1. Question of the Day  How can you change the position of 1 toothpick and leave the giraffe in exactly the same form, but possibly mirror-imaged or oriented differently, as before?

2. Question of the Day  How can you change the position of 1 toothpick and leave the giraffe in exactly the same form, but possibly mirror-imaged or oriented differently, as before?

3. CSC 212 – Data Structures

4. Using Stack

5. public interface Stack extends Collection { public E peek() throws EmptyCollectionException; public E pop() throws EmptyCollectionException; public void push(E element); }  Any type of data stored within a Stack  Generics enable us to avoid rewriting this code  Minimum set of exceptions defined by interface  Classes could throw more unchecked exceptions Stack Interface

6. Array-based Implementation  Array reference in a field  Another field tracks top element’s index  Stack size also: top  Add to next lowest index  Remove highest index Algorithm isEmpty() return top == 0 Algorithm pop() if isEmpty() then throw new EmptyCollectionException else top  top  1 return stack[top] 012 … ArrayStack stack top  0

7. Array-based Implementation  Array reference in a field  Another field tracks top element’s index  Stack size also: top  Add to next lowest index  Remove highest index 012 top … ArrayStack stack top  0 Algorithm isEmpty() return top == 0 Algorithm pop() if isEmpty() then throw new EmptyCollectionException else top  top  1 return stack[top]

8. top Array-based Implementation  Array reference in a field  Another field tracks top element’s index  Stack size also: top  Add to next lowest index  Remove highest index ArrayStack stack top  X + 1 012 … X Algorithm isEmpty() return top == 0 Algorithm pop() if isEmpty() then throw new EmptyCollectionException else top  top  1 return stack[top]

9. Array-based Implementation  Array reference in a field  Another field tracks top element’s index  Stack size also: top  Add to next lowest index  Remove highest index ArrayStack stack top  X 012 … X top Algorithm isEmpty() return top == 0 Algorithm pop() if isEmpty() then throw new EmptyCollectionException else top  top  1 return stack[top]

10.  Could fill array with data  More push() s impossible  Throw exception on error  Specific to array-based  Unchecked exception must be used Algorithm push(elem) if top == stack.length then throw new FullStackException else stack[top]  elem top  top + 1 fi ArrayStack stack top  X 012 … X top

11.  Could fill array with data  More push() s impossible  Throw exception on error  Specific to array-based  Unchecked exception must be used Algorithm push(elem) if top == stack.length then throw new FullStackException else stack[top]  elem top  top + 1 fi ArrayStack stack top  X + 1 012 … X top

12.  Could fill array with data  More push() s impossible  Throw exception on error  Specific to array-based  Unchecked exception must be used Algorithm push(elem) if top == stack.length then throw new FullStackException else stack[top]  elem top  top + 1 fi ArrayStack stack top  length 012 … top

13. Oops… My Bad  Could fill array with data  More push() s impossible  Throw exception on error  Specific to array-based  Unchecked exception must be used Algorithm push(elem) if top == stack.length then throw new FullStackException else stack[top]  elem top  top + 1 fi ArrayStack stack top  length 012 … top

14. Simple Array-based Stack Why It RocksWhy It Sucks  Easy to write & read  Simple to find bugs  Quick running times  Methods take O(1) time huge  Array must be huge  Max. possible elements  Problems occur when too many exist at once  When full, throws specific exception

15. Better Approach (Maybe?)  Implement with array, but allow resizing array  Grow as needed rather that stating full  (In Java) Requires allocating new, larger array  Copy data into new array from current array  Reassign field so that it now aliases larger array  Arrays.copyOf() does everything in single call

16. Better Approach (Maybe?)  Implement with array, but allow resizing array  Grow as needed rather that stating full  (In Java) Requires allocating new, larger array  Copy data into new array from current array  Reassign field so that it now aliases larger array  Arrays.copyOf() does everything in single call  Method call is O (1), but method takes O ( n ) time  push() now O ( n ) in worst-case, but how bad is that?

17. Ways to Grow Array  Two ways to increase array size  Constant value (e.g., 2, 4, 6, 8, 10…)  Constant factor (e.g., 2, 4, 8, 16, 32…)  Both approaches requires O(n) time…  Instantiating & copying array are slow steps average  …average (amortized) costs differ, however  Difference in how often slow step needed

18. Increase Array Size by c  To hold n elements, must grow k = n / c times  Copy array each growth, so total copies is: 1 + (c+1) + + (((k-1)*c)+1) + (k*c+1) = 1 + (k*c+1) + (c+1) + (((k-1)*c)+1) + = ((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 )  Averaged over adding n elements = O(n) each!

19. Grow by Constant Factor  To hold n elements, need to grow k = log n times  Still copy array each time, so will make 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(2n - 1) = O(n)  Average cost is O(1) ( O(n) copies ÷ n elements)

20. Your Turn  Get into your groups and complete activity

21. For Next Lecture  Read 4.1 – 4.3 before Wednesday's class  Can we do better than resizable arrays? GENIUS  Didn’t some GENIUS find too much memory expensive?  How could we grow & shrink memory demands?  Is linked list an ADT? Why or why not?  Week #7 weekly assignment due Wednesday  Programming Assignment #1 due in 2 weeks  Start before it gets too late! Planning saves time!