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Fill 2D array square and print in specific patterns

Ive been working on this assignment question for about 4 days now and I'm about to go crazy. We have specific directions to follow for this code;
"Your program should proceed as follows:
Display a welcome message.
Prompt the user for an integer which is 3. If the number entered is < 3. keep prompting the user until they enter a number 3 (use a do/while). This number will determine the size of the square array.
Fill the array as per pattern 1 and display it using printf to format the array.
Fill the same array as per pattern 2 and display it using printf t0 format the array.
Display a closing message."
Pattern:
I'm still stuck on pattern one. I tried to do first do a for loop which in it there's an if statement that checks if the column number is even or not, if it is, to print out the code backwards. The question also recommends using a while loop and a do/while loop...?
Also any tips on to how to go about the second patterns.
Here is my code.
import java.util.Scanner;
public class a3q33
{
public static void main(String[] args)
{
Scanner keyboard = new Scanner(System.in);
int n;
do
{
System.out.println("How many rows/columns do you want your array to have? (Must be at least 3)");
n=keyboard.nextInt();
} while(n < 3);
int [][] arr = new int [n][n];
int i, j, k=1;
for(i=0;i<n;i++)
{
if(i % 2 != 0)
{
for(j=n;j>0;j--)
{
k = k+n;
arr[i][j]=k;
k--;
}
}
else
{
for(j=0;j<n;j++)
{
arr[i][j]=k;
k++;
}
}
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
System.out.printf(arr[i][j]+" ");
}
System.out.printf("");
System.out.println();
}
}
}
Any help would be greatly appreciated!

I assume that you meant "row" instead of "column" for the condition check given that the convention is arr[row][column]?
Also, the for loop in your code:
for(j=n;j>0;j--)
This will decrement the row/column index down to 1 and not 0. So you will miss one element at arr[0][...] or arr[...][0].
Also, j=n will be out of bound.
Instead, try using:
for(j=n-1;j>-1;j--)
It would be a good first steps to look into.

import java.util.Scanner;
public class a3q33{
public static void main(String[] args){
Scanner keyboard = new Scanner(System.in);
int n;
do{
System.out.println("How many rows/columns do you want your array to have? (Must be at least 3)");
n=keyboard.nextInt();
} while(n < 3);
int count = 1;
int [][] arr = new int [n][n];
for(int i = 0;i<n; i++){
if(i%2==0){
for(int j = 0;j<n; j++){
arr [i][j] = count;
count++;
}
}
else{
for(int j = n-1;j>=0; j--){
arr [i][j] = count;
count++;
}
}
}
System.out.println("\nPattern 1 \n");
// print the array
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
System.out.printf("%d\t",arr[i][j]);
}
System.out.println();
}
// reset count back to 1 and fill the array in the same way only without the if/else
// for n = 5 for example this will produce [1, 2, 3, 4, 5 ]
// [6, 7, 8, 9, 10]
// [11,12, 13, 14, 15]
// [16,17, 18, 19, 20]
// [21,22, 23, 24, 25]
count = 1;
for(int i = 0;i<n; i++){
for(int j = 0;j<n; j++){
arr [i][j] = count;
count++;
}
}
// rotate arrays using arraycopy; for array at index 1 rotate by one, at index 2 by 2 and so on see
//http://stackoverflow.com/questions/31174840/how-to-rotate-an-array
//
for (int i = 1; i<n; i++){
int [] temp = new int [n];
for (int j = 0; j < n; j++) {
temp[(j + i) % n] = arr[i][j];
}
System.arraycopy(temp, 0, arr[i], 0, n);
arr[i]=temp;
}
System.out.println("\nPattern 2 \n");
// print the array
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
System.out.printf("%d\t",arr[i][j]);
}
System.out.println();
}
}
}

You can achieve this by checking the modulo:
for (int i = 0; i < n; i++) {
int start = i * n;
for (int j = 0; j < n; j++) {
arr[i][j] = i * n + ((i % 2 === 0) ? (j + 1) : (n - j));
}
}

java How do I print a reversed AND every other element in an array?

