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Highest Value Palindrome

Given a string representing the starting number and a maximum number of changes allowed, create the largest palindromic string of digits possible or the string -1 if it's impossible to create a palindrome under the contstraints.
I wrote a code who answer on the questions, but i have an error that i dont know where it is, or if even the code work.
static String highestValuePalindrome(String s, int n, int k) {
for(int i =0 ; i < n ; i++){
char[] ch =s.toCharArray();
if(n==1)
return s ;
else if ((ch[i] != ch[n-i-1]) && (k != 0) ){
ch[i] = ch[n-i-1] = 9 ;
k--;
}
}
String str = new String(ch);
return str ;
}
Output Format
Print a single line with the largest number that can be made by changing no more than digits. If this is not possible, print -1.
Sample Input
n=4, k=1
3943
Sample Output
3993
Sample Input
n=6, k=3
092282
Sample Output
992299
Sample Input
n=4, k=1
0011
Sample Output
-1

First of all there is no need to pass n as a parameter because it's just the length of the string. Secondly, this is not the complete program. I have made many changes to the given code.
public class largestPalindorme {
public static void main(String[] args) {
System.out.println(highestValuePalindrome("0011", 1));
}
static String highestValuePalindrome(String s, int k) {
char[] ch = s.toCharArray();
int n = s.length(); // which is same as n which you passed as parameter
int minChangesRequired = MinumumChangesNeeded(s);
//if the changes required to make the given string a palindrome is less then k then only it will go inside or it will return -1
if (k >= minChangesRequired) {
int diff = 0;
if (k > minChangesRequired) {
diff = k - minChangesRequired;
for (int l = 0; l < diff; l++) {
ch[l] = '9';
ch[n - l - 1] = '9';
}
}
for (int i = diff; i < n - diff / 2; i++) {
if (ch[i] != ch[n - i - 1]) {
//if checks which number is greater
int greater = Integer.parseInt(String.valueOf(ch[i])) > Integer.parseInt(String.valueOf(ch[n - i - 1])) ? Integer.parseInt(String.valueOf(ch[i])) : Integer.parseInt(String.valueOf(ch[n - i - 1]));
//replaces the smaller number from the greater number.
if (Integer.parseInt(String.valueOf(ch[i])) != greater) {
ch[i] = ch[n - i - 1];
} else {
ch[n - i - 1] = ch[i];
}
}
}
String str = new String(ch);
return str;
}
return "-1";
}
//this function returns the minimum changes we need to do to make it a palindrome.
public static int MinumumChangesNeeded(String s) {
int count = 0;
char[] ch = s.toCharArray();
int n = s.length();
for (int i = 0; i < n / 2; i++) {
if (ch[i] != ch[n - i - 1]) {
count++;
}
}
return count;}}

Java output in lines of ten

I'm in a beginners java class and I have a quick question about the output statement on my array problem for week 5. So basically I have the core of the program down, but I'm supposed to output the result in lines of ten. I for some reason can not get it to work even with looking at similar posts on here. I'm a beginner and am pretty slow at putting 2 and 2 together when it comes to programming. Once I see it I have that ah-ha! moment and that's how this whole class has gone. I know I have to use the modulus, but in my trial and error I lost my way and have probably done more damage than good. Help would be appreciated.
Here is what I have and as you can tell I was trying something without modulus:
import java.util.*;
public class ArrayLoop
{
public static void main(String args[])
{
double alpha[] = new double[50];
*//Initialize the first 25 elements of the array (int i=0; i<25; i++)//*
for(int i = 0; i < 25; i++)
{
alpha[i]= i * i;
}
*//Initialize the last 25 elements of the array (i=25; i<50; i++)//*
for(int i = 25; i < 50; i++)
{
alpha[i]= 3 * i;
}
*//Print the element of the array*
System.out.println ( "The values are: " );
for (int i = 0; i < 50; i++)
System.out.println ( alpha[i] );
}
*//Print method to display the element of the array*
void print(double m_array[])
{
for(int i = 1; i < m_array.length; i++)
{
if(i % 10 == 0){;
System.out.println();
}else{
System.out.print(" ");
}
}
if (m_array.length % 10 != 0) {
System.out.println();
}
}
}

Um .. this isn't eloquent in the least but I tried to make the fewest changes to your existing code sample.
public class ArrayLoop {
public static void main(String args[]) {
double alpha[] = new double[50];
for (int i = 0; i < 25; i++) {
alpha[i] = i * i;
}
for (int i = 25; i < 50; i++) {
alpha[i] = 3 * i;
}
System.out.println("The values are: ");
for (int i = 0; i < 50; i++) {
System.out.print(alpha[i] + " ");
}
System.out.println();
System.out.println();
for (int i = 1; i < alpha.length; i++) {
if (i != 1 && i % 10 == 0) {
System.out.print(alpha[i - 1] + " ");
System.out.println();
} else {
System.out.print(alpha[i - 1] + " ");
}
}
System.out.print(alpha[49]);
}
}
Edit: A better condition would be ...
for (int i = 0; i < alpha.length; i++) {
if (i > 0 && i % 10 == 9) {
System.out.print(alpha[i] + " ");
System.out.println();
} else {
System.out.print(alpha[i] + " ");
}
}

