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CSC 212 – Data Structures Lecture 21: IndexList a/k/a Vector, ArrayList

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Question of the Day What card(s) must you flip to verify the following statement: Cards with a vowel on one side, have an even number on the other side. A A B B 4 4 7 7

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Deque ADT Another type of Collection class Stands for Double Ended QUEue Combines ideas from Stack and Queue ADT Manipulate elements at front & rear addFront(), addLast() removeFront(), removeLast() getFront(), getLast() Can be implemented with array or linked list If using linked list, should be doubly-linked list

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List ADT Sometime want use of entire Collection Add new element before existing one Get the 3 rd element in the Collection Loop over all elements without removing them Cannot be done with Stack, Queue & Deque Lists can access all of its elements But provide different means for this access Will discuss over next several lectures

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IndexList ADT Also called a Vector Names are used interchangeably First example of a List Extends idea of an array Stores arbitrary sequence of elements Elements may appear multiple times Access elements using integer rank Like array indexes, does not imply ordering

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Ranks ArrayList organizes collection using rank Item at front of list has rank of 0 2 nd item has rank of 1 3 rd item has rank of 2 n th item has rank of n – 1 Ranks increase sequentially Merely discusses position of element Cannot skip over a rank Cannot repeat a rank

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IndexList Interface public interface IndexList extends Collection { public void add(int rank, E e) throws IndexOutOfBoundsException; public E get(int rank) throws IndexOutOfBoundsException; public E remove(int rank) throws IndexOutOfBoundsException; public E set(int rank, E newValue) throws IndexOutOfBoundsException; }

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IndexList != array Extends the idea of an array, but… IndexList does not have constant size Elements’ ranks may change over time Can implement using: Array, Singly-linked list, Doubly-linked list, Specially trained monkeys, College students

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Insertion add( r, e ) “shifts” existing elements down to make room for e For linked-list, insert node at proper location For array, must shift elements down Can take O(n) time S 012n r S 012n r S 012n e r

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Deletion remove( r ) “shifts” remaining elements up to fill hole created by removal For linked-list, happens automatically For array, must shift elements up Can also take O(n) time S 012n r S 012n r S 012n r

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IndexList’s Operations add, get, & remove similar to past operations add ≈ addFront, addLast, enqueue, push get ≈ getFront, getLast, front, top remove ≈ removeFront,removeLast,dequeue,pop But, set is a brand new operation Stores the new element at the given rank Removes (& returns) element already there Does not change rank of other elements

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IndexList’s Exceptions Any of these methods may throw IndexOutOfBoundsException Thrown when rank is illegal for operation add accepts ranks 0 - size() add(0, e) e added to front of List add(size(), e) e added at end of List get, set, & remove use ranks 0 - size()-1 Ranks start at 0, so no element at size() rank

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Implement IndexList with array Can allocating & copy into larger array Increase array length by constant, c -or- Double length of the array each time Both approaches have O( n ) complexity Cost of copying entire array But have different amortized complexities Consider cost of growth due to n calls to add()

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Constant Growth Need to grow k = n / c times Copy entire array with each growth, so total copies is: 1 + ( c +1) + (2 c +1) + + ((( k -1) *c )+1) + ( k*c +1) = (( k * c ) + 2) + (( k * c ) + 2) + + (( k * c ) + 2) = k / 2 * (( k * c ) + 2) = O( c*k 2 )= O( c * ( n / c ) 2 ) = O( n 2 * 1 / c ) = O( n 2 ) Average cost: O( n 2 ) / n = O( n )

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Doubling Growth Array grows k = log n times Still copy entire array, for this many copies: 1 + 2 + 4 + + 2 k -1 + 2 k = (2 k - 1) + 2 k = 2 k +1 - 1 = (2 * 2 k ) - 1 = (2 * 2 log n ) - 1 = O(2 n - 1) = O( n ) Average cost: O( n ) / n = O(1) But, moving elements in add() still costs O( n )!

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Your Turn Get back into groups and do activity

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Before Next Lecture… Keep up with your reading! Continue Week #9 Assignment Work on Programming Assignment #2

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