This is my first question, so sorry if I'm not clear
So I'm trying to write a void function printReverse that prints a reversed array and skips every other element in the process. This is what I have:
void printReverse(int[] array) {
for(int i = 0; i < array.length / 2; i+=2){
int temp = array[i];
array[i] = array[array.length - i - 1];
array[array.length - i - 1] = temp;
while(i < array.length){
println(array[i]);
i++;
}
}
}
Say that I had an array of {10, 20, 30, 40, 50, 60}, I'm trying to print out on different lines (println) 60 40 20
How would I do that with the code so far?

You can also start from the end of the array, it really easy to do this like that:
for(int i=array.length-1; i>=0; i=i-2)
System.out.println(array[i]);
notice that i=array.length-1 as if array.length is 3 for example, the last index will be 2. Also notice that i>=0 as i is initialize to the end of the array and 0 is the first index of the array.

Here you go:
int[] array = new int[] { 1, 2, 3, 4 };
int count = 0;
for (int i = array.length - 1; i >= 0; i--) {
if (count++ % 2 != 0) {
continue;
}
System.out.println(array[i]);
}

Besides having the system just print it, you could use a for loop to do this.
So,
for (int i = 6, i > array.length, i - 2)
{
System.out.println(array[i]);
}
Which would print out every other element in reverse order, given that there are 6 elements in the array. Of course, you could have a variable set to the length of the array, and plug in the variable instead of six, but this is just an example.
Hope this helps!

Check Image for Output Here you go in simple way,Thank you!!
import java.util.Arrays;
import java.util.Scanner;
class Demo
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
System.out.print("Enter total number of elemets: ");
int total = sc.nextInt();
int[]arr1 = new int[total];
int[]arr2 = new int[total];
System.out.print("Enter elemets: ");
for(int i = 0; i<total; i++)
{
arr1[i]= sc.nextInt();
}
System.out.print("Original Array: " +Arrays.toString(arr1));
System.out.println();
for(int a = 0 ; a< total; a++)
{
for(int j =total-a-1; j>=0; j--)
{
arr2[a]= arr1[j];
break;
}
}
System.out.print("Reversed Array: " +Arrays.toString(arr2));
}
}

How can we rotate an array to the left?

EX: I have an array {1, 2, 3, 4, 5} and an integer number 7
It will rotate 7 spaces to the right like: {4, 5, 1, 2, 3}
I also have that array {1, 2, 3, 4, 5} and an integer number -7
It will rotate 7 spaces to the left like: {3, 4, 5, 1, 2}
I have rotated the array to the right by using:
for(int i = 0; i < data.length; i++){
result[(i+n) % data.length ] = data[i];
}
But how can we rotate an array to the left?

Rotating to the left by n is the same as rotating to the right by length-n.
Rotate right (for positive n):
for(int i = 0; i < data.length; i++){
result[(i+n) % data.length ] = data[i];
}
Rotate left (for positive n):
for(int i = 0; i < data.length; i++){
result[(i+(data.length-n)) % data.length ] = data[i];
}
This way you can avoid a modulo of a negative number.
If you want to input an integer n that rotates right if n is positive and left if n is negative, you can do it like this:
int[] rotateArray(int n, int[] data)
{
if(n < 0) // rotating left?
{
n = -n % data.length; // convert to +ve number specifying how
// many positions left to rotate & mod
n = data.length - n; // rotate left by n = rotate right by length - n
}
int[] result = new int[data.length];
for(int i = 0; i < data.length; i++){
result[(i+n) % data.length ] = data[i];
}
return result;
}

In case rotate to the left, you can use this to avoid a modulo of a negative number:
int[] data = {1, 2, 3, 4, 5};
int[] result = new int[data.length];
for (int i = 0; i < data.length; i++) {
result[(i + (data.length - 2)) % data.length] = data[i];
}
for (int i : result) {
System.out.println(i);
}

