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How do I compare each corresponding index from three array of different random lengths from 1-5 without getting a possible out of bound exception

I have 3 arrays of random length. I want to create a new array that stores the largest value from comparing those 3 arrays at each index.
int size1=x.length;
int size2=y.length;
int size3=z.length;
int size=0;
if (size1>=size2 && size1>=size3)
size=size1;
else if (size2>=size1 &&size2>=size3) {
size=size2;
}
else if (size3>=size1 && size3>=size2) {
size=size3;
}
int[] largest= new int[size];
int[] x= {1, 4, 6}; // random array length from 1-5 and hypothetically each array hold these values
int[] y= {2, 4};
int[] z= {5, 6, 7, 8, 9};
// ideally after some sort of an algorithm largest[] should hold {5, 6, 7, 8, 9}
I initially thought of a for loop, but my loop will eventually throw me a out of bound exception, because of the random size length nature of the arrays and x/y/z won't hold a value at index [i]. Any other ways?
for (int i=0;i<size;i++) {
if (x[i]>y[i]) && t1[i]>t3[i]) {
largest[i]=x[i];
}
else if (y[i]>x[i]) && y[i]>z[i]) {
largest[i]=y[i];
}
else if (z[i]>x[i]) && z[i]>y[i]) {
largest[i]=z[i];
}
}

There are several ways of doing this. Here's one that avoids a ton of conditional statements at the cost of more memory.
int size = Math.max(x.length, Math.max(y.length, z.length));
int[] nooX = new int[size];
int[] nooY = new int[size];
int[] nooZ = new int[size];
// Copy over the values from x to the new array
for(int i = 0; i < x.length; i++){
nooX[i] = x[i];
}
// ... Copy paste the above and do the same for arrays nooY and nooZ
int[] largest = new int[size];
// ... Copy paste your code, using nooX, nooY, and nooZ instead of x, y, and z

A simpler approach without creating extra arrays to equalize size:
public static int[] getMaxValues(int[] x, int[] y, int[] z) {
int size = Math.max(x.length, Math.max(y.length, z.length));
int[] max = new int[size];
for (int i = 0; i < size; i++) {
int xi = i < x.length ? x[i] : Integer.MIN_VALUE;
int yi = i < y.length ? y[i] : Integer.MIN_VALUE;
int zi = i < z.length ? z[i] : Integer.MIN_VALUE;
max[i] = Math.max(xi, Math.max(yi, zi));
}
return max;
}
Test:
int[] x= {4, 4, 6}; // random array length from 1-5 and hypothetically each array hold these values
int[] y= {2, 10};
int[] z= {3, 6, 7, 8, 9};
System.out.println(Arrays.toString(getMaxValues(x, y, z)));
Output:
[4, 10, 7, 8, 9]
Update
Defining a couple of functions allows to create the following implementation using Stream API that would be able to handle non-hardcoded number of arrays:
private static int getAtIndex(int[] arr, int i) {
return i < arr.length ? arr[i] : Integer.MIN_VALUE;
}
private static int getMax(IntStream values) {
return values.max().getAsInt();
}
// use Supplier to be able to use stream of the arrays twice
public static int[] getMaxValues(Supplier<Stream<int[]>> arrs) {
return IntStream.range(0, getMax(arrs.get().mapToInt(arr -> arr.length)))
.map(i -> getMax(arrs.get().mapToInt(arr -> getAtIndex(arr, i))))
.toArray();
}
Test:
int[] maxValues = getMaxValues(() -> Stream.of(x, y, z)); // supply stream of arrays
System.out.println(Arrays.toString(maxValues));

I think we should think this way
array1 = 1, 2, 3, 4, 6, 7
array2 = 3, 4, 5, 6, 23, 4
array3 = 5, 5, 32, 3, 2, 43, 56
Like a matrix
1 2 3 4 6 7
3 4 5 6 23 4
5 5 32 3 2 43 56
We need is the greatest value in every column.
largestArr = 5, 5, 32, 6, 23, 43, 56 <-- Like this
I hope this code is the answer to your problem.
public static int[] largestColumnsArr(int arr1[], int arr2[], int arr3[]) {
int[][] arr = {arr1, arr2, arr3};
//The size of the largest sized array
int size = Math.max(arr3.length, Math.max(arr2.length, arr1.length));
int[] largestArr = new int[size];
/*
Takes the largest value in each column and assigns it to the array
If it is try catch, if the size of the arrays is exceeded, the program exit is blocked.
*/
for (int i = 0; i < size; i++) {
int largestColumnValue = 0;
try {
for (int j = 0; j < arr.length; j++) {
if (largestColumnValue < arr[j][i]) {
largestColumnValue = arr[j][i];
}
}
} catch (Exception e) {
}
largestArr[i] = largestColumnValue;
}
return largestArr;
}

