I need to write a method that will accept a timestamp and takes an array of votes, I need to return the k leading candidates with that timestamp. I came up with the following solution,

Here is the input to the methods:

votes = [{'candidate':'a', 'timestamp':2},{'candidate':'c', 'timestamp': 5},{'candidate':'c', 'timestamp': 12}]

timestamp = 5

k = 5

And the method to solve the problem,

def leading_candidates(votes, timestamp,k):

    candidates = {}
    leading_candidates = []

    for vote in votes:
        if vote['timestamp'] <= timestamp:
            if vote['candidate'] not in candidates:
                candidates[vote['candidate']] = 1
                candidates[vote['candidate']] += 1

    sorted_votes = sorted(candidates.values())[:k]

    for candidate in candidates:
        if candidates[candidate] in sorted_votes:

    return leading_candidates    

print(leading_candidates(votes, timestamp, 2))

As you can see the second solution has a time complexity of \$O(k\,n)\$ where k is the time it takes to find the index in the leading candidates sorted array, In the worst case, it can be \$O(n^2)\$ and because of sorting it may be at least \$O(n\,\log n)\$.

Is there any way we can make it work with \$O(n)\$?

  • \$\begingroup\$ You already know that everything in sorted_votes is in candidates. Why not "for sorted_votes" k times instead of "for candidates"? \$\endgroup\$ – Frank Merrow Feb 12 '20 at 3:37
  • \$\begingroup\$ @FrankMerrow Yes, but how efficient it is to find the key of particular value in the hash table? \$\endgroup\$ – kgangadhar Feb 12 '20 at 4:47
  • \$\begingroup\$ Is there any other thing I can do to change the time taken. \$\endgroup\$ – kgangadhar Feb 12 '20 at 4:48
  • \$\begingroup\$ Welcome to CodeReview@SE. What is timestamp duration? \$\endgroup\$ – greybeard Feb 12 '20 at 7:18
  • 2
    \$\begingroup\$ (I do not want to know one/the value to use: I have no idea how to interpret it just from the problem description. I think one timestamp to mark one point in time, a duration could be specified by two timestamps. Then, there is before and after.) \$\endgroup\$ – greybeard Feb 12 '20 at 8:42

So, you want to count something and afterwards get the top k? That sounds like a job for collections.Counter!

from collections import Counter

def leading_candidates(votes, timestamp, k):
    vote_counts = Counter(vote['candidate']
                          for vote in votes
                          if vote['timestamp'] <= timestamp)
    return [candidate[0] for candidate in vote_counts.most_common(k)]

if __name__ == "__main__":
    print(leading_candidates(votes, timestamp, 2))

This way you don't need to special case a candidate not yet having received a vote (something you could have also done with a collections.defaultdict(int)). And it is \$\mathcal{O}(n)\$.

Also note that if k is large, the line if candidates[candidate] in sorted_votes will become slow, as it is a linear scan. At the same time, you can iterate over the keys and values of a dictionary at the same time with candidates.items(), so you don't need to do candidates[candidate].

Python has an official style-guide, PEP8, which recommends using spaces after commas, which you forgot to do before k in the function signature.

You should always guard your code with an if __name__ == "__main__": guard to allow importing from the script without running it.

  • \$\begingroup\$ Counter.most_common() uses a heap so it is O(k log h), where h is the total number of candidates (e.g., the size of the heap). Of course for most uses, neither k nor h are large. \$\endgroup\$ – RootTwo Feb 12 '20 at 23:18
  • \$\begingroup\$ @RootTwo True. And log h < n and k is constant. But filling the Counter is still O(n). \$\endgroup\$ – Graipher Feb 12 '20 at 23:21
  • \$\begingroup\$ @RootTwo Yup, Counter.most_common() uses a heap; more specifically, it uses heapq.nlargest(). Which according to stackoverflow.com/a/23038826 and stackoverflow.com/a/33644135 it actually has a time complexity of O(h log k), where h is the total number of candidates. \$\endgroup\$ – Setris Feb 13 '20 at 1:42

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