You may also use linkedlist to achieve the same.
Integer[] arr = {1,2,3,4,5};
LinkedList<Integer> ns = new LinkedList<Integer>(Arrays.asList(arr));
int rotate=3;
if(rotate<0)
rotate += arr.length;
List<Integer> leftlist = ns.subList(0, rotate);
List<Integer> rightlist = ns.subList(rotate, arr.length);
LinkedList<Integer> result = new LinkedList<Integer>();
result.addAll(rightlist);
result.addAll(leftlist);
System.out.println(result);

If you want to rotate an array named a[n] with n as the size of an array and d as the number of rotation (in your case d=7), you can use this code:
d = d % n;
for(int i=d;i<n;i++){
System.out.print(a[i]+" ");
}
for(int i=0;i<(d);i++){
System.out.print(a[i]+" ");
}
This code will print the array elements in order after the left rotation is performed. If you want you can store it in another array, or you can show array elements as shown in the above method.

Here is the complete java program for n left rotations of an array.
public class ArrayRotation {
private static Scanner sc;
public static void main(String[] args) {
int n,k;
sc = new Scanner(System.in);
System.out.print("Enter the size of array: ");
n = sc.nextInt();
int[] a = new int[n];
System.out.print("Enter the "+n+" elements in the list: ");
for(int i=0;i<n;i++)
a[i] = sc.nextInt();
System.out.print("Enter the number of left shifts to array: ");
k = sc.nextInt();
System.out.print("Array before "+k+" shifts: ");
display(a);
solution(a,k);
System.out.println();
System.out.print("Array after "+k+" shifts: ");
display(a);
}
public static void solution(int[] a, int k){
int temp=0, j;
for(int i=0;i<k;i++){
temp = a[0];
// j=0; // both codes work i.e. for loop and while loop as well
// while(j<a.length-1){
// a[j]=a[j+1];
// j++;
// }
for(j=0;j<a.length-1;j++)
a[j]=a[j+1];
a[j]=temp;
}
}
public static void display(int[] a){
for(int i=0;i<a.length;i++)
System.out.print(a[i]+" ");
}
}

public static int[] arrayLeftRotation(int[] array, int elements, int rotations) {
// i = checks number of rotations
for (int i = 0; i < rotations; i++) {
int first = array[0];
int last = array[elements - 1];
// j = rotates each element
for (int j = 0; j < elements; j++) {
// check if at first index
if (j == 0) {
array[elements - 1] = first;
}
// check if at last index
if (j == (elements - 1)) {
// if at last index: make element in index before last = to last element
array[elements - 2] = last;
} else {
array[j] = array[j + 1];
}
}
}
return array;
}

This is the cleanest answer I could come up with.
Notice that you need to compute rotations % length so that your program is robust to a number of rotations greater than the size of your array.
static int[] rotateLeft(int[] arr, int rotations) {
int length = arr.length;
int[] result = new int[length];
rotations = rotations % length; // robust to rotations > length
int position = length - rotations; // compute outside the loop to increase performance
for (int i = 0; i < length; i++) {
result[(position + i) % length] = arr[i];
}
return result;
}

I used the below approach putting the array into a new array in its new position.
int length = a.length;
int[] newArr = new int[a.length];
for (int i = length - 1; i >= 0 ; i--){
int newPosition = i - d; // d is the number of rotation
if (newPosition < 0)
newPosition = length + newPosition;
newArr[newPosition] = a[i];
}
Initially i wanted to swap into the same array using the below approach but it did not work all the cases. For example where 2 array elements are being swapped due to rotation. So newPosition and position keep on pointing to each other. Hence rest of the array is not moved.
Not sure if it is possible to rotate array with O(n) complexity in the same array.
int length = a.length;
int position = length-1;
int temp = a[position];
for (int i = 0; i < length; i++){
int newPosition = position - d;
if (newPosition < 0)
newPosition = length + newPosition;
int cache = a[newPosition];
a[newPosition]= temp;
temp = cache;
position = newPosition;
}

