So I've been working on this lab for a while now for my programming class and so far I think I'm on the right track.
However, I'm not quite sure how to mirror the numbers. So pretty much, my code is only printing the top half of the triangle. Anyway here is the actual assignment that was given to us:
Write a program using a Scanner that asks the user for a number n between 1 and 9 (inclusive). The program prints a triangle with n rows. The first row contains only the square of 1, and it is right-justified. The second row contains the square of 2 followed by the square of 1, and is right justified. Subsequent rows include the squares of 3, 2, and 1, and then 4, 3, 2 and 1, and so forth until n rows are printed.
Assuming the user enters 4, the program prints the following triangle to the console:
1
4 1
9 4 1
16 9 4 1
9 4 1
4 1
1
For full credit, each column should be 3 characters wide and the values should be right justified.
Now here is what I have written for my code so far:
import java.util.Scanner;
public class lab6 {
public static void main(String[] args) {
Scanner kybd = new Scanner(System.in);
System.out.println(
"Enter a number that is between 1 and 9 (inclusive): ");
// this is the value that the user will enter for # of rows
int rows = kybd.nextInt();
for (int i = rows; i > 0; i--) {
for (int j = rows; j > 0; j--)
System.out.print((rows - j + 1) < i ?
" " : String.format("%3d", j * j));
System.out.println();
}
}
}
And this is what that code PRINTS when I enter 4:
Enter a number that is between 1 and 9 (inclusive):
4
1
4 1
9 4 1
16 9 4 1
As you can see, I can only get the TOP half of the triangle to print out. I've been playing around trying to figure out how to mirror it but I can't seem to figure it out. I've looked on this website for help, and all over the Internet but I can't seem to do it.
Answer is:
public static void main(String... args) {
Scanner kybd = new Scanner(System.in);
System.out.println("Enter a number that is between 1 and 9 (inclusive): ");
int rows = kybd.nextInt(); // this is the value that the user will enter for # of rows
for (int i = -rows + 1; i < rows; i++) {
for (int j = -rows; j < 0; j++)
System.out.print(abs(i) > j + rows ? " " : String.format("%3d", j * j));
System.out.println();
}
}
Try think of this as how to find points(carthesians) that are betwean three linear functions(area of triangle that lied betwean):
y = 0 // in loops i is y and j is x
y = x + 4
y = -x -4
And here is example result for 4:
And 9:
In the outer loop or stream you have to iterate from 1-n to n-1 (inclusive) and take absolute values for negative numbers. The rest is the same.
If n=6, then the triangle looks like this:
1
4 1
9 4 1
16 9 4 1
25 16 9 4 1
36 25 16 9 4 1
25 16 9 4 1
16 9 4 1
9 4 1
4 1
1
Try it online!
int n = 6;
IntStream.rangeClosed(1 - n, n - 1)
.map(Math::abs)
.peek(i -> IntStream.iterate(n, j -> j > 0, j -> j - 1)
// prepare an element
.mapToObj(j -> i > n - j ? " " : String.format("%3d", j * j))
// print out an element
.forEach(System.out::print))
// start new line
.forEach(i -> System.out.println());
See also: Output an ASCII diamond shape using loops
Another alternative :
public static void main(String args[]) {
Scanner kybd = new Scanner(System.in);
System.out.println("Enter a number that is between 1 and 9 (inclusive): ");
int rows = kybd.nextInt(); // this is the value that the user will enter for # of rows
int row = rows, increment = -1;
while (row <= rows){
for (int j = rows; j > 0; j--) {
System.out.print(rows - j + 1 < row ? " " : String.format("%3d", j * j));
}
System.out.println();
if(row == 1) {
increment = - increment;
}
row += increment;
}
}
The outer loop from 1-n to n-1 inclusive, and the inner decrementing loop from n to 0. The if condition is the absolute value of i should not be greater than n - j.
Try it online!
int n = 6;
for (int i = 1 - n; i <= n - 1; i++) {
for (int j = n; j > 0; j--)
if (Math.abs(i) > n - j)
System.out.print(" ");
else
System.out.printf("%3d", j * j);
System.out.println();
}
Output:
1
4 1
9 4 1
16 9 4 1
25 16 9 4 1
36 25 16 9 4 1
25 16 9 4 1
16 9 4 1
9 4 1
4 1
1
See also: Invert incrementing triangle pattern
Related
Recently, I had encountered an interesting programming puzzle which had some good twist and turn mentioned in the puzzle. Below the question which amazed me, I simply eager to know if any relevant solution probably in java is feasible for below scenario.
Problem statement:
There is a grid of dimension m*n, initially, a bacterium is present at the bottom left cell(m-1,0) of the grid with all the other cells empty. After every second, each bacteria in the grid divides itself and increases the bacteria count int the adjacent(horizontal,vertical and diagonal) cells by 1 and dies.
