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);
}
}
}
Related
I have a function public static void sortedlist(int[] l, int r) that takes an array int[] l and returns a new array where every non-negative element in the list would be added with every element until the rth element.
So here is an example.
Lets say we have l = {1, 2, -3, 4, 5, 4}, and r = 3. In this case, we would:
Replace l[0] with l[0] + l[1] + l[2] + l[3].
Replace l[1] with l[1] + l[2] + l[3] + l[4].
Not do anything to l[2] because it is negative.
Replace l[3] with l[3] + l[4] + l[5]. (We can't go further than the end of the array.)
Replace l[4] with l[4] + l[5].
Not change the value of a[5] because there are no values after l[5]. So the sum is l[5] itself.
Thus, the result after calling `sortedlist` would be {4, 8, -3, 13, 9, 4}.
Here is my code so far:
public class Practice2 {
public static void sortedlist(int[] l, int r) {
int[] A;
int sum = 0;
for (int i = 0; i < l.length + r; i+=r) {
sum = sum +=
}
}
}
As you can see, I'm not done with the code because I'm stumped on how am I supposed to move forward from this point.
What I'm trying to do is create a new Array A and then add the new values that I've received from sum into Array A.
Any help would be greatly appreciated. Furthermore, if you could explain the logic behind a working code would be extremely beneficial for me as well.
Thank you :)
Try this.
public static void sortedlist(int[] l, int r) {
for (int i = 0, max = l.length; i < max; ++i)
if (l[i] >= 0)
for (int j = i + 1; j <= i + r && j < max; ++j)
l[i] += l[j];
}
and
int[] a = {1, 2, -3, 4, 5, 4};
sortedlist(a, 3);
System.out.println(Arrays.toString(a));
output:
[4, 8, -3, 13, 9, 4]
Please find the solution below and i have also provided some explanation regarding the logic behind it.
Note: I have unit tested it for few cases and it seems working fine.
1) r is less than array length
2) r is equals to array length
3) r is greater than array length
public class Test {
public static void main(String[] args) {
int[] input = new int[]{1, 2, -3, 4, 5, 4};
int r = 3;
sortedlist(input,r);
}
public static void sortedlist(int[] l, int r) {
List<Integer> list = new ArrayList<>();
int itr = 0;
for(int i = itr; i < l.length ; i++){//This loop is for iterating over the given array
int temp = 0;
int itr2 = Math.min(itr + r, l.length-1);//This function takes the minimum value and it helps when the (itr+r) > l.length, it will consider the (l.length-1)
if(l[i] > 0){// checking whether the selected value is -ve or not
for(int j = i; j <= itr2 ; j++){ // This loop is for calculating the addition over the selected range
temp = temp + l[j];
}
} else {// if it is-ve, assigning the same value to temp
temp = l[i];
}
list.add(temp);// storing the calculated value in a list of integers
itr++; // incrementing the main loop iterator
}
System.out.println(list);
}
}
Output:
[4, 8, -3, 13, 9, 4]
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;
}
Given an array of ints length 3, return an array with the elements "rotated left" so {1, 2, 3} yields {2, 3, 1}.
My first attempt(I had done this in python easily so I had the same idea).
public int[] rotateLeft3(int[] nums) {
return [nums[1:] + nums[0]];
}
But as you expected I got an error, so I immediately wrote this.
public int[] rotateLeft3(int[] nums) {
int[] answer = new int[3];
answer[0] = nums[1];
answer[1] = nums[2];
answer[2] = nums[0];
return answer;
}
I feel like this is possibly the most inefficient way to answer the question, but I only did this because it says of length 3. My previous code works in python for all sizes. So I was wondering how would my previous code be written in java?
with collection:
Java - Rotating array
use Integer nums[];
if not, iterate and convert your int[] to Integer[]
or use http://commons.apache.org/lang/ :
Integer[] array2= ArrayUtils.toObject(nums);
Collections.rotate(Arrays.asList(nums), -1);
convert to array : .toArray();
Maybe...
public int[] rotateLeft(int size, int[] array) {
int temp = int[size-1]; // We're just gonna save the las value.
for (int i = size-1; i == 0; i--) {
array[i] = array[i-1]; // We move them all 1 to the left.
}
array[0] = temp; // This is why we saved the las value.
return array;
Hope it helped you!
You can do that easily with System.arraycopy:
int[] arr = { 1, 2, 3};
int[] rotated = new int[arr.length];
System.arraycopy(arr, 1, rotated, 0, arr.length - 1);
rotated[arr.length-1] = arr[0];
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.
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));