Remove last occurrence of a specific integer from an array

For some reason, my solution is not complete. I got 80/100 from hidden spec tests.
What's wrong with my solution? There is probably a certain use case that I'm not thinking of.
How would space/time complexity change using an ArrayList instead of an array?
Is there a better way to tackle this problem?
My current solution handles:
an empty input array
negative/positive integer values in the input array
duplicates in the input array
sorted/unsorted input array
Instructions:
Write a Java method removeLastOccurrence(int x, int[] arr), which removes the last occurrence of a given integer element x from a given array of integer elements arr.
The method should return a new array containing all elements in the given array arr except for the last occurrence of element x. The remaining elements should appear in the same order in the input and the returned arrays.
The code on the right shows you a code framework in which the implementation of one static method is still missing. Provide this implementation and check that it is correct by either writing more tests yourself or using the provided tests and specification tests.
My code:
class RemoveLastOccurrenceArray {
/**
* Takes the array and the last occurring element x,
* shifting the rest of the elements left. I.e.
* [1, 4, 7, 9], with x=7 would result in:
* [1, 4, 9].
*
* #param x the entry to remove from the array
* #param arr to remove an entry from
* #return the updated array, without the last occurrence of x
*/
public static int[] removeLastOccurrence(int x, int[] arr) {
// if arr == null return null;
if (arr == null || arr.length == 0) return arr;
// return a new array which will be size arr.legnth-1
int[] res = new int[arr.length - 1];
// introduce an int tracker which keep tracks of the index of the last occurrence of x
int last_index = -1;
// traverse through the array to get the index of the last occurrence
for (int i = 0; i < arr.length; i++) if (arr[i] == x) last_index = i;
int i = 0, j = 0;
// copying elements of array from the old one to the new one except last_index
while (i < arr.length) {
if (i == last_index) {
if (i++ < res.length) {
res[j++] = arr[i++];
}
} else res[j++] = arr[i++];
}
// if we pass in x which is not in the array just return the original array
if (last_index == -1) return arr;
// are there duplicates in the array? - WORKS
// does the array have negative numbers? - WORKS
// Is the array sorted/unsorted - WORKS
return res;
}
}
Passing Unit Tests
import static org.junit.Assert.*;
import org.junit.*;
public class RemoveLastOccurrenceArrayTest {
#Test
public void testRemoveArray_Empty() {
int[] array = new int[0];
assertEquals(0, RemoveLastOccurrenceArray.removeLastOccurrence(5, array).length);
}
#Test
public void testFirstSimple() {
int[] input = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int[] result = {2, 3, 4, 5, 6, 7, 8, 9, 10};
assertArrayEquals(result, RemoveLastOccurrenceArray.removeLastOccurrence(1, input));
}
#Test
public void testLastSimple() {
int[] input = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int[] result = {1, 2, 3, 4, 5, 6, 7, 8, 9};
assertArrayEquals(result, RemoveLastOccurrenceArray.removeLastOccurrence(10, input));
}
#Test
public void testPositiveInMiddleDuplicate() {
int[] input = {1, 2, 3, 3, 4, 5};
int[] result = {1, 2, 3, 4, 5};
assertArrayEquals(result, RemoveLastOccurrenceArray.removeLastOccurrence(3, input));
}
#Test
public void testNegativeFirst() {
int[] input = {-3, -1, 2, -3, 3, 4, 5, 0};
int[] result = {-3, -1, 2, 3, 4, 5, 0};
assertArrayEquals(result, RemoveLastOccurrenceArray.removeLastOccurrence(-3, input));
}
#Test
public void testLasttoRemove() {
int[] input = {1, 4, 7, 9};
int[] result = {1, 4, 7};
assertArrayEquals(result, RemoveLastOccurrenceArray.removeLastOccurrence(9, input));
}
}

Why not try iterating backwards?
for(int i = arr.length; i => 0; i--)
{
if (arr[i] == x)
{
return ArrayUtils.remove(arr, i)
}
}
Then, after you find the index, you can use the Apache Commons ArrayUtils remove command to remove the item at the

