program speed/unnecessary double-counting

Question

I am trying to find the number of pairs in a list of numbers with a specific difference. Say, with the list

1 2 3 4 5

and the difference target '2', I would want to print the number '3' because there are 3 pairs in this sequence with a difference of '2'. however, my code is super slow - it double-counts all of the pairs, and so I end up needing to divide my solutions by 2 to get the answer. Is there a way to accomplish this same task without double-counting? I appreciate any insights you might have. thanks! code is printed below

    import sys


    def main():
        solutions=0
        pairs=[]
        for i in xrange(len(numbers)):
            for j in xrange(len(numbers)):
                if i!=j:
                    pairs.append([numbers[i], numbers[j]])

        for pair in pairs:
            if abs(pair[0]-pair[1])==k:
                solutions+=1
            else:
                continue
        return solutions/2



    if __name__ == '__main__':
        lines=sys.stdin.readlines()
        n,k=map(int, lines[0].strip().split())
        numbers=map(int, lines[1].strip().split())
        print main()

DSM · Accepted Answer · 2013-01-12T06:11:42.137

For each element i in a, you want to check whether i-diff is also in a. For ~O(1) membership testing, we can use a set. Thus:

>>> a = [1,2,3,4,5]
>>> diff = 2
>>> a_set = set(a)
>>> sum(i-diff in a_set for i in a_set)
3

which is O(len(a)).

[Note that I've used the fact that i-diff in a_set, which is a bool, evaluates to 1 as an int. This is equivalent to sum(1 for i in a_set if i-diff in a_set).]

Update: it occurs to me that I've assumed that the numbers are unique. If they're not, that's okay, we could just use a collections.Counter instead to keep the multiplicity information.

score 0 · Answer 2 · answered Jan 12 '13 at 05:50

0

If you sort the array, you would be able to find all the pairs just by walking the array instead of doing an O(n^2) search. And the reason for the double counting is that you used abs so it's finding not just (1,3) but also (3,1).

answered Jan 12 '13 at 05:50

StilesCrisis

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Oh? How do you propose doing it exactly? – StilesCrisis Jan 12 '13 at 15:20
1

Fair enough, but you do need to construct an entirely separate hash table copy, so there's a memory cost to think of. – StilesCrisis Jan 14 '13 at 05:02

Ashwini Chaudhary · Answer 3 · 2013-01-12T06:17:49.077

0

Sort the array first, and then for each number(num) in the list you need to look for num-2. I guess the the fast way to do that is by binary search.

So, with binary search you'll get a O(n log(n)) solution.

edited Jan 12 '13 at 06:17

answered Jan 12 '13 at 05:52

Ashwini Chaudhary

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This is still O(n**2). If you use a `set` instead of a `list` it would be substantially faster. – Tim Jan 12 '13 at 05:53
@Tim agreed, but I also suggested him a `binary search` based solution. – Ashwini Chaudhary Jan 12 '13 at 06:15

program speed/unnecessary double-counting

3 Answers3