How many bacteria are present at the bottom right cell(m-1,n-1) after n-1 seconds?
I had taken references from
https://www.codechef.com/problems/BGH17
but failed to submit the solution
Below is the image for more insite of problem
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.Stack;
public class BacteriaProblem {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("Number of Rows: ");
int m = sc.nextInt();
System.out.println("Number of Columns: ");
int n = sc.nextInt();
int[][] input = new int[m][n];
input[m - 1][0] = 1;
Stack<String> stack = new Stack<>();
stack.push(m - 1 + "~" + 0);
reproduce(stack, input, n - 1);
System.out.println("Value at Bottom Right corner after n-1 secs: " + input[m - 1][n - 1]);
}
private static void reproduce(Stack<String> stack, int[][] input, int times) {
//exit condition
if (times < 1) {
return;
}
//bacteria after splitting
List<String> children = new ArrayList<>();
//reproduce all existing bacteria
while (!stack.isEmpty()) {
String[] coordinates = stack.pop().split("~");
int x = Integer.parseInt(coordinates[0]);
int y = Integer.parseInt(coordinates[1]);
for (int i = -1; i <= 1; i++) {
for (int j = -1; j <= 1; j++) {
if (i == 0 && j == 0) continue;
split(input, x + i, y + j, children);
}
}
input[x][y]--;
}
//add all children to stack
for (String coord : children) {
stack.push(coord);
}
//reduce times by 1
reproduce(stack, input, times - 1);
}
private static void split(int[][] input, int x, int y, List<String> children) {
int m = input.length;
int n = input[0].length;
if (x >= 0 && x < m && y >= 0 && y < n) {
input[x][y]++;
children.add(x + "~" + y);
}
}
}
Well, I was asked this question in an Online Hackerrank test and couldn't solve it at that time.
I did later try to code it and here's the soln in C++,
long countBacteriasAtBottomRight(int m, int n){
long grid[m][n];
// Set all to 0, and only bottom left to 1
for (int i=0; i<m; i++){
for (int j=0; j<n; j++){
grid[i][j] = 0;
}
}
grid[m-1][0] = 1;
// Start the cycle, do it for (n-1) times
int time = n-1;
vector<long> toBeUpdated;
while (time--){
cout << "\n\nTime: " << time;
for (int i=0; i<m; i++){
for (int j=0; j<n; j++){
while (grid[i][j] > 0){
grid[i][j]--;
// upper left
if (i > 0 && j > 0){
toBeUpdated.push_back(i-1);
toBeUpdated.push_back(j-1);
}
// upper
if (i > 0){
toBeUpdated.push_back(i-1);
toBeUpdated.push_back(j);
}
// upper right
if (i > 0 && j < n-1){
toBeUpdated.push_back(i-1);
toBeUpdated.push_back(j+1);
}
// left
if (j > 0){
toBeUpdated.push_back(i);
toBeUpdated.push_back(j-1);
}
// bottom left
if (i < m-1 && j > 0){
toBeUpdated.push_back(i+1);
toBeUpdated.push_back(j-1);
}
// bottom
if (i < m-1){
toBeUpdated.push_back(i+1);
toBeUpdated.push_back(j);
}
// bottom right
if (i < m-1 && j < n-1){
toBeUpdated.push_back(i+1);
toBeUpdated.push_back(j+1);
}
// right
if (j < n-1){
toBeUpdated.push_back(i);
toBeUpdated.push_back(j+1);
}
};
}
}
// Update all other cells
for (int k=0; k<toBeUpdated.size(); k+=2){
grid[toBeUpdated[k]][toBeUpdated[k+1]]++;
}
for (int i=0; i<m; i++){
cout << endl;
for (int j=0; j<n; j++)
cout << grid[i][j] << " ";
}
// Clear the temp vector
toBeUpdated.clear();
};
return grid[m-1][n-1];
}
The starting situation only has a value in the left-most column 0. We need to know the situation in the right-most column n-1 after time n-1. This means that we only have to look at each column once: column x at time x. What happens to column x after time x is no longer important. So we go from left to right, adding up the cells from the previous column:
1
1 8
1 7 35
1 6 27 104
1 5 20 70 230
1 4 14 44 133 392
1 3 9 25 69 189 518
1 2 5 12 30 76 196 512
1 1 2 4 9 21 51 127 323 ...
You will also notice that the result for the last cell is only influenced by two cells in the previous column, and three in the one before that, so to calculate the end result for e.g. the case n=9, you only need to calculate the values in this triangle:
1
1 4 14
1 3 9 25 69
1 2 5 12 30 76 196
1 1 2 4 9 21 51 127 323
However high the grid is, we only ever have to go up n/2 (rounded up) rows. So the total number of sums we have to calculate is n2/4, or n×m if m < n/2.