This is the answer thank you very much!
Also, if there is no x to find, yours crashes ... mine doesn't. Maybe that's where the twenty marks went?
I was already checking for this but too late in my code. So I just had to move
if (last_index == -1) return arr; before the while loop, and I got 100/100 scores.
Would your prof prefer this? Just another way, and I don't think any more efficient than your answer. But maybe they like to see the java classes used ...
Does your prof not tell you where you lost marks? You can't improve if they don't tell you what they were expecting for full marks. But here was another way ... again, no better in my opinion, and not worth twenty more marks. I'll just post it, because it is 'another way.'
public int[] removeLastOccurrence2(int x, int[] arr) {
// if arr == null return null;
if (arr == null || arr.length == 0) return arr;
// Fill an ArrayList with your initial array ...
java.util.List list = new java.util.ArrayList(arr.length);
for (int i=0; i<arr.length; i++) {
list.add(arr[i]);
}
int[] res;
// Now ... use ArrayList methods to do the work.
// Also, if there is no x to find, yours crashes ... mine doesn't.
// Maybe that's where the twenty marks went?
if ( list.lastIndexOf(x) != -1 ) { // This screens for no x found at all ...
list.remove( list.lastIndexOf(x) ); // Done!
// Make a new array to return.
res = new int[list.size()];
for (int i=0; i<list.size(); i++) {
res[i] = (int) list.get(i);
}
} else {
// No 'x' found, so just return the original array.
res = arr;
}
return res;
}

How about reverse(), remove(), reverse()? Sorry if this is already mentioned in here somewhere and I missed it.

How can I add new integers to replace the old integers to my already-existing Array?

Here is the program task:
Write a method called collapse that accepts an array of integers as a parameter and returns a new array containing the result of replacing each pair of integers with the sum of that pair.
For example, if an array called list stores the values
{7, 2, 8, 9, 4, 13, 7, 1, 9, 10}
then the call of collapse(list) should return a new array containing:
{9, 17, 17, 8, 19}.
The first pair from the original list is collapsed into 9 (7 + 2), the second pair is collapsed into 17 (8 + 9), and so on. If the list stores an odd number of elements, the final element is not collapsed.
For example, if the list had been {1, 2, 3, 4, 5}, then the call would return {3, 7, 5}. Your method should not change the array that is passed as a parameter.
Here is my currently-written program:
public static int[] collapse(int[] a1) {
int newArrayLength = a1.length / 2;
int[] collapsed = new int[newArrayLength];
int firstTwoSums = 0;
for (int i = 0; i < a1.length-1; i++) {
firstTwoSums = a1[i] + a1[i+1];
collapsed[collapsed.length-1] = firstTwoSums;
}
return collapsed;
}
I pass in an array of {7, 2, 8, 9, 4, 13, 7, 1, 9, 10} and I want to replace this array with {9, 17, 17, 8, 19}.
Note:{9, 17, 17, 8, 19} will be obtained through the for-loop that I have written.
Currently, I am having trouble with adding the integers I obtained to my "collapsed" array. It'd be a great help if you could help me or at least give me some guidance on how to do this.
Thanks in advance!

First you have to understand what is going on.
You have an array of certain size where size can either be even or odd. This is important because you are using a1.length/2 to set the size for new array, so you will also have to check for odd and even values to set the size right else it won't work for odd sized arrays. Try a few cases for better understanding.
Here's a way of doing it.
public static int[] collapseThis(int[] array) {
int size = 0;
if(isEven(array.length))
size = array.length/2;
else
size = array.length/2+1;
int[] collapsedArray = new int[size];
for(int i=0, j=0; j<=size-1; i++, j++) {
if(j==size-1 && !isEven(array.length)) {
collapsedArray[j] = array[2*i];
}
else {
collapsedArray[j] = array[2*i]+array[2*i+1];
}
}
return collapsedArray;
}
private static boolean isEven(int num) {
return (num % 2 == 0);
}

Using
collapsed[collapsed.length-1] = firstTwoSums;
The sum of your numbers will be always be put in the same index of the collapsed array, because collapsed.length - 1 is a constant value.
Try creating a new variable starting at zero, that can be incremented each time you add a sum to collapsed. For instance,
int j = 0;
for(...) {
...
collapsed[j++] = firstTwoSums;
}

