I need to arrange numbers of a given size (provided by user at runtime), to 3 dimensional array to represent these numbers in real 3D space.
For example if user enters 7 then I need to create an array of size 2,2,2 and arrange the first 7 numbers given by user into the array starting from position 0,0,0 .
The cube should always be smallest possible for example cube of size 2 can contain 2*2*2 = 8 values. And need a function that can take input numbers and return 3D integer array with values inserted from input array (input[0] will become result[0][0][0] and so on).
int input[7] ;
int[][][] result = bestFunction(int[] input) {...}
I have implemented with 3 nested for loops by checking each value at a time.
Is there a better or faster approach to implement it?
Do a Math.floor(Math.cbrt(val)) to get the dimension size.
I think we can reduce it to level 1 nesting, or using two loops, using a string input for the z-axis values.
for(int i = 0; i<size; i++)
for(int j = 0;j<size;j++)
arr[i][j]= Arrays.stream(sc.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();
where Scanner sc = new Scanner(System.in); if you're taking in user input.
Three for loops would be O(n) as long as you break right after the last element of the input is put in.
int[][][] func(int[] input) {
int size = (int) Math.ceil(Math.cbrt(input.length));
int[][][] result = new int[size][size][size];
int x = 0;
loop: for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
for (int k = 0; k < size; k++) {
result[i][j][k] = input[x++];
if (x == input.length)
break loop; //Finish processing
}
}
}
return result;
}
Related
I have n inputs.
these inputs are numbers from 1 to 100.
I want to output the number that appears less than the other ones; also if there are two numbers with the same amount of appearance, I want to output the number that is less than the other one.
I wrote this code but it doesn't work!
Scanner scanner = new Scanner(System.in);
int n=scanner.nextInt(), max=0 , ans=-1;
int[] counter = new int[n];
for(int i=0; i<n; i++)
counter[scanner.nextInt()]+=1;
for(int j=1; j<=100; j++){
if(counter[j]>max)
max=counter[j];
}
for (int i=1; i<=max; i++){
if(counter[i]>0)
if(ans==-1 || counter[ans]>counter[i] || (counter[ans] == counter[i] && i<ans))
ans=i;
}
System.out.print(ans);
There’s a couple of problems with your code, but the main one is the last for loop: You are trying to find the first (ie lowest) number whose counter is equal to max, so your loop should be from 1 to n, not 1 to max.
Another problem is if you are using the number, which is in the range 1-n, as your array index, you need an array of size n+1, not n.
I pinched this from another question regarding the title of yours:
i = input.nextInt (); while (i != 0) { counts [i]++; i = input.nextInt (); } That method increments the number at the position of the user input in the counts array, that way the array holds the number of times a number occurs in a specific index, e.g. counts holds how often 3 occurs.
counter array should contain frequency values for the numbers from 1 to 100 inclusive.
That is, either a shift by 1 should be used when counting the frequency:
int[] counter = new int[100];
for (int i = 0; i < n; i++) {
counter[scanner.nextInt() - 1]++;
}
or 101 may be used as the length of counter array thus representing values in the range [0..100], without shifting by 1.
int[] counter = new int[101];
for (int i = 0; i < n; i++) {
counter[scanner.nextInt()]++;
}
The minimal least frequent number can be found in a single loop (assuming that the counter length is 101).
int minFreq = 101, answer = -1;
for(int j = 1; j <= 100; j++) {
if (counter[j] > 0 && counter[j] < minFreq) { // check valid frequency > 0
minFreq = counter[j];
answer = j;
}
}
System.out.println(answer);
For a wider range of input values (e.g. including negative values) of a relatively small count it is better to use a hashmap instead of a large sparse array.
I have an assignment for a JAVA class I am taking. We are discussing two-dimensional arrays, however on this particular assignment, I can not figure out how to return back specific points and set those points with a specific value. Here is the assignment:
Write a method called create2DArray that will fill, create, and return
a 10 x 10 2d array with random numbers in the range of 1 to 100. Write
a method called print2DArray that will print a 10 x 10 2D array in row
column fashion. Write a method called createCoords that will search
the 2D array looking for any value that is evenly divisible by 3.
Once you have found a number you should log the row, column location.