Here is my solution in Javascript:
n is the number of elements and r is the number of rotations.
a is the array.
function rotate(a,r,n)}
for(var i = 0; i<r; i++){
var aa = array[0] //
for (var j = 0; j < n-1; j++){
array[j] = array[j+1] //
}
array[n-1] = aa //
}
return a
}
var array = [1,2,3,4,5];
console.log(rotate(array,1,5))

// Left rotation using lambda expression.
int [] ar = {1,6,7,5,9,1,34,7};
int shift =2;
int[] posAr = IntStream.range(0, ar.length).map(i->(ar[(i + (ar.length + shift)) % ar.length])).toArray();
//You can print the data
Arrays.stream(posAr).forEach(System.out::println);

enter image description here
Show code:
if(a.length==0 || a.length==1)
return a;
int r=d%a.length;
int A[]=new int[a.length];
for (int i = 0; i < a.length; i++) {
if((i+r) >= a.length)
A[i]=a[i+r-a.length];
else
A[i]=a[i+r];
}
return A;

Delete max element in array

Wrote some code to try to find the maximum element in an un-ordered array and then delete it from the array. My first loop has the logic to find the maximum element while the second loop should take in the variable from the first loop, look ahead one space and then insert into the max element.
My below code seems to work for my second idea .. but does not find the right max array element.
My array has the following values {6, 3, 5, 2, 7, 9, 4}. It is finding the max array element to be 7 .. it should be 9.
public void deleteMax() {
// set value for comparison starting from the beginning of the array
int arrayMax = arr[0];
for (int i = 0; i < nElems; i++) {
if (arr[i] > arrayMax) {
arrayMax = arr[i];
for (int k = i; k < nElems; k++) {
arr[k] = arr[k + 1];
nElems--;
break;
}
}
}
}

why not use jQuery $.grep & Math.max like:
var arr = [6, 3, 5, 2, 7, 9, 4];
var maxNum = Math.max.apply(null, arr);
var newArr = $.grep( arr, function( n ) {
return n != maxNum;
});
console.log(newArr);
Fiddle
EDIT:
Well didn't realize you're using Java as the question showed in JavaScript section...
in Java, You can find max number in array like
int maxNum = arr.get(0); // get the first number in array
for (int i = 1; i < arr.length; i++) {
if ( arr.get(i) > maxNum) {
maxNum = array.get(i);
}
}
arr.remove(maxNum);

Well,, you don't need any second loop.
Only one loop and two variables called, f.ex. MaxElementPosition and MaxElementValue, which are updated every time inside the loop if the number on this position is greater than the last MaxElementValue, update both value and position.
While the loop you do need the value for comparing. In the end you only need the position.

Your inner loop is irrelevant, you only have a 1D array, so it makes no sense to do any inner iteration.
If you insist on performing no sorting on the array, then you could do something like this:
public void deleteMax() {
// set value for comparison starting from the beginning of the array
int arrayMax = arr[0];
int maxIndex = 0;
for (int i = 0; i < nElems; i++) {
if (arr[i] > arrayMax) {
arrayMax = arr[i];
maxIndex = i;
}
}
arr.remove(maxIndex);
}
Once the for loop finishes, we remove the item at maxIndex

