Last time we looked at the QuickSort and found that it was much better, in most cases, than the selection sort, but there was still a pathological case where the QuickSort wasn’t better. Can we guarantee we are better than the O(N2) of the selection sort? Can we guarantee O(N log N)? To do this we need to find something in QuickSort we can do better. The Achilles’ heal for the QuickSort is selecting the median value of the partition. Several techniques have been developed to avoid the worst case scenarios (like selecting three values and taking the median; in other words, median of three, or just picking a random item in the partition) but none of these guarantee O(N log N). We need a more radical approach.
We need a way of preserving the comparisons; much like QuickSort does, but that doesn’t use partitions. As is often the case, when you need a radical new approach, you should try a different data structure (much like choosing a different coordinate system in calculus). If we organize the array into a binary tree, we can find the largest element in the array in O(N log N) comparisons. This is because each element only needs to be compared with its parents, which is O(log N) comparisons, and we need to do this for every element in the in the array, O(N), giving us O(N log N). This sounds promising, but how do you organize an array into a binary tree? This is easier than it sounds. If you number an array from 1 to n, where each number is the index of the elements in the array, you can treat the array as a binary tree by defining the left node of any element i to be at index 2i and the right node to be at 2i+1 (or the other way around, if you must, but I like to think of the odd children as right). To create the desired tree, we need to ensure that for every element in the array at index i, it is the greater than its children. In other words, for all elements of array a, ai ≥ a2i for 2i ≤ n and ai ≥ a2i+1 for 2i+1 ≤ n. An array that obeys this constraint is called a heap (hence the name HeapSort).
Forming a heap is fairly straight forward and follows the informal description given above; you swap each element, from 2 to n, with its parent until you come across a parent that is greater than the element. The algorithm for this might look something like,
Note, that since C# numbers the array from 0 to n-1 instead of from 1 to n, we need to subtract add one to the index j and then subtract it again while calculating the parent’s index.
Now that we have the have the greatest element in the array, we know which element should be at the end of the array. We need only swap it with the last element of the array to put it there, but doing so will violate the heap constraint above since an is guaranteed to be less than or equal to both a2 and a3. To restore order back to the heap we must push the new first element down in the heap until it is greater than or equal to both of its children or it is childless. This can be done by selecting the greatest child and swapping positions with it then repeating until it is greater than the children or it has no children. Since we are following a single line of succession, an element will only have, at most log2n generations (or levels, as it is more traditional called) below it. This means we can restore order to the heap in O(log N) comparisons. This might look like,
We now need to do this n times to completely sort the array, meaning that we are guaranteed to sort a heap in O(N log N) comparisons. This gives us O(N log N) operations to form a heap and O(N log N) operations to sort a heap which means that we can sort any array, using a heap in O(N log N) (since big O notation throws away the constant 2 in 2N log N). Putting this together gives us the HeapSort invented by J.W.J. Williams,
Here we have it. An algorithm that is guaranteed to perform on the order of O(N log N) operations to sort an array! And here, you are apt to say, “Why doesn’t everyone use a HeapSort instead of a QuickSort if HeapSort is so much better?” Good question, I am glad you asked. There are several reasons why, but I will save that for another time.