How to print out an X using nested loops - java

I have searched through to find a simple solution to this problem.
I have a method called
printCross(int size,char display)
It accepts a size and prints an X with the char variable it receives of height and width of size.
The calling method printShape(int maxSize, char display) accepts the maximum size of the shape and goes in a loop, sending multiples of 2 to the printCross method until it gets to the maximum.
Here is my code but it is not giving me the desired outcome.
public static void drawShape(char display, int maxSize)
{
int currentSize = 2; //start at 2 and increase in multiples of 2 till maxSize
while(currentSize<=maxSize)
{
printCross(currentSize,display);
currentSize = currentSize + 2;//increment by multiples of 2
}
}
public static void printCross(int size, char display)
{
for (int row = 0; row<size; row++)
{
for (int col=0; col<size; col++)
{
if (row == col)
System.out.print(display);
if (row == 1 && col == 5)
System.out.print(display);
if (row == 2 && col == 4)
System.out.print(display);
if ( row == 4 && col == 2)
System.out.print(display);
if (row == 5 && col == 1)
System.out.print(display);
else
System.out.print(" ");
}
System.out.println();
}
}
Is it because I hardcoded the figures into the loop? I did a lot of math but unfortunately it's only this way that I have been slightly close to achieving my desired output.
If the printCross() method received a size of 5 for instance, the output should be like this:
x x
x x
x
x x
x x
Please I have spent weeks on this and seem to be going nowhere. Thanks


The first thing you have to do is to find relationships between indices. Let's say you have the square matrix of length size (size = 5 in the example):
0 1 2 3 4
0 x x
1 x x
2 x
3 x x
4 x x
What you can notice is that in the diagonal from (0,0) to (4,4), indices are the same (in the code this means row == col).
Also, you can notice that in the diagonal from (0,4) to (4,0) indices always sum up to 4, which is size - 1 (in the code this is row + col == size - 1).
So in the code, you will loop through rows and then through columns (nested loop). On each iteration you have to check if the conditions mentioned above are met. The logical OR (||) operator is used to avoid using two if statements.
Code:
public static void printCross(int size, char display)
{
for (int row = 0; row < size; row++) {
for (int col = 0; col < size; col++) {
if (row == col || row + col == size - 1) {
System.out.print(display);
} else {
System.out.print(" ");
}
}
System.out.println();
}
}
Output: (size = 5, display = 'x')
x x
x x
x
x x
x x


Instead of giving a direct answer, I will give you some hints.
First, you are right to use nested for loops.
However as you noticed, you determine when to print 'x' for the case of 5.
Check that 'x' is printed if and only if row = col or row + col = size - 1


for your printCross method, try this:
public static void printCross(int size, char display) {
if( size <= 0 ) {
return;
}
for( int row = 0; row < size; row++ ) {
for( int col = 0; col < size; col++ ) {
if( col == row || col == size - row - 1) {
System.out.print(display);
}
else {
System.out.print(" ");
}
}
System.out.println();
}
}
ah, I got beaten to it xD


Here's a short, ugly solution which doesn't use any whitespace strings or nested looping.
public static void printCross(int size, char display) {
for (int i = 1, j = size; i <= size && j > 0; i++, j--) {
System.out.printf(
i < j ? "%" + i + "s" + "%" + (j - i) + "s%n"
: i > j ? "%" + j + "s" + "%" + (i - j) + "s%n"
: "%" + i + "s%n", //intersection
display, display
);
}
}


Lte's try this simple code to print cross pattern.
class CrossPattern {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
System.out.println("enter the number of rows=column");
int n = s.nextInt();
int i, j;
s.close();
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if (j == i) {
System.out.print("*");
} else if (j == n - (i - 1)) {
System.out.print("*");
} else {
System.out.print(" ");
}
}
System.out.println();
}
}
}

