Develop Reference

how to reuse elements of an array out of its for loop?

I was trying to create a code that rearranges given elements in an array -by the user- in ascending order, and I have done that, but the program requires
printing the given elements after sorting them firstly, then printing them before sorting.
I have no problem with printing the elements after sorting
the problem is with printing them before sorting
how to re-use ar[S] = in.nextInt() the given elements by the user out of its for loop
import java.util.*;
public class SortingnumbersANDswapping {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int swap;
int ar[] = new int[3]; //{8,10,5}
for (int S = 0; S < ar.length; S++) {
ar[S] = in.nextInt(); //this for loop is used to store numbers in the array
}
for (int i = 0; i < ar.length; i++) {
/* this nested for loop is used to compare the first element with the second one in the array
or the second element with the third.
*/
for (int j = i + 1; j < ar.length; j++) {
if (ar[i] > ar[j]) { //8>10-->F , 8>5 -->T , {5,10,8} the new arrangment we are going to use
swap = ar[i]; // 10>8-->T {5,8,10}
ar[i] = ar[j];
ar[j] = swap;
}
}
System.out.println(ar[i]); // to print the new order print it inside the array
}
// I wanna do something like that
// System.out.println(ar[S]);
// but of course I cant cause array S is only defined in it's loop
}
}

You can't reuse it, you need to keep the original array stored in another array that stays untouched. Additionally, Arrays are usually declared as int[] ar in Java, instead of int ar[]. Something along the following lines should work as intended:
public class SortingnumbersANDswapping {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int swap;
int[] ar = new int[3]; //{8,10,5}
for (int S = 0; S < ar.length; S++) {
ar[S] = in.nextInt(); //this for loop is used to store numbers in the array
}
System.out.println("::: Original Array :::");
int[] originalArray = Arrays.copyOf(ar, ar.length);
for (int j : originalArray) {
System.out.println(j);
}
System.out.println("::: Sorted Array :::");
for (int i = 0; i < ar.length; i++) {
for (int j = i + 1; j < ar.length; j++) {
if (ar[i] > ar[j]) { //8>10-->F , 8>5 -->T , {5,10,8} the new arrangment we are going to use
swap = ar[i]; // 10>8-->T {5,8,10}
ar[i] = ar[j];
ar[j] = swap;
}
}
System.out.println(ar[i]); // to print the new order print it inside the array
}
}
}

You can do a copy of the array using the copyOf method from the Arrays library (java.util.Arrays) before you start changing the array.
Here you can find some different approaches - https://www.softwaretestinghelp.com/java-copy-array/amp/

You can use System.out.print(Arrays.toString(arr)); to print the entire array.
public static void main(String... args) {
int[] arr = readArray(3);
System.out.print(Arrays.toString(arr));
for (int i = 0; i < arr.length; i++) {
for (int j = i + 1; j < ar.length; j++) {
if (arr[i] > arr[j]) {
swap(arr, i, j);
System.out.print(Arrays.toString(arr));
}
}
}
}
private static int[] readArray(int size) {
System.out.format("Enter %d elements:\n", size);
Scanner scan = new Scanner(System.in);
int[] arr = new int[size];
for(itn i = 0; i < arr.length; i++)
arr[i] = scan.nextInt();
return arr;
}
private static void swap(int[] arr, int i, int j) {
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}

Compressing array values into another array

I came across this problem called compress. The objective is to take an array, remove any repeated values and return a new array.
I know this would be very easy with ArrayLists, but I want to do it without them.
So far, I've just written a loop to determine the number of unique values so I can construct a new array of the appropriate length. How can I then get the unique values into the new array?
public static int[] compress(int[] array){
int length = 0;
boolean contains = false;
for (int i = 0; i < array.length; i++){
contains = false;
for (int j = 0; j < i; j++){
if (a[i] == a[j]){
contains = true;
j = i;
} else {
contains = false;
}
}
if (!contains){
length++;
}
}
int[] uniqueArray = new int[length];
}

