The following function sum_pairs(ints, s) takes two inputs: ints is a list with integer values and s is a integer number. The function sum_pairs returns the values (x, y) in ints such that \$s = x + y\$ and the index of y is the lowest for the (x, y) in ints such that \$s = x + y\$. This is a problem taken from codewars. The function passes all the test cases. However, when I attempt to run the code for a big list, the server codewars gives me the following answer: "Process was terminated. It took longer than 12000ms to complete".

I am wondering whether the code I have written to solve the problem can be optimised or should I use a different approach to give a solution.

def sum_pairs(ints, s):

    begin = 1
    end = len(ints)
    memo = []

    while begin < end:
            memo.append([begin-1, ints.index(s - ints[begin-1], begin, end)])
            begin += 1
            end = memo[-1][1]
            begin +=1

    if memo == []:
        return None
        return [ints[memo[-1][0]], ints[memo[-1][1]]]

The function sum_pairs(ints, s) searches for \$y = s - x\$ in the list ints using the function index(). If a value of y is found the index for the pair (x,y) is added to memo, then the range for the search (begin, end) are updated. Else only the lower limit begin for the search is updated.

  • \$\begingroup\$ sort and you should be able to do in in O(n logn) \$\endgroup\$ – paparazzo Aug 8 '17 at 14:58

Your code takes too long, because to check if a number is in array you use .index, effectively iterating over the whole array. However, since you need lots of these lookups, you can speed them up with more suitable data structure - set will do:

def sum_pairs_lookup(ints, s):
  waiting_for_pair = set()
  for y in ints:
    if (s - y) in waiting_for_pair:
      return [s - y, y]

Also instead of looking for corresponding y, you can look for corresponding x, starting from the array's beginning. In this case the first match would give you the result.

Run it online

Edit: changed to accomodate points raised in comments, current version passes tests

  • \$\begingroup\$ @ Daerdemandt Beautiful ! it works like a charm, as you have said. My first approach didn't perform very well not only because the data structure I used. I think the main problem was the O(^2) algorithm involved. If you need to look for a particular element on a data structure with a big number of elements do you think always it's a good idea to transform the problem in a set problem ? \$\endgroup\$ – Masaguaro Aug 10 '17 at 14:28
  • \$\begingroup\$ @ Daerdemandt is there any faster that a in set() \$\endgroup\$ – Masaguaro Aug 10 '17 at 14:41
  • \$\begingroup\$ @Masaguaro I wouldn't say set or dict are always the best solution, but quite often they are. Under the hood they are hash-based, which gives us sweet O(1) lookups most of the time. More than that, you should expect any collaborator to be familiar with them, so I think they (and their common derivatives) should be the go-to solution. However, be aware of tradeoffs, sometimes, say, Bloom filter is what you need. \$\endgroup\$ – Daerdemandt Aug 10 '17 at 17:56

Your code's complexity is \$O(n^2)\$, this means that worst case it's on-par with looping through ints in a nested loop, and returning the maximum for x, or the minimum for y. And so it's on-par with:

def sum_pairs(ints, s):
        return max(
            (x, y)
            for x in ints
            for y in ints
            if x + y = s
    except ValueError:
        return None

You could speed this up, if ints is sorted, which your code implies, then you could change max to next, and flip the x and y. However, this still has \$O(n^2)\$ time.

The simplest way to force \$O(n)\$ time is to use a deque, and consume both ends of the sorted deque. This works by removing the tail of the deque to the point that \$x + y \le s\$. After this if \$x + y = s\$, then we return that. If not we take the next item at the head of the deque. Something like:

from collections import deque

def sum_pairs(ints, s):
    q = deque(ints)
        while True:
            y = q.popleft()
            n = s - y
            while q[-1] > n:
            if q[-1] == n:
                return q[-1], y
    except IndexError:
        return None
  • \$\begingroup\$ Is normal control flow with exceptions considered normal / good style in Python? \$\endgroup\$ – Daniel Jour Aug 8 '17 at 14:40
  • \$\begingroup\$ @DanielJour It's not bad style - "What is the EAFP principle in Python?". But they can be slow, however since we go to the except one, if there are no matches, I don't think the clean-up speed matters too much here. In the code above not using exceptions would be harder to read IMO. I think using LBYL would also be slower here. \$\endgroup\$ – Peilonrayz Aug 8 '17 at 14:47
  • \$\begingroup\$ @peilonrayz is good idea to use more efficient data structures. However, If your run your script with ints = [5, 3, 7, 7] and s = 10 the result will be None, but the right result for this case is [3,7] \$\endgroup\$ – Masaguaro Aug 9 '17 at 7:46
  • \$\begingroup\$ @Masaguaro That as it assumes sorted input - "consume both ends of the sorted deque." \$\endgroup\$ – Peilonrayz Aug 9 '17 at 8:04
  • \$\begingroup\$ @Peilonrayz it needs always to take (consume) the left hand side element x of the list ints to combine with the right hand side element y of the list such as y = s - x. As you can see in the example that I gave you, for the first element 5 there is not an element y on the right hand side of the list such as y = s -x. So we need to keep those elements of the list to combine with the second element 3, and so on. \$\endgroup\$ – Masaguaro Aug 9 '17 at 8:19

To expose more the logic, this is a list version code of the @Daerdemandt idea. Two lines shorter but not as fast as as it was required:

def sum_pairs(ints, s):
    for i in range(1,len(ints)):
        if (s - ints[i]) in ints[0:i]:
            return [s - ints[i], ints[i]]
  • \$\begingroup\$ Sorry @TobySpeight, perhaps i didn't do it right. It doesn't improve the solution. It's just another version using another data structure.... I can delete if it's necessary .. \$\endgroup\$ – Masaguaro Aug 10 '17 at 15:09
  • \$\begingroup\$ The reasoning seems clear to me. \$\endgroup\$ – janos Aug 10 '17 at 15:17

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