I want to print put the elements in an array that only occur twice. So if, for example, number 2 occurs 3 or 4 times, it should not be printed. The code I've written so far is below.
The issue in my code is with the loop. For example, for the number 2, since j=i+1 is the initialization condition, the inner loop won't read the elements before the jth location- since there is a 2 at index 6, it won't count the 2s before it, making the required condition true and displaying the 2. Is there a way to fix this?
public class Problem2 {
public static void exactlytwice(int[] x) {
int count, j;
for (int i = 0; i < x.length; i++) {
count = 0;
for (j = i + 1; j < x.length; j++) {
if (x[i] == x[j])
count++;
}
if (count == 1) System.out.print(x[i] + " ");
}
}
public static void main(String[] args) {
int[] x = new int[15];
x[0] = 2;
x[1] = 2;
x[2] = 2;
x[3] = 13;
x[4] = 44;
x[5] = 44;
x[6] = 2;
x[7] = 63;
x[8] = 63;
x[9] = 90;
x[10] = 1;
x[11] = 2;
x[12] = 150;
x[13] = 150;
x[14] = 180;
exactlytwice(x);
}
}
aside from the issue you wrote, the bigger problem I see with your code is that its inefficient. You are doing a loop inside a loop here, which is very slow (o(n^2)).
My suggestion is to keep a map of numbers->count, and then only take the ones that appear only twice in the map:
public static void exactlytwice(int[] x) {
Map<Integer, Integer> counter = new HashMap<>();
for (int i = 0; i < x.length; i++) {
if (counter.contains(i)) {
counter.put(i,1);
} else {
counter.put(i,counter.get(i)+1);
}
}
for (Integer i : counter.keyset()) {
if (counter.get(i) == 2) {
System.out.println(i);
}
}
}
Think about maintaining a seperate array/list which keeps track of all the elements that has been counted/printed by which you could just skip the same number that shows again down the array.
Or you could sort the array and then perform the whole logic to check for duplicates.
Just for completeness, there is a solution without extra-map. It's still O(n^2), but uses no extra memory. And It uses kind of fun idea.
First, we only need to output first occurence of a number, every other one is not relevant, because we either have more than 2 of them, or we've already output the first one.
Second, we can then indeed continue, from i+1 element, because at this point, there are no elements equal to ith, that are earlier in array.
public static void exactlytwice(int[] x) {
int count, j;
TOP:
for (int i = 0; i < x.length; i++) {
count = 0;
for (j = 0; j < i; j++) {
if (x[j] == x[i])
// had this number earlier, so go to the next one.
continue TOP;
}
for (j = i+1; j < x.length; j++) {
if (i != j && x[i] == x[j])
count++;
}
if (count == 1) System.out.print(x[i] + " ");
}
}
In addition to the answers provided, there are two more ways you could go about this:
If array modification is permitted, change elements already encountered to a dummy value, and skip the value if encountered. This reduces the time complexity to O(n) from O(n^2), but destroys the original array. Of course, this assumes that the acceptable integers are restricted to a certain set (thanks to #Dzmitry Paulenka for reminding me that I hadn't stated this explicitly). To keep the array, you could make a copy (although then the space complexity becomes O(n) ).
If any integer is acceptable, then create an encountered boolean array, all intialized to false. Change the locations of elements encountered in the original array to true in the encountered boolean array, and if the value is already true, it can be skipped. Again, time complexity O(n), but O(n) space complexity, and unlike in the second method of 1., does not require the permissible range of numbers (ints) to be restricted.
Alternately, simply make the initialization of j as j=0, and then ensure that only those numbers which don't have that number appearing before them are printed, i.e., that the number is printed only if it occurs at j>=i (thanks to #Nir Levy for pointing the j>=i requirement out). This is (slightly) more inefficient than the code already written, but the time complexity remains the same O(n^2).