Not Tested but I think this should do the trick.
public static int[] copyArray(int [] num){
int x = 0;
int numDuplicate = 0;
int[] copy = new int[num.length]; // we use this to copy the non duplicates
HashMap<Integer, Integer> count = new HashMap<>(); //hashmap to check duplicates
for(int i = 0; i < num.length; i++){
if(count.containsKey(num[i])){
count.put(num[i], count.get(num[i])+1);
numDuplicate++; // keep track of duplicates
}else{
count.put(num[i], 1); // first occurence
copy[x] = num[i]; // copy unique values, empty values will be at end
x++;
}
}
// return only what is needed
int newSize = num.length - numDuplicate;
int[] copyNum = new int[newSize];
for(int i = 0; i < copyNum.length; i++){
copyNum[i] = copy[i];
}
return copyNum;
}
public static void main(String[] args) {
// sample elements
int[] nums = new int[20];
for(int i = 0; i < nums.length; i++){
nums[i] = (int)(Math.random() * 20);
}
System.out.println(Arrays.toString(nums));
System.out.println(Arrays.toString(copyArray(nums)));
}

2-dimensional array with given number [duplicate]

Here is the code I have so far:
public static int mode(int[][] arr) {
ArrayList<Integer> list = new ArrayList<Integer>();
int temp = 0;
for(int i = 0; i < arr.length; i ++) {
for(int s = 0; s < arr.length; s ++) {
temp = arr[i][s];
I seem to be stuck at this point on how to get [i][s] into a single dimensional array. When I do a print(temp) all the elements of my 2D array print out one a time in order but cannot figure out how to get them into the 1D array. I am a novice :(
How to convert a 2D array into a 1D array?
The current 2D array I am working with is a 3x3. I am trying to find the mathematical mode of all the integers in the 2D array if that background is of any importance.

In Java 8 you can use object streams to map a matrix to vector.
Convert any-type & any-length object matrix to vector (array)
String[][] matrix = {
{"a", "b", "c"},
{"d", "e"},
{"f"},
{"g", "h", "i", "j"}
};
String[] array = Stream.of(matrix)
.flatMap(Stream::of)
.toArray(String[]::new);
If you are looking for int-specific way, I would go for:
int[][] matrix = {
{1, 5, 2, 3, 4},
{2, 4, 5, 2},
{1, 2, 3, 4, 5, 6},
{}
};
int[] array = Stream.of(matrix) //we start with a stream of objects Stream<int[]>
.flatMapToInt(IntStream::of) //we I'll map each int[] to IntStream
.toArray(); //we're now IntStream, just collect the ints to array.

You've almost got it right. Just a tiny change:
public static int mode(int[][] arr) {
List<Integer> list = new ArrayList<Integer>();
for (int i = 0; i < arr.length; i++) {
// tiny change 1: proper dimensions
for (int j = 0; j < arr[i].length; j++) {
// tiny change 2: actually store the values
list.add(arr[i][j]);
}
}
// now you need to find a mode in the list.
// tiny change 3, if you definitely need an array
int[] vector = new int[list.size()];
for (int i = 0; i < vector.length; i++) {
vector[i] = list.get(i);
}
}

I'm not sure if you're trying to convert your 2D array into a 1D array (as your question states), or put the values from your 2D array into the ArrayList you have. I'll assume the first, but I'll quickly say all you'd need to do for the latter is call list.add(temp), although temp is actually unneeded in your current code.
If you're trying to have a 1D array, then the following code should suffice:
public static int mode(int[][] arr)
{
int[] oneDArray = new int[arr.length * arr.length];
for(int i = 0; i < arr.length; i ++)
{
for(int s = 0; s < arr.length; s ++)
{
oneDArray[(i * arr.length) + s] = arr[i][s];
}
}
}

change to:
for(int i = 0; i < arr.length; i ++) {
for(int s = 0; s < arr[i].length; s ++) {
temp = arr[i][s];

"How to convert a 2D array into a 1D array?"
String[][] my2Darr = .....(something)......
List<String> list = new ArrayList<>();
for(int i = 0; i < my2Darr.length; i++) {
list.addAll(Arrays.asList(my2Darr[i])); // java.util.Arrays
}
String[] my1Darr = new String[list.size()];
my1Darr = list.toArray(my1Darr);

