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up vote 3 down vote favorite

This is a follow up to Maximum product of 3 integers in an int array using Python

Changes to the code include renamed variables for improved readability and I added more test cases.

import unittest


def highest_product(list_of_ints):
    max_seen = [float("-inf"), float("-inf"), float("-inf")]
    min_seen = [float("inf"), float("inf")]

    for x in list_of_ints:
        if x >= max_seen[0]:
            max_seen[0], max_seen[1], max_seen[2] = x, max_seen[0], max_seen[1]
        elif x >= max_seen[1]:
            max_seen[1], max_seen[2] = x, max_seen[1]
        elif x > max_seen[2]:
            max_seen[2] = x
        if x <= min_seen[0]:
            min_seen[0], min_seen[1] = x, min_seen[0]
        elif x < min_seen[1]:
            min_seen[1] = x

    max_product_candidate_one = min_seen[0] * min_seen[1] * max_seen[0]
    max_product_candidate_two = max_seen[0] * max_seen[1] * max_seen[2]

    return max(max_product_candidate_one, max_product_candidate_two)


class TestHighestProduct(unittest.TestCase):
    def test_highest_product(self):
        self.assertEqual(highest_product([6, -1, -1, -2, 0]), 12)
        self.assertEqual(highest_product([-6, -1, -1, -2]), -2)
        self.assertEqual(highest_product([0, 0, 0]), 0)
        self.assertEqual(highest_product([0, 0, -2]), 0)


if __name__ == '__main__':
    unittest.main(verbosity=2)
share|improve this question
up vote 2 down vote accepted

This version feels much more readable, especialy since you dropped half of your comparisons.

However, I really liked @JoeWallis use of heapq. I would however use heappushpop which will provide all the comparisons you are doing at once.

But you will need to manually extract out the maximum value of max_seen since you have no guarantee anymore on its position.

You can thus write:

from heapq import heappushpop

def highest_product(list_of_ints):
    max_seen = [float("-inf")] * 3
    min_seen = [float("-inf")] * 2

    for x in list_of_ints:
        heappushpop(max_seen, x)
        heappushpop(min_seen, -x)

    # No change of sign since we changed it twice (once for each element)
    max_product_candidate_one = min_seen[0] * min_seen[1] * max(max_seen)
    max_product_candidate_two = max_seen[0] * max_seen[1] * max_seen[2]

    return max(max_product_candidate_one, max_product_candidate_two)
share|improve this answer
1  
-(-1) for heappushpop. – newToProgramming Oct 27 '16 at 13:43

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