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Why does this method only return values of 0 to a new array?

I am supposed to write a short program that takes 10 numbers, stores the values in an array, passes it to a method (eliminateDuplicates()) that creates a new array of only the unique values from the first array.
However, I am having trouble either initializing the output array, or making the eliminateDuplicates() method return the output array properly. The output array is always full of 0's and I cannot figure out why this is failing.
java.util.Arrays.parallelSort(inputNumbers); //sorts the array in ascending order
eliminateDuplicates(inputNumbers); //passes array to eliminateDuplicates method
//display each unique value in output array
System.out.print("The distinct numbers are ");
for(int i = 0; i < outputNumbers.length; i++)
System.out.print(outputNumbers[i] + " ");
}
public static int [] eliminateDuplicates(int[] list) {
int[] outputNumbers = new int [list.length];
int k = 0;
for (int i = 0; i < list.length; i++)
if(i == 0) //compares each array value against preceding value
outputNumbers[i] = list[i]; //only copies unique values to output array
else
if(list[i] != list [i-1]) {
outputNumbers[k] = list[i];
k++;
}
return outputNumbers;```

You have a local outputNumbers in eliminateDuplicates which you return. I assume you also have a redundant static outputNumbers. Option 1: Eliminate the local variable, change
int[] outputNumbers = new int [list.length];
to
outputNumbers = new int [list.length];
Option 2: Set outputNumbers on call (which is what I would likely do, and eliminate the static one)... Like,
int[] outputNumbers = eliminateDuplicates(inputNumbers);
Don't forget to remove the static one if you use option 2.

You are ignoring the array returned by your method.
Change
eliminateDuplicates(inputNumbers);
to
int[] outputNumbers = eliminateDuplicates(inputNumbers);
P.S. your output array has the same length as the input array. Therefore, since you are eliminating duplicates, it may have some 0s as its last elements. If that's not what you want, you should create the output array only after you find out how many unique numbers the input array has.

Array search using Java

So I am extremely new to Java and my assignment has me creating an array 10 indexes in size, copying it, and sorting the copy. These parts I have functioning properly. What they want me to do is prompt the user for a value they wish to search for, and if found, return the index and which array it was found in. This last search part is really messing me up. My code is very sloppy I know, but I am not very good at this yet. The book of course has an example, but it uses a driver class and I'm not entirely sure if that's required in this situation. Thanks for any replies and my apologies if I posted this incorrectly.
http://imgur.com/a/T8t3Q - This is an example of the final output required.
public class MJUnit1Ch9 {
public static void main(String[] args) {
Scanner stdIn = new Scanner(System.in);
double[] numList; //list of random numbers
double[] numListSorted; //sorted copy of array
int numSearch; //array search
int numSearchIndex; //position in array
numList = new double[10]; //creation of size 10 array
for (int i = 0; i < 10; i++) { //create random numbers between 1 and 20
numList[i] = (int) (Math.random() * 20 + 1);
}
numListSorted = Arrays.copyOf(numList, numList.length); //copy original array
Arrays.sort(numListSorted); //sort copy by API
System.out.printf("%7s%7s\n", "Unsorted Array", "Sorted Array");
for (int i = 0; i < numList.length; i++) {
System.out.printf("%7.2f%7.2f\n", numList[i], numListSorted[i]);
}
//*************************************************************************
//*************************************************************************
System.out.print("Please enter number to search for:");
}
}

You can read number using scanner object
stdIn.nextInt()
And for searching you can use Binary search method of Arrays class by passing the number which you read from scanner object

Arranging sequence / Sequence Analysis

You are given S a sequence of n integers S = s1, s2, ..., sn. Please, compute if
it is possible to split S into two parts : s1, s2, ..., si and si+1, si+2, ….., sn
(1 <= i < n) in such a way that the first part is strictly decreasing while the
second is strictly increasing one. First take n as input and then take n more
integers, output yes or no.
This is what i tried
import java.util.Scanner;
public class Sequence {
public static int c;
public static void main(String[] args) {
int n;
int count = 0;
Scanner s = new Scanner(System.in);
n = s.nextInt();
int a[] = new int [n];
for(int i = 0; i<n; i++) // loop for taking input
{
a[i] = s.nextInt();
}
for(int i = 0; i<n-2; i++) // loop for finding the minimum point
{
if(a[i]<a[i+2])
{ c = i; // associated minimum valued index to c
for( ; i<n-2; i++) /* loop for checking whether after that the array
{ is decreasing or not*/
if(a[i+1]<a[i+2])
{
count = count+1;
}
else
{
}
}
}
if(count == n-2-c)
{
System.out.println("YES");
}
else
{
System.out.println("NO");
}
}
}
This code is not passing 1 Test Case on Hackerrank.com please suggest some solution.

