Priority QueuesPriority queueA priority queue ADTEvaluating implementationsArray implementationsLinked list implementationsBinary tree implementationsHeap implementationArray representation of a heapUsing the heapThe EndPriority QueuesPriority queue•A stack is first in, last out•A queue is first in, first out•A priority queue is least-first-out–The “smallest” element is the first one removed–The definition of “smallest” is up to the programmer (for example, you might define it by implementing Comparator or Comparable)–If there are several “smallest” elements, the implementer must decide which to remove first•Remove any “smallest” element (don’t care which)•Remove the first one addedA priority queue ADT•Here is one possible ADT:–PriorityQueue(): a constructor–void add(Comparable o): inserts o into the priority queue–Comparable removeLeast(): removes and returns the least element–Comparable getLeast(): returns (but does not remove) the least element–boolean isEmpty(): returns true iff empty–int size(): returns the number of elements–void clear(): discards all elementsEvaluating implementations•When we choose a data structure, it is important to look at usage patterns–If we load an array once and do thousands of searches on it, we want to make searching fast—so we would probably sort the array–If we load a huge array and expect to do only a few searches, we probably don’t want to spend time sorting the array•For almost all uses of a queue (including a priority queue), we eventually remove everything that we add•Hence, when we analyze a priority queue, neither “add” nor “remove” is more important—we need to look at the timing for “add + remove”Array implementations•A priority queue could be implemented as an unsorted array (with a count of elements)–Adding an element would take O(1) time (why?)–Removing an element would take O(n) time (why?)–Hence, adding and removing an element takes O(n) time–This is a very inefficient representation•A priority queue could be implemented as a sorted array (again, with a count of elements)–Adding an element would take O(n) time (why?)–Removing an element would take O(1) time (why?)–Hence, adding and removing an element takes O(n) time–Again, this is very inefficientLinked list implementations•A priority queue could be implemented as an unsorted linked list–Adding an element would take O(1) time (why?)–Removing an element would take O(n) time (why?)•A priority queue could be implemented as a sorted linked list–Adding an element would take O(n) time (why?)–Removing an element would take O(1) time (why?)•As with array representations, adding and removing an element takes O(n) time–Again, this is very inefficientBinary tree implementations•A priority queue could be represented as a (not necessarily balanced) binary search tree–Insertion times would range from O(log n) to O(n) (why?)–Removal times would range from O(log n) to O(n) (why?)•A priority queue could be represented as a balanced binary search tree–Insertion and removal could destroy the balance–We need an algorithm to rebalance the binary tree–Good rebalancing algorithms require only O(log n) time, but are complicatedHeap implementation•A priority queue can be implemented as a heap•In order to do this, we have to define the heap property–In Heapsort, a node has the heap property if it is at least as large as its children–For a priority queue, we will define a node to have the heap property if it is as least as small as its children (since we are using smaller numbers to represent higher priorities)128 3Heapsort: Blue node has the heap property38 12Priority queue: Blue node has the heap propertyArray representation of a heap•Left child of node i is 2*i + 1, right child is 2*i + 2–Unless the computation yields a value larger than lastIndex, in which case there is no such child•Parent of node i is (i – 1)/2–Unless i == 01214186833 12 6 18 14 8 0 1 2 3 4 5 6 7 8 9 10 11 12lastIndex = 5Using the heap•To add an element:–Increase lastIndex and put the new value there–Reheap the newly added node•This is called up-heap bubbling or percolating up•Up-heap bubbling requires O(log n) time•To remove an element:–Remove the element at location 0–Move the element at location lastIndex to location 0, and decrement lastIndex –Reheap the new root node (the one now at location 0)•This is called down-heap bubbling or percolating down•Down-heap bubbling requires O(log n) time•Thus, it requires O(log n) time to add and remove an elementThe