#Alex is on the right track, but there is no remove function that can be called on an array in java. All that you need is the variable maxIndex that he gets, which of course will be the first occurence of this maximum, so if you need to remove every occurence of the maximum, that would be a different issue. So once you use #Alex code:
int arrayMax = arr[0];
int maxIndex = 0;
for (int i = 0; i < nElems; i++) {
if (arr[i] > arrayMax) {
arrayMax = arr[i];
maxIndex = i;
}
}
This would be much easier if instead of an array, you were to use an ArrayList, in which case the code would look like:
int arrayMax = arr.get(0);
int maxIndex = 0;
for (int i = 0; i < nElems; i++) {
if (arr[i] > arrayMax) {
arrayMax = arr[i];
maxIndex = i;
}
}
arr.remove(maxIndex);
Otherwise you would have to create a temp array to remove the value:
int[] temp = new int[arr.length-1];
for(int i = 0, index = 0; index < nElems; i++, index++)
{
if(index == maxIndex) index++;
temp[i] = arr[index];
}
arr = temp;

Printing out the least occurring elements in an array

OK, so I found this question from a few days ago but it's on hold and it won't let me post anything on it.
***Note: The values or order in the array are completely random. They should also be able to be negative.
Someone recommended this code and was thumbed up for it, but I don't see how this can solve the problem. If one of the least occurring elements isn't at the BEGINNING of the array then this does not work. This is because the maxCount will be equal to array.length and the results array will ALWAYS take the first element in the code written below.
What ways are there to combat this, using simple java such as below? No hash-maps and whatnot. I've been thinking about it for a while but can't really come up with anything. Maybe using a double array to store the count of a certain number? How would you solve this? Any guidance?
public static void main(String[] args)
{
int[] array = { 1, 2, 3, 3, 2, 2, 4, 4, 5, 4 };
int count = 0;
int maxCount = 10;
int[] results = new int[array.length];
int k = 0; // To keep index in 'results'
// Initializing 'results', so when printing, elements that -1 are not part of the result
// If your array also contains negative numbers, change '-1' to another more appropriate
for (int i = 0; i < results.length; i++) {
results[i] = -1;
}
for (int i = 0; i < array.length; i++) {
for (int j = 0; j < array.length; j++) {
if (array[j] == array[i]) {
count++;
}
}
if (count <= maxCount) { // <= so it admits number with the SAME number of occurrences
maxCount = count;
results[k++] = array[i]; // Add to 'results' and increase counter 'k'
}
count = 0; // Reset 'count'
}
// Printing result
for (int i : results) {
if (i != -1) {
System.out.println("Element: " + i + ", Number of occurences: " + maxCount);
}
}
}
credit to: https://stackoverflow.com/users/2670792/christian
for the code
I can't thumbs up so I'd just like to say here THANKS EVERYONE WHO ANSWERED.

You can also use an oriented object approach.
First create a class Pair :
class Pair {
int val;
int occ;
public Pair(int val){
this.val = val;
this.occ = 1;
}
public void increaseOcc(){
occ++;
}
#Override
public String toString(){
return this.val+"-"+this.occ;
}
}
Now here's the main:
public static void main(String[] args) {
int[] array = { 1,1, 2, 3, 3, 2, 2, 6, 4, 4, 4 ,0};
Arrays.sort(array);
int currentMin = Integer.MAX_VALUE;
int index = 0;
Pair[] minOcc = new Pair[array.length];
minOcc[index] = new Pair(array[0]);
for(int i = 1; i < array.length; i++){
if(array[i-1] == array[i]){
minOcc[index].increaseOcc();
} else {
currentMin = currentMin > minOcc[index].occ ? minOcc[index].occ : currentMin;
minOcc[++index] = new Pair(array[i]);
}
}
for(Pair p : minOcc){
if(p != null && p.occ == currentMin){
System.out.println(p);
}
}
}
Which outputs:
0-1
6-1
Explanation:
First you sort the array of values. Now you iterate through it.
While the current value is equals to the previous, you increment the number of occurences for this value. Otherwise it means that the current value is different. So in this case you create a new Pair with the new value and one occurence.
During the iteration you will keep track of the minimum number of occurences you seen.
Now you can iterate through your array of Pair and check if for each Pair, it's occurence value is equals to the minimum number of occurences you found.
This algorithm runs in O(nlogn) (due to Arrays.sort) instead of O(n²) for your previous version.

