Solution Explanation

The problem asks to find the number of elements in an array that are guaranteed to be found by a modified binary search algorithm, even if the array is unsorted. The modified binary search randomly selects a pivot, compares it to the target, and then removes elements to the left or right of the pivot based on the comparison.

The key insight is that an element is guaranteed to be found if and only if it's both the smallest element in some suffix of the array and the largest element in some prefix of the array.

The solution iterates through the array twice:

1. Forward Pass: It maintains a running maximum (mx). If an element is smaller than the current maximum, it means this element cannot be the largest in any prefix, so it's marked as not guaranteed to be found.

2. Backward Pass: It maintains a running minimum (mi). If an element is larger than the current minimum, it means this element cannot be the smallest in any suffix, so it's also marked as not guaranteed to be found.

After both passes, only elements marked as guaranteed to be found are counted.

Time and Space Complexity Analysis

• Time Complexity: O(N), where N is the length of the input array. The solution performs two linear scans of the array.
• Space Complexity: O(N) in the worst case. This is due to the ok array used to store whether each element is guaranteed to be found. However, in-place modification of the input array is possible which would make space complexity O(1).

Code Explanation (Python)

class Solution:
    def binarySearchableNumbers(self, nums: List[int]) -> int:
        n = len(nums)
        ok = [1] * n  # Initialize an array to track guaranteed elements
        mx, mi = -1000000, 1000000  # Initialize min and max to extreme values
 
        # Forward pass: Check for largest in prefix
        for i, x in enumerate(nums):
            if x < mx:
                ok[i] = 0  # Not guaranteed if smaller than previous max
            else:
                mx = x      # Update max
 
        # Backward pass: Check for smallest in suffix
        for i in range(n - 1, -1, -1):
            if nums[i] > mi:
                ok[i] = 0  # Not guaranteed if larger than previous min
            else:
                mi = nums[i]  # Update min
 
        return sum(ok)  # Count guaranteed elements

The other languages (Java, C++, Go) follow a very similar structure, only differing in syntax and specific library functions used. The core logic remains the same across all implementations.