BHSawesome AP CSA

Linear or Sequential search can be used to find a value in unsorted data. It usually starts at the first element and walks through the array or ArrayList until it finds the value it is looking for and returns its index. If it reaches the end of the array or list without finding the value, the search method usually returns a -1 to show that it didn’t find the value in the array or list. Click on Show CodeLens below to see linear search in action.

Here is the same search with an ArrayList. The same algorithms can be used with arrays or ArrayLists, but notice that size() and get(i) are used with ArrayLists instead of length and [i] which are used in arrays. Many of our examples will use arrays for simplicity since with arrays, we know how many items we have and the size won’t change during runtime. There are methods such as contains that can be used in ArrayLists instead of writing your own algorithms. However, they are not in the AP CSA Java subset.

You can also look for a String in an array or list, but be sure to use equals rather than ==. Remember that == is only true when the two references refer to the same String object, while equals returns true if the characters in the two String objects are the same.

We can also apply the linear search algorithm to data in a 2D array. We can loop through each row of the 2D array and then apply the linear search algorithm to each row to find an element. The code below demonstrates this with a 2D array of integers. Click on the Code Lens button to step through this code. Then, change it to work with a 2D array of Strings. Remember to use the equals method to compare Strings.

How do you search for something in a phone book or dictionary that is in alphabetical or numerical order? If you’re looking for something beginning with M or on page 100 in a 200 page book, you wouldn’t want to start with page 1. You would probably start looking somewhere in the middle of the book. This is the idea behind binary search.

If your array or list is already in order (sorted), binary search will on average find an element or determine that it is missing much more quickly than a linear search. But binary search can only be used if the data is sorted.

Binary search keeps dividing the sorted search space into half. It compares a target value to the value in the middle of a range of indices. If the value isn’t found it looks again in either the left or right half of the current range. Each time through the loop it eliminates half the values in the search area until either the value is found or there is no more data to look at. See the animation below from https://github.com/AlvaroIsrael/binary-search:

Binary search calculates the middle index as left + right / 2 where left starts out at 0 and right starts out at the array length - 1 (the index of the last element). Remember that integer division gives an integer result so 2.5 becomes 2. It compares the value at the middle index with the target value (the value you are searching for). If the target value is less than the value at the middle it sets right to middle minus one. If the target value is greater than the value at the middle it sets left to middle plus one. Otherwise the values match and it returns the middle index. It also stops when left is greater than right which indicates that the value wasn’t found and it returns -1.

The code for binarySearch below is from the AP CSA course description. A recursive version of this algorithm will be covered in lesson 4.17.

You can also use binary search with a String array. But, when you look for a String, be sure to use compareTo method rather than < or > which can only be used with primitive types. Remember how the String method compareTo works:

How do we choose between two algorithms that solve the same problem? They usually have different characteristics and runtimes which measures how fast they run. For the searching problem, it depends on your data.

Binary search is much faster than linear search, especially on large data sets, but it can only be used on sorted data. Often with runtimes, computer scientist think about the worst case behavior. With searching, the worst case is usually if you cannot find the item. With linear search, you would have to go through the whole array before realizing that it is not there, but binary search is much faster even in this case because it eliminates half the data set in each step. We can measure an informal runtime by just counting the number of steps.

Here is a table that compares the worst case runtime of each search algorithm given an array of n elements. The runtime here is measured as the number of times the loop runs in each algorithm or the number of elements we need to check in the worst case when we don’t find the item we are looking for. Notice that with linear search, the worst case runtime is the size of the array n, because it has to look through the whole array. For the binary search runtime, we can calculate the number of times you can divide n in half until you get to 1. So, for example 8 elements can be divided in half to narrow down to 4 elements, which can be further divided in half to narrow down to 2 elements, which can be further divided in half to get down to 1 element, and then if that is wrong, to 0 elements, so that is 4 divisions or guesses to get the answer (8->4->2->1->0). In the table below, every time we double the size of N, we need at most one more guess or comparison with binary search. It’s much faster than linear search!

Runtimes can be described with mathematical functions. For an array of size n, linear search runtime is a linear function, and binary search runtime is a function of log base 2 of n (or log n + 1 comparisons). This is called the big-O runtime function in computer science, for example O(log n) vs. O(n). You can compare the growth of functions like n and \(\log_{2}n\) as \(n\text{,}\) the data size, grows and see that binary search runs much faster for any \(n\text{.}\) You don’t need to know the log n runtime growth function for the AP exam, but you should be able to calculate how many steps binary search takes for a given n by counting how many times you can divide it in half. Or you can start at 1 and keep a count of how many times you can double it with the powers of two (1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, etc.) until you reach a number that is slightly above n.

I. The values in the array must be integers.
II. The values in the array must be in sorted order.
III. The array must not contain duplicate values.

Let’s go back to the spellchecker that we created with arrays. Here is a version of the spellchecker below that reads the dictionary file into an ArrayList. The advantage of using an ArrayList instead of an array for the dictionary is that we do not need to know or declare the size of the dictionary in advance.

In the spellchecker challenge, we used linear search to find a word in the dictionary. However, the dictionary file is actually in alphabetical order. We could have used a much faster binary search algorithm! Let’s see how much faster we can make it.

Write a linear search method and a binary search method to search for a given word in the dictionary using the code in this lesson as a guide. You will need to use size and get(i) instead of [] to get an element in the ArrayList dictionary at index i. You will need to use the equals and compareTo methods to compare Strings. Have the methods return a count of how many words they had to check before finding the word or returning.

Run your code with the following test cases and record the runtime for each word in this Google document (do File/Make a Copy) also seen below to record your answers.

What do you notice? Which one was faster in general? Were there some cases where each was faster? How fast were they with misspelled words? Record your answers in the window below.