Following the dense ranking convention, we want to get back a list containing the history of rank positions for Alice.

This is a HackerRank Algorithm Implementation problem, and I am going to show you how I solved it, using Python as implementation language.

I noticed that the first list, scores, is already sorted, we just have to get rid of duplicates to have a matching between the position and the score Alice has to get to achieve that ranking.

**Bisect solution**

I just have to do the matching. First idea jumped to my mind, was performing a binary search on scores to do it. It helps that Python provides for the job a well known library, bisect. There's just a tiny issue, bisect expects the underlying list to be sorted in natural order, so we need to reverse our scores.

It looks promising, let's implement it.

A pythonic way to get our ranking would be this:

ranking = sorted(list(set(scores)))I get the original list, convert to a set to get rid of duplicates, than back to list, so that I can sort it in natural order. Nice, but in this problem we are kind of interested in performance, since we could have up to 20 thousand items in both lists. So we want to take advantage of the fact that the list is already sorted.

So, I ended up using this rather uncool code:

ranking = [scores[-1]] for i in range(len(scores)-2, -1, -1): if scores[i] > ranking[-1]: ranking.append(scores[i])I initialize the ranking list with the last item in scores, then I range on all the other indices in the list from right to left. If the current item is bigger than the latest one pushed in ranking, I push it too.

Now I can just rely on the bisect() function in the bisect python module, that would find which position the current item should be inserted in the list. With a caveat, I have reverted the order, so I have to adjust the bisect() result to get the result I'm looking for:

results = [] last = len(ranking) + 1 for score in alice: results.append(last - bisect(ranking, score))This code pass all the tests, job done.

However. Do I really need to pay for the bisect() search for each element of alice?

**Double scan solution**

Well, actually, we don't. Since we know that both list are sorted, we can use also the ordering in alice to move linearly in ranking.

Since we are not using anymore bisect, we don't need to revert the sorting order in ranking, and the duplicate cleanup is getting a bit simpler:

ranking = [scores[0]] for score in scores[1:]: if score < ranking[-1]: ranking.append(score)

Now we compare each item in alice against the items in ranking moving linearly from bottom to head:

results = [] for score in alice: while ranking and score >= ranking[-1]: ranking.pop() results.append(len(ranking) + 1)We don't have to be alarmed by the nested loops, they don't have a multiplicative effect on the time complexity, since we always move forward on both lists, the result is a O(M + N) time complexity.

Is this a better solution than the first one? Well, it depends. We should know more on the expected input. However, for large and close values of N and M, it looks so.

I pushed the python script for both solutions and a test case to GitHub.

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