This means when your method finishes it should produce a list of
coordinates that I can use to plot my graph. This method must also
return the number of coordinates that are divisible by 3 so that I
know how many points there are to plot. I am not particular as to how
the coordinates are returned back as long as I get a list of the row,
column locations. So, I will leave it to you to work out a mechanism
for returning the values. To test that you have logged the
coordinates create another function called fillLocations that will
fill the locations in the array you have logged with
-1. So, your program should flow in this order
create2DArray
print2DArray
createCoords
fillLocations
print2DArray
I understand and have completed create2DArray and print2DArray, but I can not figure out createCoords and fillLocations. Here is what I have so far, but it does not work and there are errors present:
public int createCoords(int row1, int col1){
int[][] coords = new int[row1][col1];
int[][] count = new int[0][0];
int co = 0;
for(int row = 0; row < 10; row++)
{
for(int col = 0; col < 10; col++)
{
if(coords[row][col] % 3 == 0)
co++;
return count[row][col];
}
}
return co;}
public int fillLocations(int[][] count){
int x = 0;
int y = 0;
for(int row = 0; row < 10; row++)
{
for(int col = 0; col < 10; col++)
{
if(count[row][col] % 3 == 0)
x = row;
y = col;
break;
}
}
return (x, y);}
As a programmer you'll nearly always need to research for doing different things. This research will be easier when you divide your problem to smaller problems.
For example you need to generate random numbers? So search on google that and you'll find this: How do I generate random integers within a specific range in Java?.
You need to create and return a 2D array? Google and see Syntax for creating a two-dimensional array
And with your art, put the pieces of the puzzle together in a way that gives your desired result.
public int[][] create2DArray() {
int[][] newArray = new int[10][10];
Random random = new Random();
int range = 100;
for(int i = 0; i < 10; i++)
{
for(int j = 0;j<arr[0].length;j++)
{
newArray[i][j]= (random.nextInt(range) + 1);
}
}
return newArray;
}
This method, creates and returns a 10*10 2D array filled with random generated numbers between 1-100. I'm sure you can write the rest of your program and enjoy from it by yourself.
while (scanner.hasNextLine()) {
line = scanner.nextLine();
if (line.charAt(0) != '#') {
c++;
//looking to get dimensions of line segments from file.
dimensions = new int[c][4];
//Split my file input into an array of tokens(strings)
tokens = line.split(",");
for (int i = 0; i < tokens.length; i++) {
//Parse strings to int.
dimtoke[i] = Integer.parseInt(tokens[i]);
for (int k = 0; k < dimensions.length; k++) {
for (int j = 0; j < dimensions[0].length; j++) {
//attempt to fill 2d array with contents from file.
dimensions[k][j] = dimtoke[j];
}
}
}
}
}
I get a 2d array full of the last dimension in the file instead of all the dimensions. The dimensions come in groups of 4 in a single line of text, so i made a single array with those tokens from each line i read in from the file so i could parse them to ints. something is happening in the process of transferring the 1d array of ints to the 2d array.
You have to close {} the tokens loop like this:
for (int i = 0; i < tokens.length; i++) {
//Parse strings to int.
dimtoke[i] = Integer.parseInt(tokens[i]);
}//you need a closing here
for (int k = 0; k < dimensions.length; k++) {
for (int j = 0; j < dimensions[0].length; j++) {
//attempt to fill 2d array with contents from file.
dimensions[k][j] = dimtoke[j];
}
}
I'll try to explain it to you.
Everytime you get an int in tokens with the instruction
dimtoke[i] = Integer.parseInt(tokens[i]);
Before getting the next int in dimtoke[i+1] you get into a loop where you fill all the ints in the 2d array with that dimitoke[i].
When those loops have finished(all the 2d array has been filled with only 1 int), you go again up and fill the dimitoke with the next int, and again, fill all the 2d array with that int.
What you need to do is to first finish the for loop and fill the dimitoke with the whole set of ints, and then, fill the 2D array with the ints in dimitoke.
Your problem was that you filled the whole array everytime you put an int in dimitoke, so everytime the whole 2d array was overwritten with only 1 int.
Another thing I expect to be wrong:
When you are filling the 2D array, you put the j variable in both dimensions and dimitoke, so if dimitoke lenght is not equal than the second array lenght of dimensions, you'll fill again every row with the same if dimitoke is greater, and indexOutOfBounds Excepcion if it's lesser. Here's an example:
dimitoke dimension = 10 = {0,1,2,3,4,5,6,7,8,9}
if dimensions have dimension 4, the output of dimensions will be:
0,1,2,3
0,1,2,3
0,1,2,3
.
