Codility and the “K complementary pairs in array” challenge

Recently I had to confront myself with Codility, for those who don’t know it Codility is an online tool that provides coding challenges and it is often used by companies in the process of hiring new people when they want to test people’s coding skills.

But that’s not all, Codility also provides a great training section where you can confront yourself with a ton of coding challenges and improve your understanding of algorithms and your problem solving skills.

Yes, someone was testing my coding skills.

One of the tests I had to solve involved the concept on “K complementary pairs” in an array, basically what I was requested to do was something like this:

Given an integer array A find all the complementary pairs in A that sums up to a provided value K, so for example given the array [2, 5, -1, 6, 10, -2] and an integer K = 4 the result must be 5 because the input array contains [2, 2] [5, -1] [-1, 5] [6, -2] [-2, 6]

The solution is expected to run with a total time complexity of
O(N logN)

First approach that comes to mind, let’s write two nested for loops, sum up each element with each other and count how many times we get the input value K:

This works! Oh wait… it is not an O(N logN), this approach runs in quadratic time or something like that… shit.
This is the solution that I submitted however, because I had other tests to do and a limited amount of time so since I could not think of a better solution at the moment I just sent this one, better slower than nothing at all.

But then a few hours later I had the right idea, and I came up with an algorithm that solves the same problem in linear time.
The idea is the following:

Let’s use an HashMap to store the occurrences of each item in the input array, so each entry will be (key=item, value=occurrencies).

Now traverse the HashMap‘s keySet() and using the input value K find out which value we need in order to produce K starting from the current item.

At this point just multiply the occurrences of the current key by the number of the occurrences of the required value.

And that’s it, here’s the code.

Looking for an awesome book on algorithms? Have a look at Algorithms (4th Edition), it’s a great book, it covers a lot of topics from sorting to graphs to strings processing algorithms and can teach you a lot on the subject!

Unique opportunity! Help a fellow grow his blog!

Hi there! If you’ve read this far maybe you think this was useful, or fun, or I don’t know what but for some reason You Got Here! Great! Please consider sharing this post with your network, I am trying to get The Code Butchery to grow so I can provide more content like this, will you help me in my journey? Thank you!

Share this


  1. Hi Francesco,
    Hash map search operation has O(n) worst case comlexity. So your implementation has still O(n^2) complexity (worst case).

      1. If your hashing function and spread is good it won’t have o(n) complexity it will have o(1) complexity.

  2. I also got the same codility challenge a couple of days ago (and also couldn’t do better than the O(n^2) solution during the assigned time 🙁 ). After the test, I thought of a solution that is always O(n*log(n)) even in worst case.
    But the test also allowed to use O(n) space, which you are already using in your hash table. The space restriction can also be a factor for your hash table efficiency to deteriorate.
    So my solution is basically to sort the input array into a look-up table. This step is O(n) space and O(n*log(n)) time using qsort. Then, for each element in the original array, look-up K-arr[i] in the table using binary search. There is a modified binary search algorithm that works with duplicates also in O(log(n)) time. Search it on Google. So for all the array elements, this step is O(n*log(n)) time.
    Total time is O(n*log(n)) best, average and worst cases. No more no less!

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.