# \(K\)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.