Question

What’s the runtime for this nested for loop in big O notation?

for(i = 1 to k)
{
    for(j = i+1 to k)
    {}
}

It’s smaller than O(k^2) but I can’t figure it out.

Answer 1

Your question is closely related to the series sum S(k) = 0 + 1 + 2 + … + (k-2) + (k-1). It can be shown that S(k) = (k*(k-1))/2 = (k*k)/2 – k/2. [How? Reorder the sum as S(k) = {0+(k-1)} + {1+(k-2)} + {2+(k-3)} + …. This shows how.]

Therefore, is the algorithmic order smaller than O(k*k)? Remember that constant coefficients like 1/2 do not influence the big O notation.

Question: So it’s equivalent to replacing j = i+1 to k with j = 1 to k?

Answer: Right. This is tricky, so let’s think it through. For i == 1, how many times does the inner loop’s action run? Answer: it runs k-1 times. Again, for i == 2, how many times does the inner loop’s action run? Answer: it runs k-2 times. Ultimately, for i == k, how many times does the inner loop’s action run? Answer: it runs zero times. Therefore, over all values of i, how many times does the inner loop’s action run? Answer: (k-1) + (k-2) + ... + 0, which is just the aforementioned sum S(k).