recursive binary search function that takes in and finds an array of keys (as opposed to a single key)

Question

I need to make my binary search function search for multiple keys, not just one like I currently have. If the key is found return index of the key, else return -1.

example:

array = [ 1, 3, 5, 8, 10]
keys = [ 0, 2, 8, 5]

answer = [ -1, -1, 3, 2]

Help would be much appreciated!

def biSearch(A, low, high, key):

    mid = (high + low) // 2

    if low >= high: 
        return mid

    elif A[mid] >= key:        # we are in the left half of the array
        index = biSearch(A, low, mid, key)
        return index

    else:                   # we are in the right half of array
        index = biSearch(A, mid + 1 , high, key)
    return index


def search(A, key):
    low = 0
    high = len(A)
    index = biSearch(A, low, high, key)

    return index

Answer 1

Here is a way you might go about this:

def _binary_search(arr, low, high, key):
    mid = (high + low) // 2
    if high - low <= 1:
        return low
    elif arr[mid] > key:
        return _binary_search(arr, low, mid, key)
    else:
        return _binary_search(arr, mid, high, key)

def binary_search(arr, key):
    """
    Wraps the actual recursive search function, setting initial bounds and
    checking if the key was found or not.
    """
    index = _binary_search(arr, 0, len(arr), key)
    if arr[index] == key:
        return index
    return -1

def vectorised_binary_search(arr, key_arr):
    return [binary_search(arr, key) for key in key_arr]

print(vectorised_binary_search([1, 3, 5, 8, 10], [0, 2, 8, 5]))
print(vectorised_binary_search(range(0, 10, 2), range(10)))

which has output

[-1, -1, 3, 2]
[0, -1, 1, -1, 2, -1, 3, -1, 4, -1]

I have taken the liberty of re-writing your binary search a little.