.. c rows.
To repair that, what you can do is to change the dimitoke variable for a int n (for example) and change the code like this:
int n =0;
for (int k = 0; k < dimensions.length; k++) {
for (int j = 0; j < dimensions[0].length; j++) {
//attempt to fill 2d array with contents from file.
dimensions[k][j] = dimtoke[n];
n++;
}
}
I hope the error is out now
Not completely know what you want to do. But I think you should separate the two loop:
assign value to the dimtoke.
assign value to dimensions.
So I think apparently, you should enclose the loop here:
for (int i = 0; i < tokens.length; i++) {
//Parse strings to int.
dimtoke[i] = Integer.parseInt(tokens[i]);
}
I have made a program that outputs the number of repeats in a 2D array. The problem is that it outputs the same number twice.
For example: I input the numbers in the 2D array through Scanner: 10 10 9 28 29 9 1 28.
The output I get is:
Number 10 repeats 2 times.
Number 10 repeats 2 times.
Number 9 repeats 2 times.
Number 28 repeats 2 times.
Number 29 repeats 1 times.
Number 9 repeats 2 times.
Number 1 repeats 1 times.
Number 28 repeats 2 times.
I want it so it skips the number if it has already found the number of repeats for it. The output should be:
Number 10 repeats 2 times.
Number 9 repeats 2 times.
Number 28 repeats 2 times.
Number 29 repeats 1 times.
Number 1 repeats 1 times.
Here is my code:
import java.util.Scanner;
public class Repeat
{
static Scanner leopard = new Scanner(System.in);
public static void main(String [] args)
{
final int ROW = 10; //Row size
final int COL = 10; //Column size
int [][] num = new int[ROW][COL];
int size;
//Get input
size = getData(num);
//Find repeat
findRepeats(num, size);
}
public static int getData(int [][] num)
{
int input = 0, actualSize = 0; //Hold input and actualSize of array
System.out.print("Enter positive integers (-999 to stop): ");
//Ask for input
for(int i = 0; i < num.length && input != -999; i++)
{
for(int j = 0; j < num[i].length && input != -999; j++)
{
input = leopard.nextInt();
//Check if end
if(input != -999)
{
num[i][j] = input;
actualSize++;
}
}
}
System.out.println();
return actualSize;
}
public static void findRepeats(int [][] num, int size)
{
int findNum;
int total = 0, row = 0, col = 0;
for(int x = 0; x < size; x++)
{
//Set to number
findNum = num[row][col];
//Loop through whole array to find repeats
for(int i = 0; i < num.length; i++)
{
for(int j = 0; j < num[i].length; j++)
{
if(num[i][j] == findNum)
total++;
}
}
//Cycle array to set next number
if(col < num[0].length-1)
col++;
else
{
row++; //Go to next row if no more columns
col = 0; //Reset column number
}
//Display total repeats
System.out.println("Number " + findNum + " appears " + total + " times.");
total = 0;
}
}
}
I know why it is doing it, but I cannot figure out how to check if the number has already been checked for it to skip that number and go to the next number. I cannot use any classes or code that is not used in the code.
Since you cannot use anything other than this, lets say, basic elements of Java consider this:
Make another temporary 2D array with two columns (or just two separate arrays, personally I prefer this one). On the start of the algorithm the new arrays are empty.
When you take a number (any number) from the source 2D structure, first check if it is present in the first temporary array. If it is, just increment the value (count) in the second temporary array for one (+1). If it is not present in the first tmp array, add it to it and increase the count (+1) in the second at the same index as the newly added number in the first (which should be the last item of the array, basically).
This way you are building pairs of numbers in two arrays. The first array holds all your distinct values found in the 2D array, and the second one the number of appearances of the respective number from the first.
At the and of the algorithm just iterate the both arrays in parallel and you should have your school task finished. I could (and anyone) code this out but we are not really doing you a favor since this is a very typical school assignment.
It's counting the number two times, first time it appears in the code and second time when it appears in the code.
To avoid that keep a system to check if you have already checked for that number. I see you use check int array but you haven't used it anywhere in the code.
Do this,
Put the number in the check list if you have already found the count of it.
int count = 0;
check[count] = findNum;
count++;
Note: You can prefill you array with negative numbers at first in order to avoid for having numbers that user already gave you in input.