Also note that we don't have to store all these values at once, because we go column by column from left to right. So we only need a one-dimensional array of size n/2, and the current values in it are transformed like this (e.g. going from column 4 to 5 in the example above):
[4, 5, 3, 1] (0) -> 0 + 5 - 0 = 5
[9, 5, 3, 1] (5) -> 9 + 3 - 5 = 7
[9,12, 3, 1] (7) -> 12 + 1 - 7 = 6
[9,12, 9, 1] (6) -> 9 + 0 - 6 = 3
[9,12, 9, 4] (3) -> 4 + 0 - 3 = 1
[9,12, 9, 4, 1] (1) (additional value is always 1)
where we iterate over the values from left to right, add up the value to the left and right of the current element, subtract a temporary variable which is initialized to 0, store the result in a temporary variable, and add it to the current element.
So the theoretical time complexity is O(n2) or O(n.m) and the space complexity is O(n) or O(m), whichever is smaller. In real terms, the number of steps is n2/4 and the required space is n/2.
I don't speak Java, but here's a simple JavaScript code snippet which should easily translate:
function bacteria(m, n) {
var sum = [1];
for (var col = 1; col < n; col++) {
var temp = 0;
var height = Math.min(col + 1, n - col, m);
if (height > sum.length) sum.push(0);
for (var row = 0; row < height; row++) {
var left = row > 0 ? sum[row - 1] : 0;
var right = row < sum.length - 1 ? sum[row + 1] : 0;
temp = left + right - temp;
sum[row] += temp;
}
}
return sum[0];
}
document.write(bacteria(9, 9));
I'm newbie to programming. So as an exercise, I'm trying to print a number pattern like below
4
34
234
1234
I tried the below code
public static void main(String[] args) {
// TODO Auto-generated method stub
int n =4;
for (int i = 1; i <= n; i++) {
for (int j = i; j <= n; j++) {
System.out.print(" ");
}
int num = 4;
for (int j = 1; j <= i; j++) {
System.out.print(num);
num--;
}
System.out.println("");
}
}
but it is printing in this way.
4
43
432
4321
I think, I have to decrement the value before printing. Please correct me if i'm wrong. But I'm stuck here. Can anyone please help me?
This is the pattern you want to get:
4
34
234
1234
When you describe the pattern in words, it could look like this:
line 1 has 3 spaces, and then the digit 4
line 2 has 2 spaces, and then the digits 3 and 4
line 3 has 1 space, and then the digits 2 to 4
line 4 has 0 spaces, and then the digits 1 to 4
There is already some kind of pattern here. The last two lines look remarkably similar. Let's see whether the first two lines can be brought into the same form:
line 1 has 3 space(s), and then the digits 4 to 4
line 2 has 2 space(s), and then the digits 3 to 4
line 3 has 1 space(s), and then the digits 2 to 4
line 4 has 0 space(s), and then the digits 1 to 4
Now that looks good. The next step is to change the wording to depend on the given line:
line i has (4 - i) space(s), and then the digits (4 - i + 1) to 4
I took care not to say 5 instead of the 4 + 1, so that the 4 is still visible. Let's give this 4 another name:
line i has (max - i) space(s), and then the digits (max - i + 1) to max
Now you should be able to translate this instruction into Java code.
You only need one inner for-loop.
I use the ternary operator (also called elvis-operator because ?:) to decide whether to print the number or a blank space:
int n = 7;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
System.out.print(j > n-i ? j : " ");
}
System.out.println();
}
It prints
7
67
567
4567
34567
234567
1234567
You actually only need one nested loop in this situation:
int n = 4;
for (int i = n; i > 0; i--) {
for (int j = 1; j <= n; j++) {
if (j < i) {
System.out.print(" ");
} else {
System.out.print(j);
}
}
System.out.println();
}
So loop from 0 to n and either print a space, or print the number if the inner counter is less than i.
Output:
4
34
234
1234
How do I make this loop properly? it right now So it loops but it does not loop properly. It does this
Here are the numbers:
15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 [1]
How many positions do you want to shift?: 2
2 1 15 14 13 12 11 10 9 8 7 6 5 4 3 [3]
How many positions do you want to shift?: 4
the [] are where its suppose to ask me for my input instead of me just putting in a input
its suppose to run like this:
re are the numbers:
15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
How many positions do you want to shift?: 1
2 1 15 14 13 12 11 10 9 8 7 6 5 4 3
How many positions do you want to shift?: 4
System.out.println("Here are the numbers:");
for (i=0; i<numberArray.length; i++) {
System.out.print(numberArray[i] + " ");
}
while (x != input.nextInt()){
System.out.printf("How many positions do you want to shift?: ");
int shiftTimes=input.nextInt();
for( i = 0; i < shiftTimes; ++i)
shift.Shifter(numberArray);
for(j = 0; j < numberArray.length; j++)
System.out.printf(numberArray[j]+" ");
}
}
}
Also How Do I make it exit the program when I enter in a invalid number and how do I get get it to read a negative value and get it to shift left
Edit: heres my shifter code
public static void Shifter(int[] list)
{
int i;
if (list.length < 2) return;
int last = list[list.length - 1];
for(i = list.length - 1; i > 0; i--) {
list[i] = list[i - 1];
}
list[0] = last;
}
This should work for right shift. It should work with inputs larger then array length as well.
