CIS 121 Data Structures and Algorithms with Java Spring Stacks, Queues, and Heaps Monday, February 18 / Tuesday, February 19

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1 CIS Data Structures ad Algorithms with Java Sprig 09 Stacks, Queues, ad Heaps Moday, February 8 / Tuesday, February 9 Stacks ad Queues Recall the stack ad queue ADTs (abstract data types from lecture. Each is characterized by a specific way of removig elemets ad has a set of supported operatios. Stack Queue LIFO (last-i-first-out the most recet elemet that has bee added to the stack will be removed first. Supported operatios: push pop peek isempty size FIFO (first-i-first-out the least recet elemet that has bee added to the queue will be removed first. Supported operatios: equeue dequeue peek isempty size Implemetatio Details Stacks ad queues ca be implemeted uder the hood with almost ay data structure. I this course, we will implemet stacks ad queues usig expadable arrays. The rules we will use for icreasig or decreasig the size of a stack or queue s uderlyig array are as follows:. If the array of size is full, create a ew array of size, ad copy all elemets ito the ew array.. If the array of size has elemets i it, create a ew array of size, ad copy all elemets ito the ew array. Amortized Aalysis Amortized aalysis refers to fidig the time-averaged cost for a sequece of operatios. I other words, it is the time required to perform a sequece of operatios averaged over all the operatios performed. Sice amortized aalysis for the stack push operatio was covered i lecture, we are goig to take a closer look at the stack pop operatio. The worst case ruig time for a sigle pop operatio is O(, sice we may eed to resize the array ad copy the elemets ito it. Based o this ruig time, we might coclude that a tight boud for the worst case ruig time for pop operatios is O(, sice there are operatios ad each operatio takes worst case O( time; however, we ca fid a tighter boud through some careful aalysis. If we start from a full stack of size, what is the total cost of a sequece of pop operatios? Iitially, the array is of size ad cotais elemets. To make our aalysis simpler, let s immediately pop the first elemets. Each of these pops takes O( time. Now our array is of size but cotais oly elemets. cis/curret/lectures/08-stacks-ad-queues.pdf The aalysis for equeue ad dequeue is similar to that of push ad pop, respectively.

2 I accordace with our rules, we ca pop more elemets before resizig the array. Each of these pops takes O( time. Oce we have pop d those elemets (leavig us with elemets i our array, we must reduce the size of our array to, ad copy the remaiig elemets ito the ew array. Thus, the total cost for the first pop operatios is T ( = + ( + +. We ca apply idetical aalysis to the ew array of size that cotais elemets. We get ( = ( 8 pops for free, after which we resize the array to be of size = ad copy the remaiig ( = 8 elemets ( ito the smaller array. Thus, the total cost for the first 7 8 pop operatios is T ( 7 8 = ( Are you oticig a patter? Let s rewrite the expressio slightly ad cotiue to expad it: T ( = ( ( ( 0 + ( ( ( ( + ( + ( ( + + ( + ( ( + + ( + ( + ( We ca ow calculate the total cost of pop operatios: T ( + ( ( i + ( i + ( i i=0 = + = + = O( i=0 i (The first term i the summatio is the cost of the iitial pops, the secod term is the cost of allocatig a ew array, ad the third term is the cost of copyig the remaiig elemets ito the ew array. Thus, the amortized time complexity of a pop operatio is = O(, eve though the worst case time complexity of a sigle pop operatio is O(. Itroductio: Heaps A heap is a tree-like data structure that satisfies the heap-order property. Defiitio (Heap-Order Property. A tree has the heap-order property if for ay paret ode P with a child C, the key of P is ordered with respect to the child C. Commo examples of orderigs o a heap would be (max-heap or (mi-heap. For, the key i each ode i the heap T is greater tha or equal to the keys of all odes i its subtree.

3 A example biary max-heap. Note that the root cotais the maximum key. Notice that this defiitio immediately implies that the root must cotai either the maximum or the miimum of the orderig relatioship that we defie, sice the root is the paret/acestor of every other ode. Specializig this defiitio to keys that act like atural umbers, or keys that implemet Comparable, we have our classic mi-heap ad max-heap. This basic idea is really powerful, as the heap data structure maitais the maximum or miimum elemet wheever we add to it or remove from it. This meas that we ca retrieve the max/mi elemet quickly! Biary Heaps A biary heap is a biary tree, but with the heap-order property. A biary heap is most commoly implemeted by flatteig a tree i level order ito a array. It satisfies the followig property: Defiitio (Shape Property. A tree has the heap-shape property if the tree is a complete biary tree. That is, all levels of the tree are fully filled, except for possibly the last, where all odes are as far left as possible. With the shape property, we ca easily idex ito a biary heap, sice we will ot have to worry about gaps.

