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## Example Questions

### Example Question #1 : Deletions

Which of the following defines a method that successfully deletes an item from an array of integers?

**Possible Answers:**

public static int[] del(int[] a,int delIndex) {

if(a == null || delIndex < 0 || delIndex >= a.length) {

return null;

}

int[] ret = new int[a.length - 1];

for(int i1=0,i2 = 0; i1 < a.length; i1++) {

if(i1 == delIndex) {

delete a[i1];

}

}

return ret;

}

public static int[] del(int[] a,int delIndex) {

if(a == null || delIndex < 0 || delIndex >= a.length) {

return null;

}

for(int i = 0; i <= delIndex; i++) {

if(i == delIndex) {

delete a[i];

break;

}

}

return a;

}

public static int[] del(int[] a,int delIndex) {

if(a == null || delIndex < 0 || delIndex >= a.length) {

return null;

}

int[] ret = new int[a.length - 1];

for(int i = 0; i <= delIndex; i++) {

ret[i] = a[i];

}

return ret;

}

None of these work correctly

public static int[] del(int[] a,int delIndex) {

if(a == null || delIndex < 0 || delIndex >= a.length) {

return null;

}

int[] ret = new int[a.length - 1];

for(int i1=0,i2 = 0; i1 < a.length; i1++) {

if(i1 != delIndex) {

ret[i2] = a[i1];

i2++;

}

}

return ret;

}

**Correct answer:**

public static int[] del(int[] a,int delIndex) {

if(a == null || delIndex < 0 || delIndex >= a.length) {

return null;

}

int[] ret = new int[a.length - 1];

for(int i1=0,i2 = 0; i1 < a.length; i1++) {

if(i1 != delIndex) {

ret[i2] = a[i1];

i2++;

}

}

return ret;

}

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.

### Example Question #2 : Deletions

public static boolean remove(int[] arr, int val) {

boolean found = false;

int i;

for(i = 0; i < arr.length && !found; i++) {

if(arr[i] == val) {

found = true;

}

}

if(found) {

for(int j = i; j < arr.length;j++) {

arr[j - 1] = arr[j];

}

arr[arr.length - 1] = 0;

}

return found;

}

For the code above, what will be the content of the variable `arr`

at the end of execution, if the method is called with the following values for its parameters:

`arr = {3,4,4,5,17,4,3,1}`

`val = 4`

**Possible Answers:**

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

None of the other answers

`{3, 5, 17, 3, 1}`

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

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

**Correct answer:**

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

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,`

. Next, it starts shifting things to the left. Thus, some of the steps will look like this:**4**,4,5,17,4,3,1}

`{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}`

### Example Question #2 : Deletions

int[] arr = new int[20];

int x = 6,i=2;

for(int j = 0; j < x; j++) {

arr[j] = j * 40 + 20;

}

for(int j = x; j > i; j--) {

arr[j] = arr[j - 1];

}

arr[i] = 20;

for(int j = 0; j < x; j++) {

System.out.print(arr[j] + " ");

}

What is the output for the code above?

**Possible Answers:**

20 60 20 100 140 180

20 60 20 100 100 100

20 60 20 100 140 180 220

20 60 20 140 180 220

None of the others

**Correct answer:**

20 60 20 100 140 180

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.

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