
                if nums[mid] < target <= nums[right]:
                    left = mid + 1
                else:
                    right = mid - 1
        return -1


