Lately I’ve been looking for materials to practice for code interviews and a friend of mine pointed out Codility’s lessons to me. Their web site and application are still a bit in the early stages (imho they are not as mature as other sites like HackerRank or CodinGame) but one really nice thing that the others lack is the reading material they provide for each topic.

I plan to go through the lessons (at least 1 to 17) in the following days/weeks and I’ll write random thoughts that arise while working on them here in the blog.

I want to do it mainly to keep myself engaged and force myself to ‘distil’ a bit of what I learn in the process. Also, I want to try to be less perfectionist with my blog writing and see what happens: if I write so seldom it’s mainly because I know how long it requires and I end not writing at all.

Codility lessons

3. Time Complexity

The reading is interesting, particularly the part regarding estimates about the upper limit of the time complexity a solution should have.

Key takeaway

• An average computer can perform something like ${10}^8$ operations/s.
• The time limit for test cases is usually 1 to 10 seconds.

With both this information and those of the test (usually you are given the size of input data) you should be able to estimate the upper limit of time complexity that your solution should have.

4. Counting Elements

Ok, here I feel a bit like cheating by using Python’s Counter from the standard library’s collection module.

Exercise 4.1 of the reading material

Sometimes math allows you to literally kill algorithms’ time complexity.

Let $S_A=\sum _{i=0}^{n-1}{a}_i$ and $S_B=\sum _{i=0}^{n-1}{b}_i$.

Denote as $a_i$ and $b_j$ the elements of A and B we need (eventually) to swap.

If such elements exist that solve the problem, then it is true that: $$\begin{array}{c}{S}_A-a_i+b_j=S_B+a_i-b_j\\ \Rightarrow 2a_i-2b_j=S_A-S_B\\ \Rightarrow a_i=b_j+(S_A-S_B)/2\end{array}$$

This means that our problem admits a solution if and only if $S_A-S_B$ is divisible by 2 and if the above equality holds for some i and j.

This can be verified with an algorithm of complexity $O(n+m)$.

Solution in Python:

def solution(A, B, m):
    Sa = sum(A)
    Sb = sum(B)
    if (Sa - Sb) % 2:
        return False
    values_A = set(A)
    for v in B:
        if (Sa - Sb) // 2 + v in values_A:
            return True
    return False

5. Prefix Sums

This is a totally new concept/technique to me. It looks really powerful to avoid complexity $O(n^2)$ or $O(n\cdot m)$ and get $O(n)$ or $O(n+m)$ instead.

The idea is simple but I still need to acquire the skill to recognize problems where this technique can be used.

Question MinAvgTwoSlice

This one was hard. I wasn’t able to solve it by myself with a complexity less than $O(n^2)$ without reading “There must be some slices, with length of two or three, having the minimal average value among all the slices” in this blog post. Pretty obvious in retrospect, but I was nowhere near to getting there by myself.

Solution in Python:

def solution(A):
    N = len(A)

    partial_sums = [0] * (N + 1)
    for i in range(0, N):
        partial_sums[i + 1] = partial_sums[i] + A[i]

    averages = []
    for l in [2, 3]:
        for p in range(N + 1 - l):
            q = p + l - 1
            averages.append([p, q, (partial_sums[q + 1] - partial_sums[p]) / l])

    return min(averages, key=lambda x: x[2])[0]

That’s all for the first part of this series, I’ll add other posts as I go forward with the lessons.