# $K$th order statistic in $O(N)$¶

Given an array $A$ of size $N$ and a number $K$. The problem is to find $K$-th largest number in the array, i.e., $K$-th order statistic.

The basic idea - to use the idea of quick sort algorithm. Actually, the algorithm is simple, it is more difficult to prove that it runs in an average of $O(N)$, in contrast to the quick sort.

## Implementation (not recursive)¶

```
template <class T>
T order_statistics (std::vector<T> a, unsigned n, unsigned k)
{
using std::swap;
for (unsigned l=1, r=n; ; )
{
if (r <= l+1)
{
// the current part size is either 1 or 2, so it is easy to find the answer
if (r == l+1 && a[r] < a[l])
swap (a[l], a[r]);
return a[k];
}
// ordering a[l], a[l+1], a[r]
unsigned mid = (l + r) >> 1;
swap (a[mid], a[l+1]);
if (a[l] > a[r])
swap (a[l], a[r]);
if (a[l+1] > a[r])
swap (a[l+1], a[r]);
if (a[l] > a[l+1])
swap (a[l], a[l+1]);
// performing division
// barrier is a[l + 1], i.e. median among a[l], a[l + 1], a[r]
unsigned
i = l+1,
j = r;
const T
cur = a[l+1];
for (;;)
{
while (a[++i] < cur) ;
while (a[--j] > cur) ;
if (i > j)
break;
swap (a[i], a[j]);
}
// inserting the barrier
a[l+1] = a[j];
a[j] = cur;
// we continue to work in that part, which must contain the required element
if (j >= k)
r = j-1;
if (j <= k)
l = i;
}
}
```

## Notes¶

- The randomized algorithm above is named quickselect. You should do random shuffle on $A$ before calling it or use a random element as a barrier for it to run properly. There are also deterministic algorithms that solve the specified problem in linear time, such as median of medians.
- A deterministic linear solution is implemented in C++ standard library as std::nth_element.
- Finding $K$ smallest elements can be reduced to finding $K$-th element with a linear overhead, as they're exactly the elements that are smaller than $K$-th.