Find the subset of a set of integers that has the maximum product

Question

Let A be a non-empty set of integers. Write a function find that outputs a non-empty subset of A that has the maximum product. For example, find([-1, -2, -3, 0, 2]) = 12 = (-2)*(-3)*2

Here's what I think: divide the list into a list of positive integers and a list of negative integers:

1. If we have an even number of negative integers, multiply everything in both list and we have the answer.
2. If we have an odd number of negative integers, find the largest and remove it from the list. Then multiply everything in both lists.
3. If the list has only one element, return this element.

Here's my code in Python:

def find(xs):
    neg_int = []
    pos_int = []
    if len(xs) == 1:
        return str(xs[0])
    for i in xs:
        if i < 0:
            neg_int.append(i)
        elif i > 0:
            pos_int.append(i)
    if len(neg_int) == 1 and len(pos_int) == 0 and 0 in xs:
        return str(0)
    if len(neg_int) == len(pos_int) == 0:
        return str(0)
    max = 1
    if len(pos_int) > 0:
        for x in pos_int:
            max=x*max
    if len(neg_int) % 2 == 1:
        max_neg = neg_int[0]
        for j in neg_int:
            if j > max_neg:
                max_neg = j
        neg_int.remove(max_neg)
    for k in neg_int:
        max = k*max
    return str(max)

Am I missing anything? P.S. This is a problem from Google's foobar challenge, I am apparently missing one case but I don't know which.

Now here's actual problem:

Answer 1

from functools import reduce
from operator import mul

def find(array):
    negative = []
    positive = []
    zero = None
    removed = None

    def string_product(iterable):
        return str(reduce(mul, iterable, 1))

    for number in array:
        if number < 0:
            negative.append(number)
        elif number > 0:
            positive.append(number)
        else:
            zero = str(number)

    if negative:
        if len(negative) % 2 == 0:
            return string_product(negative + positive)

        removed = max(negative)

        negative.remove(removed)

        if negative:
            return string_product(negative + positive)

    if positive:
        return string_product(positive)

    return zero or str(removed)

Answer 2

You can simplify this problem with reduce (in functools in Py3)

import functools as ft
from operator import mul

def find(ns):
    if len(ns) == 1 or len(ns) == 2 and 0 in ns:
        return str(max(ns))
    pos = filter(lambda x: x > 0, ns)
    negs = sorted(filter(lambda x: x < 0, ns))
    return str(ft.reduce(mul, negs[:-1 if len(negs)%2 else None], 1) * ft.reduce(mul, pos, 1))

>>> find([-1, -2, -3, 0, 2])
'12'
>>> find([-3, 0])
'0'
>>> find([-1])
'-1'
>>> find([])
'1'