Solution: Order-agnostic Binary Search

class Solution:
  def search(self,arr, key):
    
    def binary_search(goleft):
      left, right = 0, len(arr) - 1

      while left <= right:
        mid = left + (right - left) // 2
        if arr[mid] == key:
          return mid
        elif goleft(key, arr[mid]):
          right = mid - 1
        else:
          left = mid + 1
      
      return -1
    

    is_ascending = arr[0] < arr[-1]

    if is_ascending:
      return binary_search(lambda key,mid: key < mid )
    else:
      return binary_search(lambda key,mid: key > mid )

class Solution:
  def search(self,arr, key):
    if len(arr) == 0:
      return -1
    if arr[0] == arr[-1]:
      return 0 if arr[0] == key else -1
    inc = arr[-1] > arr[0]
    i = 0
    j = len(arr)-1
    while i < j:
      mid = (i+j)//2
      if arr[mid] == key:
        return mid
      if arr[mid] > key and inc:
        j = mid - 1
      else:
        i = mid +1
    if arr[i] == key:
      return i
    return -1  # element not found

I have this solution, which should work. For example. for the array [1,1,1,1,1] it returns 0, but the solution says that it should return 2. I believe this is a mistake because the problem description does not specify how exactly duplicates should be handled.

A little shorter, perhaps more confusing:

if ascending == (arr[mid] < key): start = mid+1 else: end = mid-1