Develop Reference

Array not copying another array

I am trying to develop a program to delete all the median values from an array (the middle value if it has an odd number of elements, the two middle values if it has an even number of elements) until there are only two elements left, elements [0] and [1]. For example, if the user inputs 1, 2, 3, 4, 5, the program will return [1, 5]. I put down what I thought logically might help, but my array x isn't copying myArray in the for loops. I am not looking for someone to completely do the code for me, just to point out where I am wrong. Here is my code:
import java.util.*;
public class Deletion
{
public static void main(String[]args)
{
Scanner kb = new Scanner(System.in);
System.out.println("Please enter the array length:");
int [] myArray = new int[kb.nextInt()];
int [] x = new int[myArray.length - 1];
int index1 = 0;
int index2 = 0;
int index3 = 0;
if(myArray.length < 3)
{
System.out.println("Please make sure array length is greater than two. Run again.");
System.exit(0);
}
else
{
for(int i = 0; i < myArray.length; i++)
{
System.out.println("Please enter a number for position " + i + ":");
myArray[i] = kb.nextInt();
}
for(int i = 0; i < myArray.length; i++)
{
while(myArray.length > 2)
{
if(myArray.length%2 != 0)
{
index1 = (myArray.length/2);
for(int j = 0, r = 0; j < myArray.length; j++)
{
if(j != index1)
{
x[r++] = myArray[j];
myArray = x;
}
}
x = new int[myArray.length - 1];
}
else
{
index2 = (myArray.length/2);
index3 = (myArray.length/2 - 1);
for(int j = 0, r = 0; j < myArray.length; j++)
{
if(j != index2 && j != index3)
{
x[r++] = myArray[j];
myArray = x;
}
}
x = new int[myArray.length - 1];
}
}
}
}
System.out.println(Arrays.toString(myArray));
}
}

You must create the array and populate it, else it's using the same memory address, hence won't work. Use the following:
myArray = ArrayUtils.clone(x);

When you are doing do “myArray = x”, your are actually merely assigning a reference to the array. Hence, if you make any change to one array, it would be reflected in other arrays as well because both myArray and x are referring to the same location in memory.
Thus, what you need is
myArray = x.clone();

I cleaned up your code a bit. According to what you described, what really matters is pulling in the minimum and maximum values in the array - everything else will be deleted, so you simply need a single traversal through the array to find those two values.
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
boolean isValid = false;
int validLength = 0;
System.out.println("Please enter the array length:");
while (!isValid) {
int length = scanner.nextInt();
if (length < 3) {
System.out.println("Please make sure array length is greater than two. Try again.");
}
else {
isValid = true;
validLength = length;
}
}
int minimumValue = Integer.MAX_VALUE, maximumValue = Integer.MIN_VALUE;
for (int i = 0; i < validLength; i++) {
System.out.println("Please enter a number for position " + i + ":");
int nextInt = scanner.nextInt();
if (nextInt < minimumValue) minimumValue = nextInt;
else if (nextInt > maximumValue) maximumValue = nextInt;
}
System.out.println(Arrays.toString(new int[] {minimumValue, maximumValue}));
}
Edit: made another revision as using an array is unnecessary. Just keep track of the minimum and maximum values as they are being entered.

Sorting characters in a string using selection sort

What is the mistake in my code?
Given a string consisting of lowercase letters, arrange all its letters in ascending order.
Input: The first line of the input contains T, denoting number of testcases. Then follows description of each testcase. The first line of the testcase contains positive integer N denoting the length of string. The second line contains the string.
Output: For each testcase, output the sorted string.
Constraints:
1 <= T <= 100
1 <= N <= 100
import java.util.*;
class GFG {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int i = 1; i <= t; i++) {
int n = sc.nextInt();
sc.nextLine();
String S = sc.nextLine();
String sor = "";
for (int j = 0; j < n; j++) {
int min = j;
for (int k = j + 1; k < n; k++) {
if (S.charAt(k) > S.charAt(min)) {
min = k;
}
}
sor += S.substring(min, min + 1);
}
System.out.println(sor);
}
}
}
For Input:
1
5
edcab
Output:
edcbb
Expected Output:
abcde

import java.util.*;
class GFG {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int i = 1; i <= t; i++) {
int n = sc.nextInt();
sc.nextLine();
String S = sc.nextLine();
System.out.println("S: "+S);
String sor = "";
for (int j = 0; n > 0; j++) {
int min = 0;
for (int k = 0; k < n; k++) {
if (S.charAt(k) < S.charAt(min)) {
min = k;
}
}
sor += S.substring(min, min + 1);
S = S.substring(0, min) + S.substring(min + 1);
n--;
}
System.out.println(sor);
}
}
}
This code does what you want. I changed > to < and I removed every char that added to sorted String from unsorted String. This way we don't need to deal with the same char again and again.