Related

How to print a 2D array in java

Hello so am trying to create a 2D array of int with random number of rows and columns and a random starting and ending points using java to apply the A* algorithm on it.
When i add {S} and {E} to define the tow points and print it there are numbers outside of the 2D array printed.
`Random rand = new Random();
int min = 2, max = 10;
// a random number of rows and columns
int a = (int)(Math.random() * (max - min + 1)) + min;
// the location of the starting point.
int row_start = rand.nextInt(a);
int col_start = rand.nextInt(a);
// the location of the ending point.
int row_end = rand.nextInt(a);
int col_end = rand.nextInt(a);
int [][] M = new int [a][a];
public void create() {
//empty: 0; grass: 1; sand: 2; water: 3; wall: 4.
for (int i = 0; i < a; i++) {
for (int j = 0; j < a; j++) {
M[i][j] = rand.nextInt(5);
}
}
for (int i = 0; i < a; i++) {
for (int j = 0; j < a; j++) {
System.out.print(" " +M[i][j] + "\t");
if(row_start == i && col_start == j) {
System.out.print("{S}" + "\t");
}
if(row_end == i && col_end == j) {
System.out.print("{E}" + "\t");
}
}
System.out.print("\n");
}
}`
the output looks like this:
1 0 4 0
2 {S} 1 2 2
4 4 {E} 0 3
2 0 3 3
the 2 and 3 shouldn't appear there.

The problem is that you always print m[i][j].
What you need is to only print m[i][j] when i and j are not S and E positions. When i and j are S and E positions, print S or E. Otherwise, print m[i][j].
if(row_start == i && col_start == j) {
System.out.print("{S}" + "\t");
} else if(row_end == i && col_end == j) {
System.out.print("{E}" + "\t");
} else {
System.out.print(" " +M[i][j] + "\t");
}

How should I reconfigure the arrow "->" to not print when done with with my pathing?

I'm trying to create an optimal path to collect as many 1's as I can but after I execute my code, I still have an arrow pointing to nothing as there are no more places to go. How would I remove the arrow at the end of the code?
import java.util.Arrays;
import java.util.Scanner;
public class Main{
public static void main(String[] args){
Scanner s1 = new Scanner(System.in);
int n = s1.nextInt();
int m = s1.nextInt();
int mat[][] = new int[n][m];
for (int i = 0; i < mat.length; i++){
for (int j = 0; j < mat[0].length; j++){
mat[i][j] = s1.nextInt();
}
}
int path[][] = new int[n][m];
for (int i = 0; i < path.length; i++){
Arrays.fill(path[i], -1);
}
int maxCoins = util(0, 0, mat, path);
System.out.println("Max coins:" + maxCoins);
int row = 0, column = 0;
System.out.print("Path:");
while(row < mat.length && column < mat[0].length){
System.out.print("(" + (row + 1) + "," + (column + 1) + ")");
System.out.print("->");
if(row < n - 1 && column < m - 1){
int down = path[row + 1][column];
int right = path[row][column + 1];
if(down > right){
row += 1;
continue;
}
else if (right > down){
column += 1;
continue;
}
else{
row += 1;
continue;
}
}
if(row + 1 < n){
row += 1;
}
else{
column += 1;
}
}
}
private static int util(int row,int column,int mat[][], int path[][]){
if(row >= mat.length || column >= mat[0].length){
return 0;
}
if(path[row][column]!= -1){
return path[row][column];
}
int right = util(row, column + 1, mat,path);
int down = util(row + 1, column, mat,path);
path[row][column]=Math.max(right, down);
if(mat[row][column] == 1){
path[row][column] += 1;
}
return path[row][column];
}
}
My current input looks like:
5 6
0 0 0 0 1 0
0 1 0 1 0 0
0 0 0 1 0 1
0 0 1 0 0 1
1 0 0 0 1 0
And output is:
Max coins:5
Path:(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(3,5)->(3,6)->(4,6)->(5,6)->
I am just trying to remove the one at the end but unsure where to insert my code:
System.out.print("->");

Cleanest way would be using a StringJoiner.
You can use it as follows
StringJoiner joiner = new StringJoiner("->");
joiner.add("a");
joiner.add("b");
System.out.println(joiner); //prints a->b - you can use toString if you want to return a joined String
You can also define a prefix and suffix for your joined String.
Or if you are familiar with Streams, there is Collectors.joining("->") available.