4 ull A max-heap visualized as both a tree ad a array. For a elemet at idex i of A, its left ad right childre ca be foud at idices i ad i + respectively. Coversely, a elemet at idex i has its paret at idex i/. This property holds true oly if the heap begis at idex of the array (or if the array is oe-idexed. Ruig time of Operatios Ruig times are give with respect to, where is the umber of elemets i the biary heap. isert(x, k: A elemet x with key x may be iserted i O(log time. fid-mi/max(: Fidig the mi/max of a biary heap takes O( time. extract-mi/max(: Removig the root ad restorig the mi/max heap property takes O(log time. decrease/icrease-key(x, k: Chagig the key of a elemet ca be doe i O(log time. Note that the Java implemetatio of a priority queue does ot support this operatio. Partial Orderig We say that the heap-order property iduces a partial order over its elemets. Ituitively, a partial order meas that ot every pair of elemets are related. Eve though we kow that 7 is less tha, whe we isert these umbers ito the heap, we caot determie which umber is greater solely by its positio i the heap. Compare this to isertig both elemets i a biary search tree, where we ca determie the order by examiig their relative positios. We say that the biary search tree establishes a total order. For some problems, it is eough to have just a partial orderig. For example, if you wat to get the k- largest elemets of a list relatively fast, you ca use a heap to achieve this. As you ve see with mergesort ad some implemetatios of quicksort, you ca get a stroger, total orderig at the cost of a larger ruig time [Ω( log ]. However, buildig a heap oly takes time liear i the umber of elemets. Therefore, we ca get the maximum/miimum i liear time ad the partial orderig! Buildig a (Max Heap I order to build a heap, we defie the followig subroutie: max-heapify. Uder the assumptio that the left ad right subtrees of the i th vertex are valid max heaps, max-heapify esures that the subtree rooted at i is also a valid max heap. The ruig time aalysis of max-heapify is left as a discussio topic. We ca the write:

5 fuctio Max-Heapify(A, i l left(i r right(i if l A.heapsize ad A[l] > A[i] the largest l else largest i if r A.heapsize ad A[r] > A[largest] the largest r if largest i the swap(a[i], A[largest] max-heapify(a, largest Oe of childre is larger. Swap ad recurse. fuctio Build-Max-Heap(A A.heapsize A.legth for i A.legth/ dowto do max-heapify(a, i The build-max-heap algorithm starts from the last iteral ode of the biary tree represetatio of A ad coverts each subtree to a max-heap, recursig upwards. As above, the ruig time aalysis of build-max-heap is left as a discussio topic. Heapsort fuctio Heapsort(A build-max-heap(a for i A.legth dowto do swap(a[], A[i] A.heapsize A.heapsize max-heapify(a, The heapsort algorithm works by first covertig the iput array A to a max-heap. It grows the sorted subarray from right to left by swappig out the root (largest elemet at A[] to its proper place i the sorted subarray ad restorig the max-heap property o the usorted subarray. (Does this otio of dividig the iput ito a usorted/sorted regio remid you of aother sortig algorithm...? The ruig time aalysis of heapsort is also left as a discussio topic. Discussio Topics What is the worst case ruig time of max-heapify? Why? Why does costructig a heap (build-max-heap take liear time? What happes if we try to build a heap by ruig isert times istead? Give that both build-max-heap ad heapsort call max-heapify at least / times, why does heapsort ru i Θ( log time ad ot build-max-heap?

6 Discuss isertio-sort, mergesort, quicksort, ad heapsort. What are their relative advatages? Whe might oe sortig algorithm be preferred over the others? Problems Problem. You have bee hired to write a applicatio for -CIS s ew etwork router! -CIS plas to sell their routers to busiesses with large corporate etworks that eed swift detectio of etwork attacks. A etwork attack is characterized by a large amout of traffic from a sigle IP address. For the applicatio, you are parsig a stream of packets cotaiig a IP-address ad their frequecy. Routers have limited memory, ad you ca oly maitai O(k space for your applicatio, where k. Desig a O( log k time algorithm to fid the k-th most frequet IP-address, where is the total umber of IP addresses i the stream. Problem. Give: A biary tree of size Objective: Prit out the level order traversal of the biary tree Example: see below Figure : For this tree, your fuctio should prit,,, 7, 6,,. Problem. Give: A biary tree T. Objective: Prit the spiral order traversal of the tree T. Example: Hit: Try usig stacks. Figure : For this tree, your fuctio should prit,,,,, 6, 7. Problem. Cosider a idefiitely log stream of usorted itegers. We are iterested i kowig the media (i sorted order at ay give time. How would we do this i a efficiet maer? 6

7 Problem. Give: A full stack S of size ad a empty stack S of size. Objective: Sort the elemets i ascedig order i S. You may oly use the give stacks S ad S (each of size ad O( additioal space. What is the ruig time of your sortig procedure? Example: Hit: Start with a simpler example: 7