You are not exchanging the place of the min character after finding it. But Strings in java are immutable so you can't swap the places of characters in it. I suggest you convert your String to a char[] so that you can swap the characters:
public static void main (String[] args){
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for(int i=1; i<=t; i++){
int n = sc.nextInt();
sc.nextLine();
String S = sc.nextLine().toCharArray(); // convert it to char array
char[] sor = new char[S.length];
for(int j=0; j<n; j++){
int min = j;
for(int k =j+1; k<n; k++){
if(S[k]<S[min]){
min = k;
}
}
swap(S, min, j);
sor[j] = S[min]
}
System.out.println(new String(sor));// reconvert to string
}
}
public static void swap(char[] c,int x,int y){
char temp= c[x];
c[x] = c[y];
c[y] = temp;
}

You can use String.toCharArray method to iterate over the array of characters char[] of this string, sort their decimal values and return back the string that contains the characters of the sorted array:
public static void main(String[] args) {
String str = "edcab";
String sorted = selectionSort(str.toCharArray());
System.out.println(sorted); // abcde
}
public static String selectionSort(char[] arr) {
// iterate over all subsets of the array
// (0-last, 1-last, 2-last, 3-last, ...)
for (int i = 0; i < arr.length; i++) {
// assume the min is
// the first element
char min = arr[i];
// index of the
// min element
int min_i = i;
// check the elements
// after i to find
// the smallest
for (int j = i + 1; j < arr.length; j++) {
// if this element
// is less, then it
// is the new min
if (arr[j] < min) {
min = arr[j];
min_i = j;
}
}
// if min element is not
// equal to the current
// one, then swap them
if (i != min_i) {
char temp = arr[i];
arr[i] = arr[min_i];
arr[min_i] = temp;
}
}
return String.valueOf(arr);
}
You can use String.codePoints method to iterate over int values of the characters of this string, sort them and collect another sorted string:
String str = "edcab";
String sorted = str.codePoints()
.sorted()
.mapToObj(Character::toString)
.collect(Collectors.joining());
System.out.println(sorted); // abcde
See also:
• How do I not take Special Characters in my Password Validation (without Regex)?
• Java Selection Sort

removing the duplicates from array

I have homework about arrays in Java and I am stuck on this question.
Fill in the body of the program below, which removes duplicate values from the sorted array input. Your solution should set the variable result to the number of values remaining after the duplicate values have been removed. For example, if input is (0,1,1,2,3,3,3,4,4), the first five values of input after removing the duplicates should be (0,1,2,3,4), and the value of result should be 5.
Here's my code:
import java.util.Scanner;
public class RemoveDups {
public static void main (String[] args) {
Scanner scan = new Scanner(System.in);
int[] input = 0,1,1,2,3,3,3,4,4;
int result;
int count = input[0];
result++;
String count1="";
int result2=0;
count1=count1+count;
input[0]=Integer.parseInt(count1);
count1="";
for (int j = 1; j <input.length-1;j++ ) {
if (count != input[j+1] && result2 == 0||count != input[j-1] &&result2==0 ) {
input[j] = count;
result++;
count = input[j + 1];
count1=count1+count;
input[j]=Integer.parseInt(count1);
count1="";
}
}
for (int i = 0; i < result; i++) {
System.out.println(input[i]);
}
}
}
}
I can't do this exercise. i have left always the last cell in array that is different from all another cells and this code not working for me.

public static int removeDuplicateElements(int arr[], int n){
if (n==0 || n==1){
return n;
}
int j = 0;
for (int i=0; i < n-1; i++){
if (arr[i] != arr[i+1]){
arr[j++] = arr[i];
}
}
arr[j++] = arr[n-1];
return j;
}
public static void main(String args []) {
int arr[] = {0,1,1,2,3,3,3,4,4};
int length = arr.length;
length = removeDuplicateElements(arr, length);
for (int i=0; i<length; i++)
System.out.print(arr[i]+" ");
}
Answer will be 0 1 2 3 4
Please refer following link.
Remove Duplicate Element in Array using separate index

I am not sure if you needed a filtered array or just the result value. The below will give you result value.
Since this is homework, I suggest you work on the below logic to create the non duplicate array.
int result = 1;
if(input == null || input.length == 0){
result = 0;
}
else{
for(int i = 1; i < input.length; i++){
if(input[i-1] != input[i]){
result++;
}
}
}

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));
}
}