# Counting big enough elements

Let $$v$$ be a vector of positive integers of length $$n$$. I want to find the highest $$k$$ such as there are at least $$k$$ elements of $$v$$ that are greater or equal to $$k$$.

What would be the best algorithm for finding $$k$$?

Examples:

• $$v = (1, 2, 3, \ldots, 100) \rightarrow k = 50$$, because there are at least 50 elements that are greater or equal to 50 (there are even 51 of those, fifty through a hundred, but that doesn’t matter); but there are only 50 (so—less than 51) elements greater or equal to 51.
• $$v = (100, 101, 102, 103, 104) \rightarrow k = 5$$, because all five elements are greater or equal to five, and there cannot possibly be six elements greater or equal to six.
• $$v = (1, 1, 1, 100, 100, 100) \rightarrow k = 3$$

Any ideas how to approach this?

My initial idea is this:

• Sort the vector in descending order in $$O(n \log n)$$
• Starting with the median (it seems, intuitively, as a good place to start), check if a number on a set position $$i$$ is greater or equal than $$i$$—since everything to the left of it is greater to equal to it (the vector is sorted), we know there are at least that many elements that are greater or equal to $$v[i]$$.
• Go right or left accordingly to what was found—if we found that there are not enough values greater to $$v[i]$$, let’s try with the smaller-or-equal number, $$v[i+1]$$. If it was big enough, go the opposite direction, checking $$v[i-1]$$.
• Our work is done when there is some $$i_m$$ for which the requirement holds, but it doesn’t hold for $$i_{m} + 1$$. That would require at most $$O(n)$$ checks.
• Thus, the slowest part is the sort itself, making the algorithm $$O(n \log n)$$ total.

I’m curious if this is the best we can do, or can we possibly make it even faster.

• What's the best algorithm you've been able to come up with so far? Are you able to think of any algorithm at all? – D.W. Dec 30 '19 at 18:21
• I thought about sorting the vector (so the later lookup can be done faster), finding the median and seeing if that works or not and going from there, but I cannot think of any reason why it should necessarily end quickly and if not, I’m going to do $O(n)$ lookups each taking $O(\log n)$. – Althorion Dec 30 '19 at 18:50
• Cool. I suggest editing your question to include your best algorithm and its overall running time, so we know what we need to beat to do better. Do you have any reason to expect anything better? What's the context in which you encountered this problem? And welcome to CS.SE, by the way! – D.W. Dec 30 '19 at 18:55
• Thanks! It came up as a part of statistical analysis of some data, and I was just curious if I can figure out some better solution for that—I feel like it could be done linearly, though it’s just a feeling. – Althorion Dec 30 '19 at 19:08

The running time satisfies the recurrence $$T(n)=T(n/2)+O(n)$$, since you can find the median and partition the array in $$O(n)$$ time. The solution to this recurrence is $$T(n)=O(n)$$, so the total running time is $$O(n)$$. This is faster than your $$O(n \log n)$$ time algorithm.