I know its already been answered but here is my take. This function will take a 2d array input and return a 1d array output.
public int[] output(int[][] input){
int[] out = new int[input.length * input[0].length]
for (int i = 0; i < input.length; i++) {
for (int j = 0; j < input[i].length; j++) {
out[i + (j * input.length)] = input[i][j]
}
}
return out;
}

System.arraycopy should be faster than anything we can write. Also use the built-in java iterator on rows. Here is an example for double arrays. You should be able to use any type or class. If your rows are all the same length, then totalNumberElements = array2D.length * array2D[0].length;
static double[] doubleCopyToOneD(double[][] array2D, int totalNumberElements) {
double[] array1D = new double[totalNumberElements];
int pos = 0;
for (double[] row: array2D) {
System.arraycopy(row, 0, array1D, pos, row.length);
pos += row.length;
}
return array1D;
}

import java.util.*;
public class Main {
public static int A[][] = new int[3][3];
public static int B[] = new int[9];
public static void main(String[] args) {
int temo = 0,t;
Scanner s = new Scanner(System.in);
System.out.println("Enter No for Matrix A");
for (int row = 0; row < A.length; row++) {
for (int col = 0; col < A.length; col++) {
A[row][col] = s.nextInt();
}
System.out.print("\n");
}
for (int row = 0; row < A.length; row++) {
for (int col = 0; col < A.length; col++) {
t= A[row][col];
B[temo]= t;
temo++;
}
System.out.print("\n");
}
System.out.print("After Converted to one d \n");
for(int i =0;i<B.length;i++) {
System.out.print(" "+B[i]+" ");
}
}
}

Finding multiple modes in an array of integers with 1000 elements

So I need a way to find the mode(s) in an array of 1000 elements, with each element generated randomly using math.Random() from 0-300.
int[] nums = new int[1000];
for(int counter = 0; counter < nums.length; counter++)
nums[counter] = (int)(Math.random()*300);
int maxKey = 0;
int maxCounts = 0;
sortData(array);
int[] counts = new int[301];
for (int i = 0; i < array.length; i++)
{
counts[array[i]]++;
if (maxCounts < counts[array[i]])
{
maxCounts = counts[array[i]];
maxKey = array[i];
}
}
This is my current method, and it gives me the most occurring number, but if it turns out that something else occurred the same amount of times, it only outputs one number and ignore the rest.
WE ARE NOT ALLOWED TO USE ARRAYLIST or HASHMAP (teacher forbade it)
Please help me on how I can modify this code to generate an output of array that contains all the modes in the random array.
Thank you guys!
EDIT:
Thanks to you guys, I got it:
private static String calcMode(int[] array)
{
int[] counts = new int[array.length];
for (int i = 0; i < array.length; i++) {
counts[array[i]]++;
}
int max = counts[0];
for (int counter = 1; counter < counts.length; counter++) {
if (counts[counter] > max) {
max = counts[counter];
}
}
int[] modes = new int[array.length];
int j = 0;
for (int i = 0; i < counts.length; i++) {
if (counts[i] == max)
modes[j++] = array[i];
}
toString(modes);
return "";
}
public static void toString(int[] array)
{
System.out.print("{");
for(int element: array)
{
if(element > 0)
System.out.print(element + " ");
}
System.out.print("}");
}

Look at this, not full tested. But I think it implements what #ajb said:
private static int[] computeModes(int[] array)
{
int[] counts = new int[array.length];
for (int i = 0; i < array.length; i++) {
counts[array[i]]++;
}
int max = counts[0];
for (int counter = 1; counter < counts.length; counter++) {
if (counts[counter] > max) {
max = counts[counter];
}
}
int[] modes = new int[array.length];
int j = 0;
for (int i = 0; i < counts.length; i++) {
if (counts[i] == max)
modes[j++] = array[i];
}
return modes;
}
This will return an array int[] with the modes. It will contain a lot of 0s, because the result array (modes[]) has to be initialized with the same length of the array passed. Since it is possible that every element appears just one time.
When calling it at the main method:
public static void main(String args[])
{
int[] nums = new int[300];
for (int counter = 0; counter < nums.length; counter++)
nums[counter] = (int) (Math.random() * 300);
int[] modes = computeModes(nums);
for (int i : modes)
if (i != 0) // Discard 0's
System.out.println(i);
}