With Java 8 you can achieve this using streams like this :
public static void main(String[] args)
{
List<Integer> list = Stream.of(12,1,3,4,2,3,7,6,7,3,1,8,4,12,33,45,78,36,8)
.collect(Collectors.groupingBy(x->x, Collectors.summingInt(x->1)))
.entrySet().stream().filter(x->x.getValue()==2)
.collect(ArrayList<Integer>::new,(x,y)->x.add(y.getKey()),ArrayList<Integer>::addAll);
System.out.println(list);
}
The result is :
[1, 4, 7, 8, 12]
Code:
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Random;
public class Writer {
private int counter;
private Random generator = new Random();
private List<Integer> data = new ArrayList<Integer>();
private Map<Integer,Integer> map = new HashMap<Integer,Integer>();
public Writer(int n){
populate(n);
}
private final void populate(int n){
for(int i = 0; i != n; i++){
data.add(generator.nextInt(10));
}
}
private final void reset(){
this.counter = 0;
}
public final void occurence(){
for(int dp : data){
for(int i = 0; i < data.size(); i++){
if(dp == data.get(i)){
counter += 1;
}
}
map.put(dp, counter);
reset();
}
}
public final void filter(int d){
for(int key : map.keySet()){
if(map.get(key) == d){
System.out.println("Key: " + key + " Value: " + map.get(key));
}
}
}
public void rawData(){
System.out.println(data.toString());
}
public void output(){
System.out.println(map.toString());
}
Initiate:
// Create instance of Writer class and generate '100' random numbers
Writer writer = new Writer(100);
// Output raw data
writer.rawData();
// Process data
writer.occurence();
// Filter numbers with occurence '10'
writer.filter(10);
// Output processed data
writer.output();
Output (from from calling filter(10)):
Key: 3 Value: 10
Key: 8 Value: 10
Related
I have an atomic integer array of size 10. I am using this array to organize numbers 1-10 sent in by threads. This 1-10 will eventually be able to change to be a range of numbers larger than 10 and the list is to contain the 10 greatest numbers in that range. I can see the numbers going into the loops and recognizing that they are greater than a number currently there. However, there is never more than 2 numbers in the array when it is printed out. I have tried to trace my code in debug mode, however, it looks as if it is working as intended to me. I feel like there may be a simple error to my logic? I am completely sure all values are entering in the function as I have triple checked this. I start at the end of the array which should contain the highest value and then swap downwards once the slot has been determined. I would appreciate the assistance. This is just a simple experiment I am doing in order to grasp the basics before I try to tackle a homework assignment.
Here an example of my code:
public class testing{
static AtomicIntegerArray maxList = new AtomicIntegerArray(10);
final static int n = 10;
static void setMax(int value)
{
for(int i = 9; i >= 0; i--)
{
if(value > maxList.get(i))
{
int temp = maxList.get(i);
maxList.set(i,value);
if(i == 0)
{
maxList.set(i, value);
}
else
{ for(int j = i-1; j > 0; j--)
{
maxList.set(j, temp);
temp = maxList.get(j-1);
}
}
break;
}
}
public static void main(String[] args)
{
for (int i = 0; i < n; i++)
{
setMax(i);
}
}
}
Here is an example of how it is being called:
Brooke, there is a small bug in your 'j' loop. You had saved the state of a variable (temp), however your logic in the j loop lost the state. This new logic preserves the state of the previous element in the list.
Try this:
for (int j = i - 1; j >= 0; j--) {
int t2 = maxList.get(j);
maxList.set(j, temp);
temp = t2;
}
I would like to use list.negativeNumbers(); to call out the part of a code, which counts how many negative numbers are in a list.
public void negativeNumbers(){
int negative = 0;
for (int i = 0; i <= size; i++){
if (i < 0) {
negative = negative + 1;
}
System.out.println("There are "+ negative +" negative elements in the list!");
}
}
Can you help me in creating a method, that could count negative numbers in the list the correct way?
public void negativeNumbers(){
int negative = 0;
for (int i = 0; i <= size; i++){
if (i < 0) {
negative = negative + 1;
}
System.out.println("There are "+ negative +" negative elements in the list!");
}
}
You will learn better by working out the solution for yourself.
Your code as presented is a good start but lacks:
a defined list of values to test.
a definition of the size of the list (use list.size()).
proper indexing of the list to access values to test (use list.get(i) or list[i]).
a test of each element in the list to determine its negativity. Your code tests whether the list increment variable is < 0.
negative = negative + 1 is ok, but simpler to write ++negative.