One good way to do that is by binary search:
You will have three variables : lowbound,middle,upperbound and
you start from the middle of your array and lowbounf=0,upperbound =n-1.
Next you will check with a linear passing of the array if s1,s2,...smiddle are strictly decreasing and smiddle,....,sn are strictly increasing .If yes then middle is your solution.
If s1,s2,...smiddle is not strictly decrasing and smiddle,....,sn is not strictly increasing you have no solution.
If s1,s2,...smiddle is not strictly decrasing and smiddle,....,sn is strictly increasing then uperbound=middle,middle=(upperbound+lowbound)/2 and try again.
If s1,s2,...smiddle is strictly decrasing and smiddle,....,sn is not strictly increasing then lowbound=middle,middle=(upperbound+lowbound)/2 and try again.
This until you find a solution ,or find that there is no solution or until lowbound=upperbound.
Example:
sequence: 7 8 5 1 2 3 4 5 6 7 8
middle=5 (the element 3),lowbound=0,upperbound=10, 7,8,5,4,1,2,3 is not strictly decreasing ,while 4,5,6,7,8 strictly increasing.
so: upperbound=5,middle=2 (the element array[middle]=2),7,8,5 are strictly decreasing ,1,2,3,4,5,6,7,8 are stricty increasing so the solution is middle = 2 .(Note middle=2 means that is the third element of array, the first is array[0] ,second is array[1] and third is array[2]=array[middle]=5 ).
The above solution is trying log n times (due to binary search) to linearly check the array (every linear check is O(n)) .So this solution is O(n log n).

import java.util.*;
public class Main {
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();int f=0;
int arr[]=new int[n];
for(int i=0;i<n;i++)
{
arr[i]=sc.nextInt();
}
int i=0;
for(i=0;i<n-1;i++)
{
if(arr[i]<arr[i+1])
{
break;
}
}
for(int j=i+1;j<n-1;j++)
if(arr[j]>arr[j+1])
f=1;
if(f==1)
System.out.println("false");
else
System.out.println("true");
}
}

finding the number of pairs of numbers in an array that add up to a number

I am trying to come up with a program that will search inside of an array that is given a length by the user that picks out whether there is a pair of numbers that sum to 7. The idea is that if there is k amount of dice being thrown, how many pairs of numbers out of those dice thrown add up to 7. So far this is all that I could come up with but I am very stuck.
This is the driver class for the program. I have to write a class that will make this driver function properly.
import java.util.Scanner;
public class SevenDriver{
public static void main(String[] args){
System.out.println("Enter number of dice to toss");
Scanner s = new Scanner(System.in);
int diceCount = s.nextInt();
SevenTally t = new SevenTally(diceCount);
int experiments = 1000000;
int wins = 0;
for(int j = 0; j < experiments; j++)
if(t.experiment()) wins++;
System.out.println((double)wins/experiments);
}
}
This is what I have so far. It does not currently work or compile. I am just looking for some ideas to get me going. Thanks!
public class SevenTally{
private int diceCount;
public SevenTally(int die){
diceCount = die;
}
public int genDice(){
return 1 + (int)(Math.random()*6);
}
public boolean experiment(){
boolean[] nums = new boolean[diceCount];
int ranNum;
int sum = 7;
for(int i = 0; i < nums.length; i++){
ranNum = genDice();
if (nums[ranNum] == sum){
return true;
}
}
int left = 0;
int right = nums.length - 1;
while(left<right){
int tempSum = nums[left] + nums[right];
if(tempSum == 7){
return true;
}
else if(tempSum>7){
right--;
}
return false;
}
}

First populate your array of length k with random int in [1;6]
The number of possible pairs in an array of length k is the number of 2-combinations in the array, which is (k-1)*k/2 (http://en.wikipedia.org/wiki/Combination)
You can test all the possible pairs (i,j) in your array like so:
int win = 0;
int tally = 7;
for(int i=0; i<k-1; i++){
for(int j=i+1; j<k; j++){
if(array[i]+array[j] == tally){
win++;
}
}
}
What this does is that it sets the first element of the pair to be the first element of the array, and sums it with the other elements one after the other.
It pairs array[0] with array[1] to array[k-1] at the first pass of the i for loop, that's k pairs.
Then k-1 pairs at second pass, and so on.
You end up with (k)+(k-1)+(k-2)+...+1 pairs, and that's exactly (k-1)*k/2 pairs.
done =]
edit: sorry, haven't read the whole thing. the method experiment() is supposed to return a boolean. you can return win>0?true:false; for example...

This Wiki page has some algorithms to do that. Its not a trivial problem...

