As usual before going into the details of efficient implementation let's check what is wrong with naive implementations? For example, what if we store the contents of a priority queue just in an unsorted array or in an unsorted list? In this example on the slide, we use a doubly linked list. Well, in this case inserting a new element is very easy. We just append the new element to the end of our array or list. For example, as follows, if our new element is seven we can just put it to the next available cell in our array where we can just append it to the end of the list. So we put 7 to the end. We say that the previous element of 7 is 2 and that there is no next element, right? So it is easy and it takes constant time, okay? Now, what about extracting the maximum element in this case? Well, unfortunately we need to scan the whole array to find the maximum element. And we need to scan the whole list to find the maximum element which gives us a linear running time. That is we O(n), right? In our previous naive implementation using an unsorted array or list, the running time of the extract max operation is linear. Well, a reasonable approach to try to improve this, is to keep the contents of our array, for example array, sorted. Well, what are the advantages of this approach? Well, of course, in this case, extract max is very easy. So, the maximum element, is just the last element of our array. Right? Which means that the running time of ExtractMax in this case is just constant. However, the disadvantage is that now the insertion operation takes linear time, and this is why. Well, to find the right position for the new element we can use the binary search. This is actually good, well it can be done in logarithmic time. For example, if we need to insert 7 in our priority queue, then in logarithmic time we will find out that it should be inserted between 3 and 9 in this for example. However unfortunately after finding this right position, we need to shift everything to the right of this position by one. Right just to create a vacant position for 7. For this we need to first shift 16 to this cell. Then we move 10 then to this cell, then we move 9 to this cell, and finally we put 7 in to this cell, and we get it sorted already. So in the worst case, we need to shift a linear number of cells, a linear number of elements, which gives us a linear running time for the insertion operation. As we've just seen, inserting an element into a sorted array is expensive because to insert an element into the middle we need to shift all elements to the right of this position by one. So, this makes the running time of the insertion procedure linear. However if we use a doubly linked list, then inserting into the middle of this list is actually constant time operation. So let's try to use a sorted list. Well, the first advantage is that the extract max operation still takes constant time. Well this is just because, well, the maximum element in our list is just the last element, right? So for this reason, we have a constant time for extract max. Also, another advantage is that inserting in the middle of this list actually takes a constant amount of work, not linear, and this is why. Again let's try to insert 7 into our list. Well, this can be done as follows. We know that inserting 7 should be done between 3 and 9. So we do just the following. We will remove this. We remove this pointer and this pointer. We'll say that now the next element after 3 is 7 and the previous element before 7 is 3. And also the next element after 3, after 7 is 9, and previous element before 9 is 7. So inserting an element just involves changing four pointers. Right? So it is a constant time operation. However everything is not so easy, unfortunately. And this is because just finding the right position for inserting this new element takes a linear amount of work. And this in particular because we cannot use binary search for lists. Given the first element of this list and the last element of this list, we cannot find the position of the middle element of this list because this is not an arry. We cannot just compute the middle index in this array. So for this reason, just finding the right position for the new element I mean to keep the list sorted takes already a linear amount of work. And for this reason, inserting into a sorted list still takes a linear amount of work. Well to conclude if you implement a priority queue using a list or an array sorted or not then one of the operations insert and extract max takes a linear amount of work. In the next video we will show a data structure called binary heap which allows to implement a priority queue so that both of these operations can be performed in logarithmic amount of work.
