Recursive Binary Search Function In Python With Issues

Question

I'm working on a recursive binary search for school. Because of the test code, I'm limited to my input being lyst and target and output should only be a bool asserting whether the target was found or not. I've found examples of other questions kind of like mine, but they usually resort to having the upper and lower bounds as part of the input. I can't do that apparently. I'm also unsure as to whether or not I can use a helper function. I will try to find out. This is what I have so far:

def binary_search(lyst, target):
    left = 0
    right = len(lyst) - 1

    while left <= right:
        mid = (right + left) // 2
        if len(lyst) == 0:
            return False
        elif lyst[mid] < target:
            left = mid
            return binary_search(lyst[left: right], target)
        elif lyst[mid] > target:
            right = mid
            return binary_search(lyst[left: (right + 1)], target)
        elif lyst[mid] == target:
            return True
    return False

It works for some cases, but not all. For example, it will find target of 3 in in the following list my_list = [1,2,3,4,5,6,7,8,9,10], but will not find three in a list of 1-15. I feel like I'm close, but could use a little help.

Answer 1

Since recursion is not a requirement, you can just keep track of the start and end indices and stop when they've squeezed together. This avoids the slices while also allowing the function signature to remain unchanged:

def binary_search(lyst, target):
    start = 0
    end = len(lyst)
    
    while start < end:
        mid = (end - start) // 2 + start
        if lyst[mid] == target:
            return True
        if target < lyst[mid]:
            end = mid
        else:
            start = mid + 1
    return False


my_list = [1,2,3,4,5,6,7,8,9,10, 11, 12, 13, 14, 15]

print(binary_search(my_list, 3))
# True

# Some edge cases:
print(binary_search([1], 3))
# False

print(binary_search([], 3))
# False

print(binary_search([3], 3))
# True

print(binary_search([1, 3], 3))
# True

If you still want to use recursion with slices you don't need the loop, the recursion takes the place. Everything else is very similar, but in this case you need to deal with the edge case where recursion should stop:

def binary_search_rec(lyst, target):
    if not lyst:
        return False

    mid = len(lyst) // 2
    
    if lyst[mid] == target:
        return True
    if target < lyst[mid]:
        return binary_search_rec(lyst[:mid], target)
    else:
        return binary_search_rec(lyst[mid + 1:], target)

of course, as mentioned in the comments, you loose the efficiency because you need to create new lists with each call.