for (int i = shiftTimes%numberArray.length; i > 0; i--) {
System.out.print(numberArray[numberArray.length - i] + " ");
}
for (int i = 0; i < numberArray.length - shiftTimes%numberArray.length; i++) {
System.out.print(numberArray[i] + " ");
}
Reversing this logic should produce a left shift approach.
An invalid input would be the length of the array (because the result will be the same) or 0 because that doesn't do anything:
if (shiftTimes == numberArray.length || shiftTimes == 0) {
// present error to user
}
UPDATE: Putting the logic in your function. Also updated the invalid input check.
public static void Shifter(int[] list, int input)
{
for (int i = input%list.length; i > 0; i--) {
System.out.print(list[list.length - i] + " ");
}
for (int i = 0; i < list.length - input%list.length; i++) {
System.out.print(list[i] + " ");
}
}
The function call would be:
Shifter(numberArray, shiftTimes);
Pasted below is a program to print Pascal's triangle. The way to compute any given position's value is to add up the numbers to the position's right and left in the preceding row. For instance, to compute the middle number in the third row, you add 1 and 1. the sides of the triangle are always 1 because you only add the number to the upper left or the upper right (there being no second number on the other side).
int pascal[][]=new int[50][50]; int j;
for(int i=0;i<m;i++)
{
pascal[i][i]=1;
for(j=1;j<i;j++)
{
pascal[i][j]=pascal[i-1][j-1]+pascal[i-1][j];
}
for(int n=1;n<=m-i;n++)
{
System.out.print(" ");
}
for(int k=1;k<=i;k++)
{
System.out.print(pascal[i][k]);
System.out.print(" ");
}
System.out.println(" ");
}
Is there any way to accomplish this without using arrays?
I'm trying this combination without arrays:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
You can easily compute it using Combination.
You can compute combination as:
|n| = (n!) / ((n-k)!*k!)
|k|
So if you want to print the image above you would start as:
int size = 5;
for(int i = 0; i < size; i++){
for(int k = 0; k < (size - i)/2; k++)
System.out.print(" "); // print the intendation
for(int j = 0; j <= i; j++){
System.out.print(combination(i,j));
}
System.out.println("");
}
How will do a program that displays a multiplication table based on the size that the user inputs? And will add each row and each column? Something like this:
Enter a number: 4
1 2 3 4 10
2 4 6 8 20
3 6 9 12 30
4 8 12 16 40
10 20 30 40
I tried this:
Scanner s = new Scanner(System.in);
System.out.print("Enter a number: ");
int x = s.nextInt();
for(int i = 1; i <= x; i++)
{
for (int j = 1; j <=x; j++)
{
System.out.print((i*j) + "\t");
}
System.out.println();
}
Sample Output:
Enter a number: 4
1 2 3 4
2 4 6 8
3 6 9 12
4 8 12 16
How I will do to add each row and each column?
Since this seems like homework, I wouldn't feel comfortable writing your code for you. However, keep the following things in mind.
Your matrix will always be a square, as the user only enters a single number, of n x n numbers.
Since these numbers increment by one along the row and column, the sum of each row and column pair will be the same. In other words, the total of row[n] will equal the total of column[n].
Using that, you can create a single array of size n to store the sum of each row. For example:
Enter a number: 3
1 2 3 x
2 4 6 y
3 6 9 z
x y z
When you're looping through each row, you can store the row total in the array.
Row 0: Add 1 + 2 + 3 and store in array[0]
Row 1: Add 2 + 4 + 6 and store in array[1]
Row 2: Add 3 + 6 + 9 and store in array[2]
At the end of each row you can simply display the total in array[row]. When you finish drawing all rows, you'd simply loop through array and display each total value.
Hope this points you in the right direction!
public static void main(String[] args){
Scanner s = new Scanner(System.in);
System.out.print("Enter size of table: ");
int x = s.nextInt();
int r = 0;
int l = 0;
int f = 0;
for(int i=1;i<=x;i++){
for (int j=1; j <=x; j++)
{
r = r + j;
System.out.print(i*j+"\t");
}
System.out.print(r);
System.out.println();
System.out.println();
l=l+i;
}
for(int k = 1; k<=x;k++)
{
f=f+l;
System.out.print(f + "\t");
}