You're generating a random number in ranNum, and then using it as an index into the array nums. Meanwhile, nums never gets filled, so no matter which box you index into, it never contains a 7.
What you want to do, if I understand your problem correctly, is fill each space in the array with the result of a die roll, then compare every two positions (rolls) to see if they sum to seven. You can do that using a nested for loop.
Essentially, you want to do this: (written in pseudocode as I'm not a java programmer)
int[] results[numrolls]
for (count = 0 to numrolls-1) { results[numrolls]=dieRoller() }
for (outer = 0 to numrolls-2)
for (inner = outer+1 to numrolls-1)
if (results[outer] + results[inner] == 7) return true
return false;
However, in this case there's an even easier way. You know that the only ways to get a sum of 7 on 2d6 are (1,6),(2,5),(3,4),(4,3),(5,2),(6,1). Set up a 6-length boolean array, roll your dice, and after each roll set res[result] to true. Then return (1-based array used for simplicity) ( (res[1] && res[6]) || (res[2] && res[5]) || (res[3] && res[4]) ).

ArrayIndexOutOfBoundsException means you are trying to access an element of the array that hasn't been allocated.
In your code, you create a new array d of length diceCount, but then you genDice() on always 6 elements.

Java: how do I initialize an array size if it's unknown?

I'm asking the user to enter some numbers between 1 and 100 and assign them into an array. The array size is not initialized since it is dependent on the number of times the user enters a number.
How should I assign the array length?
If user enters 5 6 7 8 9 (5 numbers), then
int[] list;
becomes
int[] list = new int[5];
I'm trying to use a loop, but it won't stop.
int[] integers;
int j = 0;
do {
integers = new int[j + 1];
integers[j] = in.nextInt();
j++;
} while((integers[j-1] >= 1) ||(integers[j-1]) <= 100);

You should use a List for something like this, not an array. As a general rule of thumb, when you don't know how many elements you will add to an array before hand, use a List instead. Most would probably tackle this problem by using an ArrayList.
If you really can't use a List, then you'll probably have to use an array of some initial size (maybe 10?) and keep track of your array capacity versus how many elements you're adding, and copy the elements to a new, larger array if you run out of room (this is essentially what ArrayList does internally). Also note that, in the real world, you would never do it this way - you would use one of the standard classes that are made specifically for cases like this, such as ArrayList.

I think you need use List or classes based on that.
For instance,
ArrayList<Integer> integers = new ArrayList<Integer>();
int j;
do{
integers.add(int.nextInt());
j++;
}while( (integers.get(j-1) >= 1) || (integers.get(j-1) <= 100) );
You could read this article for getting more information about how to use that.

I agree that a data structure like a List is the best way to go:
List<Integer> values = new ArrayList<Integer>();
Scanner in = new Scanner(System.in);
int value;
int numValues = 0;
do {
value = in.nextInt();
values.add(value);
} while (value >= 1) && (value <= 100);
Or you can just allocate an array of a max size and load values into it:
int maxValues = 100;
int [] values = new int[maxValues];
Scanner in = new Scanner(System.in);
int value;
int numValues = 0;
do {
value = in.nextInt();
values[numValues++] = value;
} while (value >= 1) && (value <= 100) && (numValues < maxValues);

If you want to stick to an array then this way you can make use. But its not good as compared to List and not recommended. However it will solve your problem.
import java.util.Scanner;
public class ArrayModify {
public static void main(String[] args) {
int[] list;
String st;
String[] stNew;
Scanner scan = new Scanner(System.in);
System.out.println("Enter Numbers: "); // If user enters 5 6 7 8 9
st = scan.nextLine();
stNew = st.split("\\s+");
list = new int[stNew.length]; // Sets array size to 5
for (int i = 0; i < stNew.length; i++){
list[i] = Integer.parseInt(stNew[i]);
System.out.println("You Enterred: " + list[i]);
}
}
}

String line=sc.nextLine();
int counter=1;
for(int i=0;i<line.length();i++) {
if(line.charAt(i)==' ') {
counter++;
}
}
long[] numbers=new long[counter];
counter=0;
for(int i=0;i<line.length();i++){
int j=i;
while(true) {
if(j>=line.length() || line.charAt(j)==' ') {
break;
}
j++;
}
numbers[counter]=Integer.parseInt(line.substring(i,j));
i=j;
counter++;
}
for(int i=0;i<counter;i++) {
System.out.println(numbers[i]);
}
I always use this code for situations like this. beside you can recognize two or three or more digit numbers.

int i,largest = 0;
Scanner scan = new Scanner(System.in);
System.out.println("Enter the number of numbers in the list");
i = scan.nextInt();
int arr[] = new int[i];
System.out.println("Enter the list of numbers:");
for(int j=0;j<i;j++){
arr[j] = scan.nextInt();
}
The above code works well. I have taken the input of the number of elements in the list and initialized the array accordingly.

**input of list of number for array from single line.
String input = sc.nextLine();
String arr[] = input.split(" ");
int new_arr[] = new int[arr.length];
for(int i=0; i<arr.length; i++)
{
new_arr[i] = Integer.parseInt(arr[i]);
}