Here's a simple example:
import java.util.ArrayList;
public class negnum {
public static void main(String [] args) {
ArrayList<Integer> nums = new ArrayList();
nums.add(0);
nums.add(10);
nums.add(-10);
nums.add(20);
nums.add(-20);
nums.add(30);
nums.add(-30);
nums.add(40);
nums.add(-40);
nums.add(50);
int negs = 0;
for (int i = 0; i < nums.size(); i++){
int n = nums.get(i);
if (n < 0) {
++negs;
System.out.println(n);
}
}
System.out.println(negs +" negatives");
}
}
c:\dev>java negnum
-10
-20
-30
-40
4 negatives
If it is a list of integers you should not be doing "i < 0" but rather the number at index of i. If you were to do that, you would also want to do "< size" rather than "<= size" or else you would run into an IndexArrayOutOfBounds.
It depends on how your double linked list is implemented, but if it extends the normal List<Integer> interface it would look like:
final Integer ZERO = Integer.valueOf(0);
int countNegativeElements() {
return stream().filter(MyList::isNegative).count();
}
static boolean isNegative(Integer i) {
return i.compareTo(ZERO) < 0;
}
with streams. More traditionally with an collection for-each (or an iterator for-each):
int countNegativeElements() {
int count = 0;
for(Integer i : this) { // or iterator()
if (isNegative(i)) count++;
}
return count;
}
This does not expect concurrent modifications and is optimized for collections where iterating is the fastest access. If you have a simple type list then you can replace isNegative with a simple < 0.
This article here talks about different ways to iterate a collection.
In my code I assumed you will add the method directly to your list implementation class. Replace this or this.iterator() or this.stream() with your actual list instance if it is external.
Update:
I just saw your link to the actual linked list you are using, this would look like this (hand made iteration):
int countNegativeElements() {
Node n = start;
int count = 0;
while(n != null) {
if (n.getData() < 0) count++;
n = n.getLinkNext();
}
return count;
}
Using null since there is no hasLinkNext() which would be typical for homework :)
I don't think it is a particular good idea to work with implementations which do not fit in the Collections framework, but maybe it is required to keep the lessons simple.
I want to add numbers in sorted way before entering vector. But the result is not right and I am confused where the problem is ? Output is shown below.
I want to sort using some algorithm without any inbuilt methods.
import java.util.Vector;
public class Test {
public static void main(String ar[]){
//Numbers to enter in vector
int[] number = {5,2,98,3,10,1};
Vector<Integer> v = new Vector<Integer>();
v.add(number[0]);
for(int i=1;i<number.length;i++){
for(int j=v.size();j>0;j--){
System.out.println("Entered: "+number[i]);
if(number[i] <= v.get(j-1)){
v.add(j-1,number[i]);
break;
}else{
v.add(j,number[i]);
break;
}
}
}
for(int s:v)
System.out.print(s + " ");
}
}
OUTPUT:
Entered: 2
Entered: 98
Entered: 3
Entered: 10
Entered: 1
2 5 3 10 1 98
You have a second (inner) for loop based on the variable j, but that "loop" will only execute exactly one time. Both conditions inside the j loop cause the loop to exit (break;).
When you're adding each number, the only possibilities are last or next to last.
Your inner for loop doesn't actually loop.
Regardless of the condition number[i] <= v.get(j-1),
the loop will exit after one step.
What you want to do is,
iterate from the beginning of the vector,
and when you find an element that's bigger than the one you want to insert,
then insert it, and break out of the loop.
This is opposite of what you did so far, which is iterating from the end of the vector.
If the end of the loop is reached without inserting anything,
then append the value.
The program badly needs some other improvements too:
If you don't need the vector to be thread-safe, then you don't need Vector. Use ArrayList instead.
The special treatment for the first number is unnecessary.
The outer loop can be written in a more natural way using the for-each idiom.
No need to loop to print the elements, the toString implementation of Vector is already easy to read.
The variable names are very poor and can be easily improved.
The indentation is inconsistent, making the code very hard to read.