I think this is a convenient answer.
public static void main(String[] args){
int[] numbers = {1,2,3,4,5};
int[] newList = collapse(numbers);
System.out.println(Arrays.toString(newList));
}
public static int[] collapse(int[] data){
int[] newList = new int[(data.length + 1)/2];
int count = 0;
for (int i = 0; i < (data.length / 2); i++){
newList[i] = data[count] + data[count + 1];
System.out.println(newList[i]);
count = count + 2;
}
if (data.length % 2 == 1){
newList[(data.length / 2)] = data[data.length - 1];
}
return newList;
}

i would combine the cases for the array with either odd or even elements together as below:
public static int[] collapse(int[] a1) {
int[] res = new int[a1.length/2 + a1.length % 2];
for (int i = 0; i < a1.length; i++)
res[i/2] += a1[i];
return res;
}

public static int[] collapse(int[] a1) {
int newArrayLength = a1.length / 2;
int[] collapsed;
if(a1.length%2 == 0)
{
collapsed = new int[newArrayLength];
}
else
{
collapsed = new int[newArrayLength+1];
collapsed[newArrayLength] = a1[a1.length-1];
}
int firstTwoSums = 0;
for (int i = 0; i < newArrayLength; i++) {
firstTwoSums = a1[i*2] + a1[i*2+1];
collapsed[i] = firstTwoSums;
}
return collapsed;
}
I modified your code and you may try it first.

How to add an element to Array and shift indexes?

I need to add an element to Array specifying position and value.
For example, I have Array
int []a = {1, 2, 3, 4, 5, 6};
after applying addPos(int 4, int 87) it should be
int []a = {1, 2, 3, 4, 87, 5};
I understand that here should be a shift of Array's indexes, but don't see how to implement it in code.

The most simple way of doing this is to use an ArrayList<Integer> and use the add(int, T) method.
List<Integer> list = new ArrayList<Integer>();
list.add(1);
list.add(2);
list.add(3);
list.add(4);
list.add(5);
list.add(6);
// Now, we will insert the number
list.add(4, 87);

This should do the trick:
public static int[] addPos(int[] a, int pos, int num) {
int[] result = new int[a.length];
for(int i = 0; i < pos; i++)
result[i] = a[i];
result[pos] = num;
for(int i = pos + 1; i < a.length; i++)
result[i] = a[i - 1];
return result;
}
Where a is the original array, pos is the position of insertion, and num is the number to be inserted.

Jrad solution is good but I don't like that he doesn't use array copy. Internally System.arraycopy() does a native call so you will a get faster results.
public static int[] addPos(int[] a, int index, int num) {
int[] result = new int[a.length];
System.arraycopy(a, 0, result, 0, index);
System.arraycopy(a, index, result, index + 1, a.length - index - 1);
result[index] = num;
return result;
}

You must make a new array, use System.arraycopy to copy the prefix and suffix, and set that one slot to the new value.

If you prefer to use Apache Commons instead of reinventing the wheel, the current approach is this:
a = ArrayUtils.insert(4, a, 87);
It used to be ArrayUtils.add(...) but that was deprecated a while ago. More info here: 1

I smell homework, so probably an ArrayList won't be allowed (?)
Instead of looking for a way to "shift indexes", maybe just build a new array:
int[] b = new int[a.length +1];
Then
copy indexes form array a counting from zero up to insert position
...
...
//edit: copy values of course, not indexes

Unless I'm missing something, the question is not about increasing the array size. In the example the array size remains the same. (Like a bit shift.)
In this case, there is really no reason to create a new array or to copy it. This should do the trick:
static void addPos(int[] array, int pos, int value) {
// initially set to value parameter so the first iteration, the value is replaced by it
int prevValue = value;
// Shift all elements to the right, starting at pos
for (int i = pos; i < array.length; i++) {
int tmp = prevValue;
prevValue = array[i];
array[i] = tmp;
}
}
int[] a = {1, 2, 3, 4, 5, 6};
addPos(a, 4, 87);
// output: {1, 2, 3, 4, 87, 5}

Here is a quasi-oneliner that does it:
String[] prependedArray = new ArrayList<String>() {
{
add("newElement");
addAll(Arrays.asList(originalArray));
}
}.toArray(new String[0]);

org.apache.commons.lang3.ArrayUtils#add(T[], int, T) is deprecated in newest commons lang3, you can use org.apache.commons.lang3.ArrayUtils#insert(int, T[], T...) instead.
Deprecated this method has been superseded by insert(int, T[], T...) and may be removed in a future release. Please note the handling of null input arrays differs in the new method: inserting X into a null array results in null not X
Sample code:
Assert.assertArrayEquals
(org.apache.commons.lang3.ArrayUtils.insert
(4, new int[]{1, 2, 3, 4, 5, 6}, 87), new int[]{1, 2, 3, 4, 87, 5, 6});

Have a look at commons. It uses arrayCopy(), but has nicer syntax. To those answering with the element-by-element code: if this isn't homework, that's trivial and the interesting answer is the one that promotes reuse. To those who propose lists: probably readers know about that too and performance issues should be mentioned.