Next time in your for loop skip checking that number which you have already found a count for
for(int x = 0; x < size; x++) {
findNum = num[row][col];
if(check.containsNumber(findNUm)) { //sorry there is no such thing as contains for array, write another function here which checks if a number exists in the array
//skip the your code till the end of the first for loop, or in other words then don't run the code inside the for loop at all.
}
}
Frankly speaking I think you have just started to learn coding. Good luck! with that but this code can be improved a lot better. A piece of advice never create a situation where you have to use 3 nested for loops.
I hope that you understood my solution and you know how to do it.
All answers gives you some insight about the problem. I try to stick to your code, and add a little trick of swap. With this code you don't need to check if the number is already outputted or not. I appreciate your comments, structured approach of coding, and ask a question as clear as possible.
public static void findRepeats(int [][] num, int size)
{
int findNum;
int total = 1, row = 0, col = 0;
int [] check = new int[size];
while(row < num.length && col < num[0].length)
{
//Set to number
findNum = num[row][col];
//Cycle array to set next number
if(col < num[0].length-1)
col++;
else
{
row++; //Go to next row if no more columns
col = 0; //Reset column number
}
//Loop through whole array to find repeats
for(int i = row; i < num.length; i++)
{
for(int j = col; j < num[i].length; j++)
{
if(num[i][j] == findNum) {
total++;
//Cycle array to set next number
if(col < num[0].length-1)
col++;
else
{
row++; //Go to next row if no more columns
col = 0; //Reset column number
}
if(row < num.length - 1 && col < num[0].length -1)
num[i][j] = num[row][col];
}
}
}
//Display total repeats
System.out.println("Number " + findNum + " appears " + total + " times.");
total = 1;
}
}
you can use a HashMap to store the result. It Goes like this:
// Create a hash map
HashMap arrayRepeat = new HashMap();
// Put elements to the map
arrayRepeat.put(Number, Repeated);
I'm having difficulty understand how to write this array. I need it to out-print 10x5 (50 elements total), and have the first 25 elements equal to the sqrt of the index that it is in, and the last 25 to equal 3 * the index. Yes, this is homework but I'm not asking for you to do it for me, I just need help! I'm getting errors when using Math saying that I cant use double and the double array together. Here is what I have so far:
public class snhu4 {
public static void main(String args[]) {
double alpha[][] = new double[10][5];
double[] sum, sum2;
for (int count=0; count<=25;count++) {
alpha[count]= Math.sqrt(count);
}
for (int count=26; count<=50;count++) {
alpha[count]= count *3;
}
for (int count=0; count<=50;count++) {
System.out.print(alpha[count]);
}
}
}
Because alpha is a multidimensional array, you can't refer to its elements like a normal array.
int myarray[][] = new int[2][2];
In the above example, the array myarray is multidimensional. If I wanted to access the second element in the first array, I would access it like this:
int myint = myarray[0][1];
You are trying to access a multidimensional array by using the access for a normal array. Change
alpha[count]
to
alpha[0][count]
or similar.
Read here for more information on multidimensional arrays.
you defined alpha as a 2D array with lets say 10 items in the first dimension and 5 in the second, and 5x10 is 50 elements.
When using your array to assign values to these elements, u must call upon the array using 2 indices, one for each dimension:
alpha[i][j] = /*double value*/; //with 0<=i<=9 and 0<=j<=4
So the first 25 elements going from left to right in dimension order is going to be:
[0to9][0] and [0to9][1] and [0to4][2]
the next 25 will be
[4to9][2] and [0to9][3] and [0to9][4]
from then on i cannot give you the answers to your homework, but the loops should look like this:
int j;
for(int i = 0; i<25; i++)
{
j=i/10; //integer division will return 0 for i<10, 1 for 10<i<20, etc..
alpha[i%10][j] = Math.sqrt(i);
}
and you can figure out the rest
The 10x5 appears to be an output constraint, not a design constraint.
You are using Java, so use Java constructs, not C-language constructs;
specifically store the values in a List not an array.
Here are some hints:
List<Integer> valuesList = new ArrayList<Integer>();
for (int index = 0; index < 25; ++index)
Integer currentValue = Math.sqrt(index);
valuesList.add(currentValue);
for (int index = 25; index < 50; ++index)
Integer currentValue = index * 3;
valuesList.add(currentValue)
int count = 1;
for (Integer current : valuesList)
if ((count % 5) == 0) // write a newline.
System.out.print(current);
++count