With the problem fixed and the suggestions applied:
List<Integer> list = new ArrayList<>();
for (int current : numbers) {
boolean inserted = false;
for (int j = 0; j < list.size(); j++) {
if (current <= list.get(j)) {
list.add(j, current);
inserted = true;
break;
}
}
if (!inserted) {
list.add(current);
}
}
System.out.println(list);
Last but not least, instead of searching for the insertion point by iterating over the list,
you could achieve much better performance using binary search,
especially for larger sets of values.
Another simple solution would be:
import java.util.Vector;
public class Test {
public static void main(String ar[]){
//Numbers to enter in vector
int[] number = {5,2,98,3,10,1};
Vector<Integer> v = new Vector<Integer>();
v.add(number[0]);
for(int i=1, j;i<number.length;i++){ //j declared here for better scope
for(j=v.size();j>0 && v.get(j-1)>number[i] ;j--); //<-- some changes here,
v.add(j,number[i]); //<-- and here
}
}
for(int s:v)
System.out.print(s + " ");
}
}
The inner for loop is simply used to find the right index for an element to be inserted.
Your inner loop seems to not looping more than one time. That's why the key is not being inserted into right place.
A more concise solution would be
public class Test {
public static void main(String ar[]){
//Numbers to enter in vector
int[] number = {5,2,98,3,10,1};
Vector<Integer> v = new Vector<Integer>();
v.setSize(number.length);
v[0] = number[0];
for(int i=1, vSize = 1; i < number.length; i++, vSize++){
int j = 0, k = 0;
for(j = 0; j < vSize; j++) {
if(v[j] < number[i]) {
break;
}
for(k = vSize; k > j; k--) {
v[k] = v[k -1];
}
v[k] = number[i];
}
for(int s:v)
System.out.print(s + " ");
}
}
I'm playing around with double arrays and am trying to set all the even elements of an array to 0 and all of the odd elements of the array to 1. Everything looks okay to me, but when I run it I get a bunch of errors. Not sure what's wrong; I've been looking at it for a while with no luck. Any advice on how to fix the errors it gives would be great, thanks!
Code:
public class SetOf0and1 {
public static void main(String[]args)
{
int [][] numbers1 = {{4,2,5}, {2,4,1}, {1,3}};
System.out.println("Before setting elements between 0 and 1: ");
displayArray(numbers1);
setEvenRowsTo0OddRowsTo1 (numbers1);
System.out.println("After setting the elements between 0 and 1");
displayArray(numbers1);
}
public static void setEvenRowsTo0OddRowsTo1(int [][]array)
{
for(int i=0; i<array.length;i++)
{
for(int j=0; j<array[i].length;j++)
{
if(i%2 == 0)
array[i][j]=0;
else
array[i][j]=1;
}
}
}
public static void displayArray(int [][]array)
{
for(int i=0;i<array.length;i++)
{
for( int j=0; j<array.length;j++)
{
System.out.println(array[i][j] + " " );
}
System.out.println();
}
}
}
Errors given:
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 2
at SetOf0and1.displayArray(SetOf0and1.java:38)
at SetOf0and1.main(SetOf0and1.java:10)
public static void displayArray(int [][]array)
{
for(int i=0;i<array.length;i++)
{
for( int j=0; j<array.length;j++)
^^^^^^^^^
{
System.out.println(array[i][j] + " " );
}
System.out.println();
}
Your inner loop should stop at array[i].length.
In the method displayArray, the line:
for( int j=0; j<array.length;j++)
Should be:
for( int j=0; j<array[i].length;j++)
array.length does not return the length you thing it does! You have a 2 dimentional array. So if we say you have array[x][y] then array.length will be x and array[i].length (for 0 <= i < x) will be y. This could be different depending on the length of the array on that index. (so the formula does not exactly apply like that)
int [][] numbers1 = {{4,2,5}, {2,4,1}, {1,3}};
this statement initializes an array with three arrays of the legthes 3, 3 and 2!!!
(in the third block you have only two elements !!! - {1,3})
In your displayArray-method you use two times ...
array.length
... to distinct the size of the loop
that sets the number of loops to 3 ... But last block is only two elements long -> errror.
Use this instead for the second loop:
for( int j=0; j<array[i].length;j++)
If you want to check if a number is odd, you can do it this way:
int answer = thenumber % 2;
'thenumber' is the integer to check if it is even.
Then 'answer' would be 0 if the number was even.