This algorithm is recording the values having the least number of occurrences so far (as it's processing) and then printing all of them alongside the value of maxCount (which is the count for the value having the overall smallest number of occurrences).
A quick fix is to record the count for each position and then only print those whose count is equal to the maxCount (which I've renamed minCount):
public static void main(String[] args) {
int[] array = { 5, 1, 2, 2, -1, 1, 5, 4 };
int[] results = new int[array.length];
int minCount = Integer.MAX_VALUE;
for (int i = 0; i < array.length; i++) {
for (int j = 0; j < array.length; j++) {
if (array[j] == array[i]) {
results[i]++;
}
}
if (results[i] <= minCount) {
minCount = results[i];
}
}
for (int i = 0; i < results.length; i++) {
if (results[i] == minCount) {
System.out.println("Element: " + i + ", Number of occurences: "
+ minCount);
}
}
}
Output:
Element: 4, Number of occurences: 1
Element: 7, Number of occurences: 1
This version is also quite a bit cleaner and removes a bunch of unnecessary variables.

This is not as elegant as Iwburks answer, but I was just playing around with a 2D array and came up with this:
public static void main(String[] args)
{
int[] array = { 3, 3, 3, 2, 2, -4, 4, 5, 4 };
int count = 0;
int maxCount = Integer.MAX_VALUE;
int[][] results = new int[array.length][];
int k = 0; // To keep index in 'results'
for (int i = 0; i < array.length; i++) {
for (int j = 0; j < array.length; j++) {
if (array[j] == array[i]) {
count++;
}
}
if (count <= maxCount) {
maxCount = count;
results[k++] = new int[]{array[i], count};
}
count = 0; // Reset 'count'
}
// Printing result
for (int h = 0; h < results.length; h++) {
if (results[h] != null && results[h][1] == maxCount ) {
System.out.println("Element: " + results[h][0] + ", Number of occurences: " + maxCount);
}
}
Prints
Element: -4, Number of occurences: 1
Element: 5, Number of occurences: 1

In your example above, it looks like you are only using ints. I would suggest the following solution in that situation. This will find the last number in the array with the least occurrences. I assume you don't want an object-oriented approach either.
int [] array = { 5, 1, 2, 40, 2, -1, 3, 2, 5, 4, 2, 40, 2, 1, 4 };
//initialize this array to store each number and a count after it so it must be at least twice the size of the original array
int [] countArray = new int [array.length * 2];
//this placeholder is used to check off integers that have been counted already
int placeholder = Integer.MAX_VALUE;
int countArrayIndex = -2;
for(int i = 0; i < array.length; i++)
{
int currentNum = array[i];
//do not process placeholders
if(currentNum == placeholder){
continue;
}
countArrayIndex = countArrayIndex + 2;
countArray[countArrayIndex] = currentNum;
int count = 1; //we know there is at least one occurence of this number
//loop through each preceding number
for(int j = i + 1; j < array.length; j++)
{
if(currentNum == array[j])
{
count = count + 1;
//we want to make sure this number will not be counted again
array[j] = placeholder;
}
}
countArray[countArrayIndex + 1] = count;
}
//In the code below, we loop through inspecting each number and it's respected count to determine which one occurred least
//We choose Integer.MAX_VALUE because it's a number that easily indicates an error
//We did not choose -1 or 0 because these could be actual numbers in the array
int minNumber = Integer.MAX_VALUE; //actual number that occurred minimum amount of times
int minCount = Integer.MAX_VALUE; //actual amount of times the number occurred
for(int i = 0; i <= countArrayIndex; i = i + 2)
{
if(countArray[i+1] <= minCount){
minNumber = countArray[i];
minCount = countArray[i+1];
}
}
System.out.println("The number that occurred least was " + minNumber + ". It occured only " + minCount + " time(s).");