This is a programming problem where the given array A is of infinite length and we have to find the position of a value x in it. The first n values are sorted and after n-th number, all remaining values in the array are None. For example: A= [1, 3, 5, 100, 102, 1050, 1061, None, None, None, …]. But the catch is that we do not know the value of n here to directly run binary search in an O(log n) time complexity. This is an interesting Divide and Conquer problem.

So in my solution below, I gradually take a wider step up from checking just one element from the beginning of this infinite array and check if the last value in that window is big enough to capture x, which indicates we have covered enough length to capture the first n-th values. At each step, I double the window size.

Then when we have found the window withing which either x should reside, or we hit None, we run binary search within A[i,j]. The example below assumes x is guaranteed to be in the array.

<pre class="wp-block-code">```
 def binary_search(i, j, x):    
     mid = int((i+j) / 2)
     mid_val = A[mid]
      
     ind = None

     if mid_val == x:
         ind = mid    
     elif mid_val is None or mid_val > x:
         ind = binary_search(i, mid, x) 
     elif mid_val < x:
         ind = binary_search(mid, j, x)
     return ind
        
def find_x_in_infinite_array(A, x):
    i = 0
    k = 0
    while True:
        j = i + 2 ** k
        j_val = A[j]
        
        if j_val is None or j_val > x:
            index = binary_search(i, j, x)
            break
        i = j
        k = k + 1

```