You have to print the number first then decide whether to print space or newline by checking the modulus:
int arr[] = new int[50];
// Initialize array here
for (int i = 0; i < arr.length; i++) {
System.out.print(arr[i]);
if (i > 0 && (i + 1) % 10 == 0) {
System.out.println();
} else {
System.out.print(" ");
}
}

You have a couple of % 10 snippets in your code so I'm not entirely certain how that's "trying something without modulus" :-)
Having said that, modulus is exactly what you need, as per the following psuedo-code:
count = 0
for each item in list:
if count > 0 and (count % 10) == 0:
print end of line
print item
print end of line
In Java, you would use something like:
public class Test {
static public void main(String args[]) {
for (int i = 0; i < 24; i++) {
if ((i > 0) &&((i % 10) == 0)) {
System.out.println();
}
System.out.print ("" + i * 3 + " ");
}
System.out.println();
}
}
In other words, immediately before you print an item, check to see if it should be on the next line and, if so, output a newline before printing it.
Note that arrays in Java are zero based, so you need to start with an index of zero rather than one in your loops.
Now that's pretty close to what you have so you're on the right track but, for the life of me, I cannot see in your print() method where you actually print the item! That should be number one on your list of things to look into :-)
I urge you to try and work it out from the above text and samples but, if you're still having troubles after more than half an hour or so, the below code shows how I'd do it.
public class Test {
static void print (double m_array[]) {
for (int i = 0; i < m_array.length; i++) {
if ((i > 0) && ((i % 10) == 0))
System.out.println();
System.out.print (m_array[i] + " ");
}
System.out.println();
}
static public void main(String args[]) {
double[] x = new double[15];
for (int i = 0; i < x.length; i++)
x[i] = i * 3;
print (x);
}
}

Find the Last Digit In An Array

I need to find the last digit in a array and see if it is equal to zero. Here is the code I'm using;
import java.util.Scanner;
public class NrOccurrence
{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
System.out.print("Enter the integers between 1 and 100: ");
int[] numbers = new int[100], times = new int[100];
boolean zero = false;
while (zero == false)
{
for (int a = 0; a <= numbers.length; a++)
{
numbers[a] = scan.nextInt();
times[a]++;
if (numbers.equals(0))
{
zero = true;
}
}
}
for (int b = 0; b <= numbers.length; b++)
{
System.out.println(numbers[b] + " occurs " + times[b] + " times");
}
scan.close();
}
}

Create a method like this:
private boolean isLastItemZero(int[] numbers)
{
boolean isLastItemZero = false;
if ((numbers != null) && (numbers.length > 0))
{
isLastItemZero = numbers[numbers.length - 1] == 0;
}
return isLastItemZero;
}
And call it once you're done reading in all of the numbers from the user.

First of all for (int a = 0; a <= numbers.length; a++) will give youIndexOutOfBoundsException .Java uses 0 bases indexing which means that an array of size n has indices up to and including n-1. Change it tofor (int a = 0; a < numbers.length; a++) . Same thing here for (int b = 0; b <= numbers.length; b++)
Second i am not sure what you are trying to check here :
if (numbers.equals(0))
{
zero = true;
}
but you could simply do :
if(numbers[i] == 0);
Now if you wanna check if the last element in the array is 0you can simply do:
if(numbers[numbers.length - 1] == 0)
//do something

By definition, if the remainder of a number divided by 10 is 0, then the last digit must be 0. So you just need;
if(numbers[i] % 10 == 0) { zero = true; }
Hope this helps.

Nested loop construction

This is part of my homework. All I need is a bit of advice. I need to write some nested loop constructs, to print the following:
"122333444455555"
"+**+++****+++++"
"--***++++-----******+++++++"
Here is my code to print the first set of symbols
public static void main (String[] args)
{
int i,j;
for(i=1;i<6;++i)
{
for(j=1;j<i+1;++j)
{
System.out.print(i);
}
}
}
This works perfectly fine. I'm just having trouble figuring out the second and third set of symbols. Apologies for my lack of experience, I'm fairly new to Java.