int[] b = new int[a.length +1];
System.arraycopy(a,0,b,0,4);
//System.arraycopy(srcArray, srcPosition, destnArray, destnPosition, length)
b[4]=87;
System.arraycopy(a,4,b,5,2);
b array would be created as {1, 2, 3, 4, 87, 5,6};

Try this
public static int [] insertArry (int inputArray[], int index, int value){
for(int i=0; i< inputArray.length-1; i++) {
if (i == index){
for (int j = inputArray.length-1; j >= index; j-- ){
inputArray[j]= inputArray[j-1];
}
inputArray[index]=value;
}
}
return inputArray;
}

System.arraycopy is more performant but tricky to get right due to indexes calculations. Better stick with jrad answer or ArrayList if you don't have performance requirements.
public static int[] insert(
int[] array, int elementToInsert, int index) {
int[] result = new int[array.length + 1];
// copies first part of the array from the start up until the index
System.arraycopy(
array /* src */,
0 /* srcPos */,
result /* dest */,
0 /* destPos */,
index /* length */);
// copies second part from the index up until the end shifting by 1 to the right
System.arraycopy(
array /* src */,
index /* srcPos */,
result /* dest */,
index + 1 /* destPos */,
array.length - index /* length */);
result[index] = elementToInsert;
return result;
}
And JUnit4 test to check it works as expected.
#Test
public void shouldInsertCorrectly() {
Assert.assertArrayEquals(
new int[]{1, 2, 3}, insert(new int[]{1, 3}, 2, 1));
Assert.assertArrayEquals(
new int[]{1}, insert(new int[]{}, 1, 0));
Assert.assertArrayEquals(
new int[]{1, 2, 3}, insert(new int[]{2, 3}, 1, 0));
Assert.assertArrayEquals(
new int[]{1, 2, 3}, insert(new int[]{1, 2}, 3, 2));
}

public class HelloWorld{
public static void main(String[] args){
int[] LA = {1,2,4,5};
int k = 2;
int item = 3;
int j = LA.length;
int[] LA_NEW = new int[LA.length+1];
while(j >k){
LA_NEW[j] = LA[j-1];
j = j-1;
}
LA_NEW[k] = item;
for(int i = 0;i<k;i++){
LA_NEW[i] = LA[i];
}
for(int i : LA_NEW){
System.out.println(i);
}
}
}

Following code will insert the element at specified position and shift the existing elements to move next to new element.
public class InsertNumInArray {
public static void main(String[] args) {
int[] inputArray = new int[] { 10, 20, 30, 40 };
int inputArraylength = inputArray.length;
int tempArrayLength = inputArraylength + 1;
int num = 50, position = 2;
int[] tempArray = new int[tempArrayLength];
for (int i = 0; i < tempArrayLength; i++) {
if (i != position && i < position)
tempArray[i] = inputArray[i];
else if (i == position)
tempArray[i] = num;
else
tempArray[i] = inputArray[i-1];
}
inputArray = tempArray;
for (int number : inputArray) {
System.out.println("Number is: " + number);
}
}
}

frequency of items in int[] as opposed to List<Integer>?

I'm trying to figure out how to get the frequency of items within a list. When I approach this problem I typically, in the past, did:
int occurrences = Collections.frequency(list, 0);
It works when my list is a List<Integer> list. Is there a way to do this if I'm using int[] list? When I try collections, my list gets converted and then my code breaks. I can convert my code if needed, but was wondering, if there was a way to get the frequency from int[] instead.

You can (1) write your own linear-time frequency method, or (2) convert to an array of boxed int types and use Arrays.asList with Collections.frequency.
int[] arr = {1, 2, 3};
Integer[] boxedArr = new Integer[arr.length];
for(int i = 0; i < arr.length; i++)
boxedArr[i] = arr[i];
System.out.println(Collections.frequency(Arrays.asList(boxedArr), 1));

You could create a List from the int[], but otherwise, you just have to write your own.
int[] l = //your data;
List<Integer> list = new List<Integer>();
for(int i : l)
list.add(i);
int o = Collections.frequency(list, 0);
Or Arrays.asList(l); to make it shorter.

int occurrences = Collections.frequency(Arrays.asList(list), 0);
Or if you are against converting it to a list:
int occurrences = 0;
for (int i = 0; i < list.length; i++)
{
if(list[i] == X) // X being your number to check
occurrences++;
}

You can do this way as well.
List<Integer> intList = Arrays.asList(new Integer [] {
2, 3, 4, 5, 6,
2, 3, 4, 5,
2, 3, 4,
2, 3,
2
});
System.out.println(" count " + Collections.frequency(intList, 6));