Algorithm Optimization [Python]

[Jamison Lahman]

Homework Statement

Given a list of integers and a single sum value, return the first two values (parse from the left please) in order of appearance that add up to form the sum.

sum_pairs([11, 3, 7, 5],         10)
#              ^--^      3 + 7 = 10
== [3, 7]

sum_pairs([4, 3, 2, 3, 4],         6)
#          ^-----^         4 + 2 = 6, indices: 0, 2 *
#             ^-----^      3 + 3 = 6, indices: 1, 3
#                ^-----^   2 + 4 = 6, indices: 2, 4
#  * entire pair is earlier, and therefore is the correct answer
== [4, 2]

sum_pairs([0, 0, -2, 3], 2)
#  there are no pairs of values that can be added to produce 2.
== None/nil/undefined (Based on the language)

sum_pairs([10, 5, 2, 3, 7, 5],         10)
#              ^-----------^   5 + 5 = 10, indices: 1, 5
#                    ^--^      3 + 7 = 10, indices: 3, 4 *
#  * entire pair is earlier, and therefore is the correct answer
== [3, 7]

Negative numbers and duplicate numbers can and will appear.

NOTE: There will also be lists tested of lengths upwards of 10,000,000 elements. Be sure your code doesn't time out.This is a problem off of codewars.com. I have made two versions that satisfy the eight examples but there is a 12000ms time restraint that I need to overcome. I am not versed in CS so I don't know much about algorithms or how to optimize them.

2. Homework Equations
The examples are:

l1= [1, 4, 8, 7, 3, 15]
l2= [1, -2, 3, 0, -6, 1]
l3= [20, -13, 40]
l4= [1, 2, 3, 4, 1, 0]
l5= [10, 5, 2, 3, 7, 5]
l6= [4, -2, 3, 3, 4]
l7= [0, 2, 0]
l8= [5, 9, 13, -3]

print(sum_pairs(l1, 8) == [1, 7]) #"Basic: %s should return [1, 7] for sum = 8" % l1)
print(sum_pairs(l2, -6) == [0, -6]) #"Negatives: %s should return [0, -6] for sum = -6" % l2)
print(sum_pairs(l3, -7) == None) #"No Match: %s should return None for sum = -7" % l3)
print(sum_pairs(l4, 2) == [1, 1]) #"First Match From Left: %s should return [1, 1] for sum = 2 " % l4)
print(sum_pairs(l5, 10) == [3, 7]) #"First Match From Left REDUX!: %s should return [3, 7] for sum = 10 " % l5)
print(sum_pairs(l6, 8) == [4, 4]) #"Duplicates: %s should return [4, 4] for sum = 8" % l6)
print(sum_pairs(l7, 0) == [0, 0]) #"Zeroes: %s should return [0, 0] for sum = 0" % l7)
print(sum_pairs(l8, 10) == [13, -3]) #"Subtraction: %s should return [13, -3] for sum = 10" % l8)

They should all return True

The Attempt at a Solution

My first attempt is simple and straight forward:

def sum_pairs(ints, s):
    prev_list = []
    for num in ints:
        for test in prev_list:
            if num+test == s:
                return ([test, num])
        prev_list += [num]

My second attempt is the same structure however I store and organize previous test numbers by either positive or negative:

def sum_pairs(ints, s):
    neg_list = []
    pos_list = []
    for num in ints:
        if num > s:
            for neg in neg_list:
                if neg == s - num:
                    return [neg,num]
        else:
            for pos in pos_list:
                if pos == s - num:
                    return [pos,num]
        if num < 0:
            neg_list += [num]
        else:
            pos_list += [num]

With this new method, I went from 50 total possible returns to 31. I am not sure if this is the right direction to go. It very well may be I'm doing more work by organizing than I am saving.

From here, my plan was to make four lists: two negatives with one where abs(number) >= abs(s) and the other abs(number) < abs(s) with a similar pattern for positive numbers. I think it might be helpful to find the sign of 's' at the very beginning but I haven't figured out how to use that to my advantage..

 

[willem2]

You just need to look at all the numbers once, and store them is such a way while doing this, you can immediately look up if the new number forms a pair with any number that you have already seen.
This can be done in a lot of ways: keeping the list sorted, storing the list as a hashtable or tree, or use an array of all the possible values that can occur.

 

[Jamison Lahman]

You just need to look at all the numbers once, and store them is such a way while doing this, you can immediately look up if the new number forms a pair with any number that you have already seen.
This can be done in a lot of ways: keeping the list sorted, storing the list as a hashtable or tree, or use an array of all the possible values that can occur.

Ahhh I think I was trying to do this intuitively but lacked the programming experience. I think I will try to sort the list first:

if prev_num not in prev_list:
    prev_list += [prev_num]
    prev_list = sorted(prev_list)

After that, I will be able search the list backwards or forwards until (s-(num+test)) changes signs.

If that doesn't do the trick, I will try hashtables. The book I used to learn python briefly mentioned how fast hashtables were but didn't go over them in-depth.
Thanks