One solution is:
final String[] arr = {"*", "+"};
And in your inner loop:
System.out.print(arr[i % 2]);
The % (Modulo) operator is responsible of the switches between * and + symbols:
For even i it'll be *, otherwise it'll be +.
Output: "+**+++****+++++".
(Regarding the second output, I'll not show you the solution, but it's very similar to this one once you understand it).

public static void main(String[] args) throws IOException {
int i, j;
for (i = 1; i < 6; ++i) {
for (j = 1; j < i + 1; ++j) {
System.out.print(i);
}
}
System.out.println();
for (i = 1; i < 6; i++) {
if (i % 2 == 1) {
for (j = 1; j < i + 1; ++j){
System.out.print("+");
}
} else {
for (j = 1; j < i + 1; ++j){
System.out.print("*");
}
}
}
System.out.println();
for (i = 2; i < 8; i++) {
if (i % 3 == 1) {
for (j = 1; j <= i; ++j){
System.out.print("+");
}
} else if (i % 3 == 2) {
for (j = 1; j <= i; ++j){
System.out.print("-");
}
} else {
for (j = 1; j <= i; ++j){
System.out.print("*");
}
}
}
}
Cycle #1:
You have to print out numbers from one to five and each number N has to be printed out N times.
for (i = 1; i < 6; ++i) { // this would set `i` to numbers from 1-5
for (j = 1; j < i + 1; ++j) { // for each cycle (next number) it prints
//it out N times where N is the cycle number. 1 is the first cycle,
//2 is the second and so on.
Cycle #2:
Same problem but instead of printing out number of the cycle you have to print out + or * based on if the cycle number is odd or even.
To check if the number is even you can use:
int number = 1;
if(number % 2 == 0){ // is true if the number is even
This checks whats the remainder from the division of number by two.
Cycle #3:
Same as #2 but you start from the second cycle, not from the first and you check for the remainder after division by 3.

If I understand, the third set is composed by sequence of "-*+" so:
String sequence = "-*+";
String s = "+**+++****+++++";
int seqI = 0;
for(i=0; i != s.size(); ++i) {
for(j=0; j < i+2; ++j) {
System.out.print(sequence[seqI]);
}
if(seqI < sequence.size()) {
++seqI;
} else {
seqI = 0;
}
}

You can define a function like this:
public static char Output(int i, int mode)
{
if (mode == 1)
{
return (char) i;
}
else if (mode == 2)
{
if (i % 2 == 0)
{
return '+';
}
else
{
return '*';
}
}
else if (mode == 3)
{
if (i % 3 == 0)
{
return '-';
}
else if (i % 3 == 1)
{
return '*';
}
else
{
return '+';
}
}
}
And use it just like:
for (int mode = 1 ; mode < 4 ; mode++)
{
for (int i = 1 ; i < 6 ; i++)
{
for (int j = 0 ; j < i + (int)(mode / 3) ; j++)
{
System.out.println(Output(i, mode));
}
}
}
Note: Yes! Actually my code is hard to read, but if you try to read it, you will learn something more than other answers!

Array java help needed

I have this program that takes user input and displays the number of times each integer is entered. I pretty much have it down pat but need another loop to omit the shown occurrence of 0. In other words any number with 0 in it cannot be read, also for some reason i am getting two outputs from the same number in my program. For example, if I enter 3,3 I will get 3 occurs 1 time and 3 occurs 2 times as output. The 2 times one being correct and the first one being incorrect.
public class Six_Three {
public static void main(String[] args) {
Scanner input = new Scanner (System.in);
System.out.print("enter integers between 1 and 100: ");
int[] num = new int[100];
int data = input.nextInt();
while ((data = input.nextInt()) != 0) {
num[data]++;
}
for (int i = 1; i < 100; ++i) {
if (num[i] > 0)
System.out.println(i + " occurs " + num[i] + " times ");
}
}

You need two separate loops: the first to gather the information, and the second to print the results:
int data = 0;
while ((data = input.nextInt()) != 0)
{
num[data]++;
}
for (int i = 0; i < 100; ++i)
{
if (num[i] != 0) { /* print num[i] */ }
}

Just loop over the num array after your while loop to print the counts.
for (int index = 0; index < num.length; index++) {
if (num[index] != 0)
System.out.println(data + " occurs " + num[data] + " time(s).");
}

You are printing an output every time an integer is read. Your program is behaving as expected.
To get what you want, you need to scan all the input before you produce any output.
Try this instead:
while (data != 0){
data = input.nextInt();
num[data]++;
}
for (int i = 1; i < 100; ++i) { // your version is 0...99, else array index out of bounds
if (num[i] > 0)
System.out.println(i + " occurs " + num[i] + " times ");
}

The way you write it the last number has to be 0 to make the scanning stop. It might be a good idea to check if there's another int available and use that as a condition for the scanning loop. That way your program can accept any integer.
while (input.hasNextInt()){
num[input.nextInt()]++;
}

it's so simple
int data = 0;
int[] num = new int[100];
int i = 0;
while (i < num.length) {
if ((data = input.nextInt()) == 0)
break;
num[i] = data;
i++;
}
for (i = 0; i < 100; ++i) {
int times = 0;
if (num[i] != 0) {
for (int j = 0; j < 100; j++) {
if (num[j] == 0) {
break;
} else if (num[i] == num[j]) {
times++;
}
}
System.out.println(num[i] + " occurs " + times + " times ");
} else {
break;
}
}