Write better code: Day 1 - Highest Product

Updated on Dec 07, 2018

I'm going to spend 1-2 hours/day solving different algorithms. More about this here.

Day 1: Question 2

Given an array of integers, find the highest product you can get from three of the integers.

The input array_of_ints will always have at least three integers.

Question from InterviewCake

If you want to follow along, feel free to post your answers below.

My answers are in the comments.

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DISCUSSION (9)

This was the logic:

  • Get three highest inputs - multiply
  • I could sort and choose highest 3 - but sort is worse than O[n] - so .. Trying something else.
  • Go thru each int one by one. - o[n]
  • See if the current int is greater than previous max_1 : then with max_2: then with max_3
  • Return product.
  • O[1] space as storing fixed.
  • O[n] time

Examples:

[10, 7, 5, 8, 11, 9] -> 990 ;
[10, 7, 100] -> 7000 ;
[-10, 7, 100] -> -7000;
[-10, 7, 100, 1] -> 700;

code
Github

I feel like much less comparisons can be done by keeping track of the lowest number:

def get_product_of_3(array_of_ints):
    maxes = [0, 0, 0]
    min_val = 0
    min_i = 0
    for int in array_of_ints:
        if int > min_val:
            maxes[min_i] = int
            min_i, min_val = min(enumerate(maxes), key=lambda x: x[1])

    return maxes[0]*maxes[1]*maxes[2]

Sorry, I replaced one language I don't know with a similar language I don't know because I didn't want to install ruby.

Thanks Mihail. This looks much more neat. Will test it out later.

But did you test the case if two negative numbers are passed?
For e.g [-5, -4, 1, 3, 2] -> should give 60.

Yes, I see the problem.
So far I see three cases:

  • The biggest product is all positives.
  • The biggest product is two smallest negatives and one biggest positive.
  • The biggest product is negative itself(!), which can happen in at least these two distinct cases:
    • [n,p,p]: There are only 3 inputs, one of them negative. For any length 3 input we should just return the input anyway, implicitly or explicitly.
    • [n,n,n...]: There are 3 or more inputs, none of which are positive. In this case we should return the 3 negative numbers closest to 0 (3 biggest numbers, same as the all-positive happy case)

I'll post a fix later, let's see how messy.

Okay, I think I got it:

import math


def get_product_of_3(array_of_ints):
    if len(array_of_ints) < 3:
        return 0

    min_max = -math.inf
    min_max_i = 0
    maxes = [min_max, min_max, min_max]
    max_neg = 0
    max_neg_i = 0
    negs = [0, 0]
    for int in array_of_ints:
        if int > min_max:
            maxes[min_max_i] = int
            min_max_i, min_max = min(enumerate(maxes), key=lambda x: x[1])
        if int < max_neg:
            negs[max_neg_i] = int
            max_neg_i, max_neg = max(enumerate(negs), key=lambda x: x[1])

    prod_maxes = maxes[0]*maxes[1]*maxes[2]
    prod_negs = negs[0]*negs[1]
    # This condition is only needed for `[n,p,p]`, to not return the invalid product of `0` when the product of maxes is negative. 
    if prod_negs != 0:
        return max(prod_negs*max(maxes), prod_maxes)
    return prod_maxes

If we instead explicitly return the product of 3 inputs, this condition is no longer needed:

import math


def get_product_of_3(array_of_ints):
    if len(array_of_ints) < 3:
        return 0
    if len(array_of_ints) == 3:
        return array_of_ints[0]*array_of_ints[1]*array_of_ints[2]

    min_max = -math.inf
    min_max_i = 0
    maxes = [min_max, min_max, min_max]
    max_neg = 0
    max_neg_i = 0
    negs = [0, 0]
    for int in array_of_ints:
        if int > min_max:
            maxes[min_max_i] = int
            min_max_i, min_max = min(enumerate(maxes), key=lambda x: x[1])
        if int < max_neg:
            negs[max_neg_i] = int
            max_neg_i, max_neg = max(enumerate(negs), key=lambda x: x[1])

    prod_maxes = maxes[0]*maxes[1]*maxes[2]
    prod_negs = negs[0]*negs[1]
    return max(prod_negs*max(maxes), prod_maxes)

Another possible optimization, for input such as [...-100,-99,-98,-97,1,-96,-95,-94...] is to stop expecting the "all negatives" case. This can easily be done by resetting all negative maxes to 0, so that incoming negatives are always smaller:

import math


def get_product_of_3(array_of_ints):
    if len(array_of_ints) < 3:
        return 0
    if len(array_of_ints) == 3:
        return array_of_ints[0]*array_of_ints[1]*array_of_ints[2]

    awaiting_not_negatve = True
    min_max = -math.inf
    min_max_i = 0
    maxes = [min_max, min_max, min_max]
    max_neg = 0
    max_neg_i = 0
    negs = [0, 0]
    for int in array_of_ints:
        if int > min_max:
            maxes[min_max_i] = int
            if awaiting_not_negatve and int >= 0:
                awaiting_not_negatve = False
                for i, v in enumerate(maxes):
                    if v < 0:
                        maxes[i] = 0
            min_max_i, min_max = min(enumerate(maxes), key=lambda x: x[1])
        if int < max_neg:
            negs[max_neg_i] = int
            max_neg_i, max_neg = max(enumerate(negs), key=lambda x: x[1])

    prod_maxes = maxes[0]*maxes[1]*maxes[2]
    prod_negs = negs[0]*negs[1]
    return max(prod_negs*max(maxes), prod_maxes)

Just realised that my original attempt was wrong !

For two negative integers e.g [-5, -4, 1, 3, 2] the correct answer is 60.
But the code above gives 6.

So.. need to change.

Ended up doing this by checking if the product of current number with the previous stored two numbers is greater than the the current max.

Had to store 5 variables:

  1. Highest number
  2. Lowest number
  3. Highest product of two numbers.
  4. Lowest product of two numbers.
  5. Max product of previous 3 numbers.

Then with each new number I check if the max product is greater than previous, and also update each. As there are negative numbers, the lowest product of two numbers can suddenly give the max product when multiplied with the current.

Code

This is O[n] which is still better than sort which is O[nlogn]

Will you believe me if I tell you that I didn't see this comment thread until now (well after posting my fixed solution)?

Ofcourse :) Only then would you have tried solving it twice and with a different logic.

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Arjun Rajkumar
I’m a software developer primarily building web apps using Ruby on Rails. Working on a personal project + freelancing. I'm looking for additional freelance work so hit me up if you need help!
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