Your first approach is promising, you can expand it as follows:
for (int i = 0; i < array.length; i++)
{
counts[array[i]]++;
if (maxCounts < counts[array[i]])
{
maxCounts = counts[array[i]];
maxKey = array[i];
}
}
// Now counts holds the number of occurrences of any number x in counts[x]
// We want to find all modes: all x such that counts[x] == maxCounts
// First, we have to determine how many modes there are
int nModes = 0;
for (int i = 0; i < counts.length; i++)
{
// increase nModes if counts[i] == maxCounts
}
// Now we can create an array that has an entry for every mode:
int[] result = new int[nModes];
// And then fill it with all modes, e.g:
int modeCounter = 0;
for (int i = 0; i < counts.length; i++)
{
// if this is a mode, set result[modeCounter] = i and increase modeCounter
}
return result;

THIS USES AN ARRAYLIST but I thought I should answer this question anyways so that maybe you can use my thought process and remove the ArrayList usage yourself. That, and this could help another viewer.
Here's something that I came up with. I don't really have an explanation for it, but I might as well share my progress:
Method to take in an int array, and return that array with no duplicates ints:
public static int[] noDups(int[] myArray)
{
// create an Integer list for adding the unique numbers to
List<Integer> list = new ArrayList<Integer>();
list.add(myArray[0]); // first number in array will always be first
// number in list (loop starts at second number)
for (int i = 1; i < myArray.length; i++)
{
// if number in array after current number in array is different
if (myArray[i] != myArray[i - 1])
list.add(myArray[i]); // add it to the list
}
int[] returnArr = new int[list.size()]; // create the final return array
int count = 0;
for (int x : list) // for every Integer in the list of unique numbers
{
returnArr[count] = list.get(count); // add the list value to the array
count++; // move to the next element in the list and array
}
return returnArr; // return the ordered, unique array
}
Method to find the mode:
public static String findMode(int[] intSet)
{
Arrays.sort(intSet); // needs to be sorted
int[] noDupSet = noDups(intSet);
int[] modePositions = new int[noDupSet.length];
String modes = "modes: no modes."; boolean isMode = false;
int pos = 0;
for (int i = 0; i < intSet.length-1; i++)
{
if (intSet[i] != intSet[i + 1]) {
modePositions[pos]++;
pos++;
}
else {
modePositions[pos]++;
}
}
modePositions[pos]++;
for (int modeNum = 0; modeNum < modePositions.length; modeNum++)
{
if (modePositions[modeNum] > 1 && modePositions[modeNum] != intSet.length)
isMode = true;
}
List<Integer> MODES = new ArrayList<Integer>();
int maxModePos = 0;
if (isMode) {
for (int i = 0; i< modePositions.length;i++)
{
if (modePositions[maxModePos] < modePositions[i]) {
maxModePos = i;
}
}
MODES.add(maxModePos);
for (int i = 0; i < modePositions.length;i++)
{
if (modePositions[i] == modePositions[maxModePos] && i != maxModePos)
MODES.add(i);
}
// THIS LIMITS THERE TO BE ONLY TWO MODES
// TAKE THIS IF STATEMENT OUT IF YOU WANT MORE
if (MODES.size() > 2) {
modes = "modes: no modes.";
}
else {
modes = "mode(s): ";
for (int m : MODES)
{
modes += noDupSet[m] + ", ";
}
}
}
return modes.substring(0,modes.length() - 2);
}
Testing the methods:
public static void main(String args[])
{
int[] set = {4, 4, 5, 4, 3, 3, 3};
int[] set2 = {4, 4, 5, 4, 3, 3};
System.out.println(findMode(set)); // mode(s): 3, 4
System.out.println(findMode(set2)); // mode(s): 4
}

There is a logic error in the last part of constructing the modes array. The original code reads modes[j++] = array[i];. Instead, it should be modes[j++] = i. In other words, we need to add that number to the modes whose occurrence count is equal to the maximum occurrence count

how to get the most common character in an array?

