Binary Searchable Numbers in an Unsorted Array

lcpy gen -n 1966   # by problem number
lcpy gen -s binary_searchable_numbers_in_an_unsorted_array   # by problem name

Input: nums = [7]
Output: 1
Explanation:
Searching for value 7 is guaranteed to be found.
Since the sequence has only one element, 7 will be chosen as the pivot. Because the pivot equals the target, the function will return true.

Input: nums = [-1,5,2]
Output: 1
Explanation:
Searching for value -1 is guaranteed to be found.
If -1 was chosen as the pivot, the function would return true.
If 5 was chosen as the pivot, 5 and 2 would be removed. In the next loop, the sequence would have only -1 and the function would return true.
If 2 was chosen as the pivot, 2 would be removed. In the next loop, the sequence would have -1 and 5. No matter which number was chosen as the next pivot, the function would find -1 and return true.

Searching for value 5 is NOT guaranteed to be found.
If 2 was chosen as the pivot, -1, 5 and 2 would be removed. The sequence would be empty and the function would return false.

Searching for value 2 is NOT guaranteed to be found.
If 5 was chosen as the pivot, 5 and 2 would be removed. In the next loop, the sequence would have only -1 and the function would return false.

Because only -1 is guaranteed to be found, you should return 1.

class Solution:
    # Time: O(n)
    # Space: O(n)
    def binary_searchable_numbers(self, nums: list[int]) -> int:
        n = len(nums)
        searchable = [True] * n
        left_max = -(10**5 + 1)
        for i, value in enumerate(nums):
            if value < left_max:
                searchable[i] = False
            else:
                left_max = value
        right_min = 10**5 + 1
        total = 0
        for i in range(n - 1, -1, -1):
            if nums[i] > right_min:
                searchable[i] = False
            else:
                right_min = nums[i]
            if searchable[i]:
                total += 1
        return total