**[Answer to the problem in Top k.]**

This was the problem:

Given a list of

nnumbers, get the topkones in O(n) time. (Partial credit for o(nlogn) solutions.)

*n* log *k* solution

The easiest way to solve this problem in o(*n* log *n*) time is to scan the *n* elements, keeping track of the greatest *k* elements seen up to the current element. Using a heap to store the top *k* elements, we get the following algorithm:

#define SWAP(type, a, b) {type swap_temp = a; a = b; b = swap_temp;} static void minheap_pushdown(int *h, size_t hs, size_t i) { size_t j = 0; if (2 * i + 2 < hs) j = (h[2 * i + 1] < h[2 * i + 2]) ? 2 * i + 1 : 2 * i + 2; else if (2 * i + 1 < hs) j = 2 * i + 1; if (j != 0 && h[j] < h[i]) { SWAP(int, h[i], h[j]); minheap_pushdown(h, hs, j); } } static void minheap_raise(int *h, size_t i) { if (i == 0) return; if (h[i] < h[(i - 1) / 2]) { SWAP(int, h[i], h[(i - 1) / 2]); minheap_raise(h, (i - 1) / 2); } } void top_k_nlogk_cp(int *top_k, size_t k, const int *l, size_t n) { size_t i; for (i = 0; i < k; i++) { top_k[i] = l[i]; minheap_raise(top_k, i); } for (i = k; i < n; i++) { if (l[i] > top_k[0]) { top_k[0] = l[i]; minheap_pushdown(top_k, k, 0); } } }

As *minheap_pushdown()* and *minheap_raise()* have O(log *k*) cost, the total cost of this solution will be O(*n* log *k*). This is better than O(*n* log *n*) if *k* grows slower than any power of *n* (if *k* is constant or O(log *n*), for example).

The previous solution is enough to get partial credit, but we can do better ðŸ˜€

##### Linear time solution

Quicksort is based on a linear time partition operation. This operation starts with an array and an element of this array, called pivot, and ends with a partially sorted array, having the pivot in his sorted position, smaller elements before it and bigger ones after it. Then doing a single partition operation would suffice… if we could guarantee that the chosen pivot is the (*n*–*k*+1)-th element of the sorted array ðŸ˜€

We cannot do that, but we can do something that is asymptotically equivalent: doing a quicksort-style recursion, though only in the partition that contains the desired pivot. The result is this (converting the tail recursion to iteration):

int *top_k_n_ip(int *l, size_t n, size_t k) { size_t lo, hi, pos, i, j; int pivot; lo = 0; hi = n; pos = n - k; while (hi - lo > 1) { i = lo + 1; j = hi - 1; pivot = l[lo]; while (1) { while (i < hi && l[i] <= pivot) i++; while (j > lo && l[j] > pivot) j--; if (i > j) break; SWAP(int, l[i], l[j]); } SWAP(int, l[lo], l[j]); if (j < pos) lo = j + 1; else if (j > pos) hi = j; else break; } return l + n - k; }

Now we need to check that this gives us a linear-time algorithm. The execution time of the body of the `while (1)`

loop is bounded from above by *C* times *hi* – *lo*, where *C* is a constant. If we assume that the chosen pivot divides the *l* [ *lo* .. *hi* ] array section in two roughly equal parts (this assumption can be made rigorous by choosing the pivot carefully), we get

where *A*, *B* and *C* are constants.

Summing all the terms containing *B* and taking out all the *C* factors we get:

As we know that any correct algorithm must employ time (it needs to inspect all elements), we also know that this algorithm has time complexity. Now let’s check that against reality ðŸ˜€

##### Benchmarks

We will benchmark the following algorithm implementations:

it just sorts a copy of the input array and copies the top*n*log*n*(copying):*k*elements.scans the input array, keeping the top*n*log*k*(copying):*k*elements in a heap.applies the O(*n*(copying):*n*) in-place algorithm over a copy of the input array.sorts the input array in-place and returns a pointer to the top*n*log*n*(in-place):*k*elements.applies the O(*n*(in-place):*n*) quicksort-derived algorithm that was described in the previous section.Miles Macklin‘s solution, just calling*n*(in-place, C++):*std::nth_element()*ðŸ™‚

As *k* has a fixed value in the previous plot, the O(*n* log *k*) algorithm is effectively an O(*n*) algorithm. In fact, it performs better than the truly linear algorithms, probably due to cache effects.

Another thing that can be seen in these plots is the difficulty of seeing the difference between *n* and *n* log *n* in a log-log plot, but that is something to be expected.

Here the qualitative behavior of the O(*n* log *k*) solution is the expected one. The C implementation of the O(*n*) algorithm looks surprisingly good, with a performance competitive to the C++ implementation included as part of the standard library.