Three solutions that come to mind:
Add another check inside the loop, and put your sysout -> thingy after that check.
Usually code would generate some kind of list or similar data about the results and return it. It's a lot simpler to print lists, because you know the length etc.
Another common solution is to use StringBuilder and correct it before generating the output with toString()

You could just do something like this:
if (!(row == mat.length - 1 && column == mat[0].length - 1)) {
System.out.print("->");
}
Or a little cleaner:
if (arrowIsNotAtTheEnd(mat, row, column)) {
System.out.print("->");
}
// ...
private static boolean arrowIsNotAtTheEnd(int[][] mat, int row, int column) {
return !(row == mat.length - 1 && column == mat[0].length - 1);
}

For java 8 and above, the String class already has a convenient join method.
CharSequence[] path=new CharSequence[]{
"(1,1)","(2,1)","(2,2)","(2,3)","(2,4)","(3,4)","(3,5)","(3,6)","(4,6)","(5,6)"};
String output=String.join("->",path);
System.out.println(output);
//output: (1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(3,5)->(3,6)->(4,6)->(5,6)

how to print this pattern in java i have no idea how

i cant print this pattern :
1 2 3 4 *
1 0 0 * 5
1 0 * 0 5
1 * 0 0 5
* 2 3 4 5
i tried to print but only on row :
public class Pattern {
public static void main(String[] args) {
for(int j=1;j<=5;j++)
{
if(j>4) {
System.out.print("*");
}
else {
System.out.print(j);
}
}
}
}

Let's try to consider the logic of this output:
You have an NxN matrix, with rows and columns numbered 1..N.
If you're on the secondary diagonal (i.e. row+column=N+1), print a *
Else, if you're on the borders (i.e., either the row or the column is 1 or N), print the column number
Else, print a 0
Now, you just need to convert this logic to Java:
int size = 5;
for (int i = 1; i <= size; ++i) {
for (int j = 1; j <= size; ++j) {
char ch;
if (i + j == size + 1) {
ch = '*';
} else if (i == 1 || i == size || j == 1 || j == size) {
ch = (char) ('0' + j);
} else {
ch = '0';
}
System.out.print(ch + " ");
}
System.out.println();
}

Java 2D Array manipulation

I have to modify the below method:
private final static int NUM = 6;
public void fun(int[][] grid) {
for(int row = 0; row < NUM; row++) {
for(int col = 0; col < NUM; col++) {
if((grid[row][col] % 2) == 0) {
grid[row][col] = 0;
}
}
}
}
This method checks if it's a even number and if so it replaces its value with 0. Simple.
I now need to modify it so that it directs each cell to simultaneously replace its value with its number of diagonal neighbors that hold a value of 0.
I've thought about this for about an hour and tried many different solutions, most of which resulted in an out of bounds exception. I'm stumped and don't know how to accomplish this.
If the code is right, using the integers for the grid array below, it will reproduce the numbers shown in the bottom of the picture.