Suppose I have an integer array like this:
{5,3,5,4,2}
and I have a method which returns the most common character
public int highestnumber(String[] num) {
int current_number = Integer.parseInt(num[0]);
int counter = 0;
for (int i = 1; i < num.length; ++i) {
if (current_number == Integer.parseInt(num[i])) {
++counter;
} else if (counter == 0) {
current_number = Integer.parseInt(num[i]);
++counter;
} else {
--counter;
}
}
return current_number;
}
but if I have multiple common character then i need to get the number which is closest to one(1), like if i have an array like this:
{5,5,4,4,2};
then the method should return 4, what should I do for this?

As per what I understand your question,
What you have to done is,
1. Create ArrayList from your int[]
2. Use HashMap for find duplicates, which one is unique
3. Sort it as Ascending order,
4. First element is what you want..
EDIT: Answer for your question
int[] arr = {5, 4, 5, 4, 2};
ArrayList<Integer> resultArray = new ArrayList<Integer>();
Set<Integer> set = new HashSet<Integer>();
for (int i = 0; i < arr.length; i++)
{
if (set.contains(arr[i]))
{
System.out.println("Duplicate value found at index: " + i);
System.out.println("Duplicate value: " + arr[i]);
resultArray.add(arr[i]);
}
else
{
set.add(arr[i]);
}
}
Collections.sort(resultArray);
for (int i = 0; i < resultArray.size(); i++)
{
Log.e("Duplicate Values:", resultArray.get(i) + "");
}
Your need is,
int values = resultArray.get(0);

Sort the array then count runs of values.

Fast way.
Create a counter int array one element for each number. Go through the array once and increment corresponding counter array for each number. Set highest number to first counter element then go through and change highest number to current element only if it is bigger than highest number, return highest number.
public int highestNumber(String[] num){
int[] count = new int[10];
int highest_number = 0;
int highest_value = 0;
for(int i = 0; i < num.length; i++)
count[Integer.parseInt(num[i])]++;;
for(int i = 0; i < count.length; i++)
if(count[i] > highest_value){
highest_number = i;
highest_value = count[i];
}
return highest_number;
}
10x slower but without other array.
Create three ints one for number and two for counting. Go through the array once for each int and increment current counting each time it shows up, if bigger that highest count, set to highest count and set highest number to current count. Return highest number.
public int highestNumber(String[] num){
int highest_number = 0;
int highest_value = 0;
int current_value = 0;
for(int i = 0; i < 10; i++){
for(int j = 0; j < num.length; j++)
if(i == Integer.parseInt(num[j]))
current_value++;
if(current_value > highest_value){
highest_value = current_value;
highest_number = i;
}
current_value = 0;
}
return highest_number;
}
The first is obviously much faster but if for whatever reason you don't want another array the second one works too.

You can also try this:
import java.util.TreeMap;
public class SmallestFrequentNumberFinder {
public static int[] stringToIntegerArray(String[] stringArray) {
int[] integerArray = new int[stringArray.length];
for (int i = 0; i < stringArray.length; i++) {
integerArray[i] = Integer.parseInt(stringArray[i]);
}
return integerArray;
}
public static int getSmallestFrequentNumber(int[] numbers) {
int max = -1;
Integer smallestFrequentNumber = null;
TreeMap<Integer, Integer> frequencyMaper = new TreeMap<Integer, Integer>();
for (int number : numbers) {
Integer frequency = frequencyMaper.get(number);
frequencyMaper.put(number, (frequency == null) ? 1 : frequency + 1);
}
for (int number : frequencyMaper.keySet()) {
Integer frequency = frequencyMaper.get(number);
if (frequency != null && frequency > max) {
max = frequency;
smallestFrequentNumber = number;
}
}
return smallestFrequentNumber;
}
public static void main(String args[]) {
String[] numbersAsString = {"5", "5", "4", "2", "4", "4", "2", "2"};
final int[] integerArray = stringToIntegerArray(numbersAsString);
System.out.println(getSmallestFrequentNumber(integerArray));
}
}