// APS101, Winter 2009: Lecture 36 (Apr. 7)
//
// Review: last time we covered the basics of Exception Handling.

// throwing an exception:
throw new Exception();
throw new NullPointerException("A message that describes the problem.");

// we wrote our own custom exception:
TTTException e = new TTTException();
e
e.getMessage()

// catching an exception:
// use the try-catch block that we learned in lecture last time.

// Now, we will consider two search algorithms.

// first, assume that the list (array of ints) is in no particular order.
// Ex. int[] list = {11, 2, 8, 9, 0, 3}.
// we want to search for an element x in the list. 
// if x is found in the list, return the index of x in list.
// if x is not found in the list, return -1.
// Ex. if x = 8, then the method should return 2 (8's position in list).

// can you think of a simple algorithm to perform this task?
// (go through list sequentially and compare x with every element). 
// this is called Linear Search... see method linearSearch in Search.java

Search.linearSearch(new int[]{1, 5, 99, 54, 43, 3}, 54) // returns 3
Search.linearSearch(new int[]{1, 5, 99, 54, 43, 3}, 55) // returns -1
// if there are duplicate elements in the list, the method returns the 
// index of the first element that is found.
Search.linearSearch(new int[]{1, 5, 99, 54, 54, 43, 3}, 54) // still returns 3

// however, this is not the most efficient algorithm... 
// what if x is found at the very end of the list? or not found at all?
// in the worst case, the algorithm would have to make N comparisons,
// where N is the number of elements in the list.
// this is bad, because there could be millions of elements in the list.

// there is a better way to search.
// first, the list must be sorted in ascending order.
// then, we can use the sorted property of the list to search for x faster.

// (see the PDF file with the Binary Search example)
// (also see method binarySearch in Search.java)

Search.binarySearch(new int[]{5, 23, 45, 99, 200}, 45) // returns 2
Search.binarySearch(new int[]{5, 23, 45, 99, 200}, 46) // returns -1
Search.binarySearch(new int[]{5, 23, 45, 99, 200}, 200) // returns 4
Search.binarySearch(new int[]{5, 23, 45, 99, 200}, 5) // returns 0
Search.binarySearch(new int[]{5, 23, 45, 99, 200}, 1) // returns -1
Search.binarySearch(new int[]{5, 23, 45, 99, 200}, 250) // returns -1