What is the problem, you just need to put if statements, like there can be maximum 4 possible neighbors so check that how many are equal to 0. But this is not enough you just need to add one more condition in each of the if statements. The condition would be that the neighbor you are trying to check is possible or not.
That is: Total 4 neighbors. If the coordinates of your main cell are x, y then:
1st Diagonal neighbor: x-1, y-1
2nd Diagonal neighbor: x-1, y+1
3rd Diagonal neighbor: x+1, y+1
4th Diagonal neighbor: x+1, y-1
These are all the 4 diagonal neighbors' coordinates but the last thing you need to check is whether they go out of bonds or not. For example for checking the 1st Diagonal neighbor I would do:
if((x-1)>0 && (y-1)>0){
//and then check here if that block is = `0`
}
and for other having say x+1 or y+1 you will need to check whether or not they are less than the NUM. Like if I want to check the 3rd Diagonal Neighbor:
if((x+1)<NUM && (y+1)<NUM){
//and then check here if that block is = `0`
}
Update: What do you mean by check here if that block is = 0?
If you want to check that is the diagonal neighboring blocks are equal to 0 or not then you will need to do it in a loop. Here is how:
public void fun(int[][] grid) {
for(int row = 0; row < NUM; row++) {
for(int col = 0; col < NUM; col++) {
if((grid[row][col] % 2) == 0) {
grid[row][col] = 0;
}
}
}
for(row = 0; row< NUM; row++){
for(int col = 0; col < NUM; col++) {
int count = 0;
// To check for the 1st Diagonal Neighbor
if((row-1)>0 && (col-1)>0){
if(grid[row-1][col-1]==0){
count++;
}
}
//Similarly for 2nd, 3rd and 4th Diagonal Neighbors
//and then
grid[row][col]=count;
}
}
}
Update 2:
For say the 3rd diagonal neighbor the code block would be like this:
if((row+1)<NUM && (col+1)<NUM){
if(grid[row+1][col+1]==0){
count++;
}
}
Answer
final private static int NUM = 6;
public void fun(int[][] grid) {
for(int row = 0; row < NUM; row++) {
for(int col = 0; col < NUM; col++) {
int counter = 0;
if((row - 1) > 0 && (col - 1) > 0) {
if(grid[row - 1][col - 1] == 0) {
counter++;
}
}
if((row - 1) > 0 && (col + 1) < NUM) {
if(grid[row - 1][col + 1] == 0) {
counter++;
}
}
if((row + 1) < NUM && (col - 1) > 0) {
if(grid[row + 1][col - 1] == 0) {
counter++;
}
}
if((row + 1) < NUM && (col + 1) < NUM) {
if(grid[row + 1][col + 1] == 0) {
counter++;
}
}
grid[row][col] = counter;
}
}
}

My assignment is to make a house using for loops. The house should look like this

*
*****
*********
*********
**** ***
**** ***
so far i only have this
for (int i=1; i<10; i += 4)
{
for (int j=0; j<i; j++)
{
System.out.print("*");
}
System.out.println("");
}
}
}

The simplest decision will be:
for (int y = 0; y < 6; y++) {
int shift = y < 2 ? 4 / (y + 1) : 0;
for (int x = 0; x < 9 - shift; x++) System.out.print(x >= shift && (y < 4 || (x < 4 || x > 5)) ? "*" : " ");
System.out.println();
}

I think Andre's answer is the most concise one, but if you want to have configurable home building you can use next one(try to change HEIGHT/WIDTH to see effect):
public class House {
public static void main(String[] args) {
final int HEIGHT = 6;
final int WIDTH = 9;
for (int i = 0; i < HEIGHT * 2; i += 2) {
for (int j = 0; j < WIDTH; j++) {// check for roof
if ((i + (i % 2) + (WIDTH) / 2) < j // right slope
|| (i + (i % 2) + j) < (WIDTH) / 2)// left slope
{
System.out.print(" ");
} else {
if ((i / 2 >= HEIGHT * 2 / 3) && (j >= WIDTH / 2) && j < WIDTH / 2 + HEIGHT / 3) {// check for door
System.out.print(" ");
} else {// solid then
System.out.print("*");
}
}
}
System.out.println();
}
}
}
EDIT - answer to comment:
Try to run next two example and compare output:
public static void main(String[] args) {
final int SIZE = 9;
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
System.out.print(i < j ? "+" : "-");
}
System.out.println();
}
}
and
public static void main(String[] args) {
final int SIZE = 9;
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
System.out.print(i < SIZE - j - 1 ? "+" : "-");
}
System.out.println();
}
}
First one will give you right slope and second left one. It all come from geometric properties of points. In first case all points will have bigger value on on x axis than on y axis. In second both x and y in sum won't exceed SIZE.
You can try to modify boolean expression inside of if() statement and see what happens, but I'd encourage you to get piece of paper and try to play with paper and pen and see what properties certain points have. Let me know if you need more explanation.

You could use a two-dimensional array like this:
char matrice [][]= {{' ',' ',' ',' ' '*', ' ',' ',' ',' '},
{' ',' ','*','*', '*', '*','*',' ',' '}};
(And so on). You basically draw your house using your array indexes.
Now you can parse each line using System.out.print() when you have to print a character, and System.out.println("") between each row.
It would look like this:
for(char[] line : house){
for(char d : line){
System.out.print(d);
}
System.out.println("");
}
You should take a look at the for-each statement documentation if you're not familiar with it.

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