And if you want to loop through the array and do it:
for (int i = 0; i < numbers1.length(); i++)
{
if (numbers1[i] % 2 == 0) {
//EVEN
numbers1[i] = 0;
}
else if (numbers1[i] % 2 == 1) {
//ODD
numbers1[i] = 1;
}
}
And, even more compact:
for (int i = 0; i < numbers1.length(); i++)
{
numbers1[i] %= 2;
}
Edit: I forgot that it was an array you had! I was thinking about ArrayList! Fixed.
I don't know if this is a stupid question, but I need to dynamically change the number of for-loops without using recursion.
For example, if n=3, I need 3 nested for-loops.
for(int i=0; i<size; i++){
for(int j=0; j<size-1; j++){
for(int k=0; k<size-2; k++){
//do something
}
}
}
If n=5:
for(int i=0; i<size; i++){
for(int j=0; j<size-1; j++){
for(int k=0; k<size-2; k++){
for(int l=0; l<size-3; l++){
for(int m=0; m<size-4; m++){
//do something
}
}
}
}
}
Is there any way to achieve this without recursion?
Another question: what is the use of Multiple Dispatch in Java? I'm trying to code something in ONE METHOD, and it should run different events in different cases of the parameter. NO IF STATEMENTS / TERNARY OPERATORS / CASES.
NOTE: I can ONLY have one method (part of the problem), and cannot use recursion. Sorry.
Think about how many times you run through this loop. It looks like (size!) / (size - n)!:
int numLoops = 1;
for (int i = 0; i < n; i++) {
numLoops*= (size - i);
}
for (int i = 0; i < numLoops; i++) {
//do something
}
It depends what exactly you're trying to do. Recursion can always be replaced with iteration (see this post for examples using a Stack to store state).
But perhaps the modulo (%) operator could work here? i.e. Have a single loop that increments a variable (i) and then the other variables are calculated using modulo (i % 3 etc). You could use a Map to store the values of the variables indirectly, if there are a varying number of variables.
You have to create array of loop counters and increment it manually.
Quick and dirty example:
public static void nestedFors(int n, int size) {
assert n > size;
assert size > 0;
int[] i = new int[n];
int l = n - 1;
while(l >= 0) {
if(l == n - 1) {
System.out.println(Arrays.toString(i));
}
i[l]++;
if(i[l] == size - l) {
i[l] = 0;
l--;
} else if(l < n - 1) {
l++;
}
}
}
Replace System.out.println(Arrays.toString(i)) with your own code.
You can check it here: http://ideone.com/IKbDUV
It's a bit convoluted, but: here is a way to do it without recursion, in one function and without ifs.
public static void no_ifs_no_recursion(int n){
int[] helper = new int[n-1];
int[] pointers = new int[n]; //helper for printing the results
int totalsize = 1;
for (int loops = 2; loops <= n; loops++){
helper[n - loops] = totalsize;
totalsize*=loops;
}
for (int i=0; i<totalsize; i++){
int carry = i;
for (int loops = 0; loops < n-1; loops++){
pointers[loops] = carry/helper[loops];
carry = carry - (pointers[loops]*helper[loops]);
}
System.out.println(Arrays.toString(pointers));
//or do something else with `i` -> `pointers[0]`, `j` -> `pointers[1]`, `k` -> `pointers[2]` etc..
}
}
I think you need a backtracking algorithm.
But then you would replace your nested loops with recursion.
I don't want to post links here as seems moderators don't like that.
Look at "eight queens puzzle" (you can Google it), you will get my idea.
I know this idea works as I've posed this same question (which you have) to myself on many occasions, and I've applied it several times successfully.
Here is a small example (I changed it as the previous one was a bit complex).
public class Test001 {
public static void main(String[] args) {
loop(0, 5, 10);
}
/**
* max_level - the max count of nesting loops
* size - the size of the collection
*
* 0 - top most level
* level 1 - nested into 0
* level 2 - nested into 1
* ...
* and so on.
*/
private static void loop(int level, int max_level, int size){
if (level > max_level) return;
for (int i=0; i<size-level; i++){
System.out.println("Now at level: " + level + " counter = " + i);
loop(level + 1, max_level, size);
}
}
}
But this still uses recursion.