Standard Operations & Algorithms - AP Computer Science A

Selection sort is comprised of outer and inner for loops that swap elements of the unsorted array into a sorted array. The largest possible number of times each loop can run is the number of elements in the array. Thus, the worst possible run time is .

MergeSort is has a running time of O(N). Selection sort has a running time of O(N2). Selection sort has O(N2) comparisons due to the swap in the algorithm.

The first call is in the declaration. Because 6 > 1, it calls recur, which makes the total 2.

Next, it calls recur(6/2). Because 3 > 1, it calls recur again, which makes the total 3.

Next, it calls recur(3/2). Because it's dividing integers, this evaluates to recur(1).

Because 1 <= 1, it doesn't call recur anymore, so that total number of calls is 3.

In this case, the names help to identify the different types of traversals. In order traversal processes the binary tree "in order", meaning it will go through the left subtree, process the node, then go on to the right. Why is this in order? Because remember, a binary sort tree lists any value less than the node to the left, and any value greater to the right, so in order traversal will actually go from greatest to smallest.

Preorder traversal is the same except that it processes the root node first, hence the "pre" order.

Insertion Sort is a sorting algorithm that starts at the beginning of an array and with each iteration of the array it sorts the values from smallest to largest.

Therefore, after four iterations of Insertion Sort, the first four numbers will be in order from smallest to largest.

All of the choices are important when choosing a sorting algorithm. Space and time complexity are the characteristics by which we measure the performance of an algorithm. the array size directly affects the performance of an algorithm. In addition, the language which the algorithm is written in can also affect the performance (for example, insertion sort may run faster in one language versus another, due the way the language was designed).

Mergesort consists of breaking down an unsorted list into subarrays; sorting each sub-array, and recombining the sub-arrays into larger arrays.

In merge sort, a list is broken up into sublists containing 1 element. Each element is then compared to another element and sorted. Each 2-element group is then combined with other 2-element groups, comparing the first value of each group and deciding how to four elements. The larger group of four is compared to another group of four, until the process ends and the list is sorted.

Mergesort is the most efficient among the choices. Both selection sort and insertion sort use O(N2) time. Bubble Sort may seem like a good answer but uses O(N2) time most of the time and can be adapted to use O(N) time however only when the list is nearly sorted, so it's a gamble. Mergesort always uses O(NlogN) time and thus is always the most efficient among the four choices.

Selection sort is comprised of outer and inner for loops that swap elements of the unsorted array into a sorted array. The largest possible number of times each loop can run is the number of elements in the array. Thus, the worst possible run time is .

MergeSort is has a running time of O(N). Selection sort has a running time of O(N2). Selection sort has O(N2) comparisons due to the swap in the algorithm.

What we are looking for in this problem is a standard traversal of a 2D array. When you do this, you need to make sure that you iterate through both the rows and the columns. In order to do this, you first must set up a loop like:

for(int i = 0; i < vals.length;i++) {...

The value of vals.length indicates the number of rows in the 2D array.

Now, for each row, you have a certain number of columns. (A 2D array is like an "array of arrays".) Thus, for each row, you need to run through the entire set of columns for that row:

for(int j = 0; j < vals\[i\].length; j++) { ...

Notice that none of the incorrect questions use vals\[i\].length in this way. Thus, none of them iterate through the two dimensions of the array correctly.

Of course, this is an inefficient way to do such a delete, but arrays are rather "locked" data structures in that their size cannot change without a reassignment. (You could, of course keep track of the last "used" index. However, that is a different implementation, not reflected here.) The correct answer is the one that carefully goes through the original array, copying those contents into the new array skipping the one value that is not wanted.

This code simulates the removal of a value from an array by shifting all of the elements after that one so that the array no longer contains the first instance of that value. So, for instance, this code takes the original array {3,4,4,5,17,4,3,1} and notices the first instance of 4: {3,**4**,4,5,17,4,3,1}. Next, it starts shifting things to the left. Thus, some of the steps will look like this:

{3,4,4,5,17,4,3,1}

{3,4,5,5,17,4,3,1}

{3,4,5,17,17,4,3,1}

...

{3,4,5,5,17,4,1,1}

Then, at the very end, it sets the last element to 0:

{3,4,5,17,4,3,1,0}

Use a for loop to traverse the list. The .size() method is used for Lists as opposed to the .length method for arrays. The .get() method is used for Lists as opposed to accessing the index for arrays.

It is easiest to understand this by a bit of code parsing:

First, there is the loop: for(int j = 0; j < x; j++) { // ...

This will fill the array with 20 + 0, 20 + 40, 20 + 80, ... Thus, it will be:

20, 60, 100, 140, 180, 220

Next, there is the array: for(int j = x; j > i; j--) { // ...

This basically shifts everything after index i backward by 1. Thus you have:

20, 60, 100,100, 140, 180, 220

Next, you set arr\[2\] equal to 20:

20, 60, 20,100, 140, 180, 220

Finally you output each of the first 6 elements. Be careful here. Notice that it is from j = 0 to x - 1!

Thus, the answer is:

20 60 20 100 140 180

This algorithm basically implements a simple kind of array deletion.

A quesiton like this is most easily understood by doing a careful reading of the code in question. Let's consider each major section of code:

int[] ret = new int[arr.length + 1];

This line of code creates a new array, one that is 1 longer than the array arr.

for(int i = 0; i < index; i++) . . .

This loop copies into ret the values of arr up to index - 1.

(This is because of the i < index condition)

Then, the code stores the value val in ret[index]

_for(int i = index + 1; i < ret.length; i++) . . ._

It then finishes copying the values into ret.

Thus, what this does is insert a value into the original array arr, returning the new array that is one size larger. (This is necessary because of the static sizing of arrays in Java.)

The most obvious possible error is that the array arr might be a null value. You need to check for these kinds of values before using the variables. (If you do arr[0] on a null value, an exception will be thrown.) In addition, it is possible that a value for index might be given that is too large. Consider if index = 100 but arr is only 4 elements long. Then, you will have ret be a 5 value array. When the first loop starts to run, you will go all the way to 99 (or at least attempt to do so) for the index value i_; h_owever, once you get to ret[4] = arr[4], there will be an out of bounds error on arr, which only has indices 0, 1, 2, and 3. Of course, there could be other problems later on with ret, but the question only asks about this first loop.

Let us presume that the array arr looks like the following list of integers:

{4,51,41,0,0,0,0,0,0,0}

Presume that our new value is 77.

A stack could either push on to the beginning and move everything back one, giving you something like:

{77,4,51,41,0,0,0,0,0,0}

However, none of the options do this. (No, not even the one that uses the index 0. This one does not add on to the array so much as replace the first element of it.)

So, the other option is to put it on the "end" of the list. The variable arrFill is being used for this purpose. If it is 3, this means that it is the value of the fourth element. Thus, you can set arr\[4\] = 77 (where 4 really is arrFill).

This will give you:

{4,51,41,77,0,0,0,0,0,0}

You also need to add one to the value of arrFill.

The other options do not correctly address the array. They either are too large or too small by one.