I want to print the pyramid pattern using word "Stream" by using for loop in Java. Please anyone can help me with this. I have printed pyramid with "*". I have also attached the program below:
Desired Result:
S
S t
S t r
S t r e
S t r e a
S t r e a m
What I have so far:
public class Pyramid
{
public static void main(String[] args)
{
System.out.println("-----Pyramid------");
int n = 5;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n - i; j++)
System.out.print(" ");
for (int k = 1; k <= 2 * i - 1; k++)
System.out.print("S");
System.out.print("\n");
}
}
}
This should work with any word.
As a hint I'd recommend to start loops with 0 instead of 1.
public static void main(String[] args) {
System.out.println("-----Pyramid------");
String word = "Stream";
int n = word.length();
for (int i = 0; i < n+2; i++) {
for (int j = 0; j <= n - i; j++)
System.out.print(" ");
for (int k = 0; k < i - 1; k++)
System.out.print(word.charAt(k) + " ");
System.out.print("\n");
}
}
Output:
-----Pyramid------
S
S t
S t r
S t r e
S t r e a
S t r e a m
You can use the charAt method in order to extract from the word Stream the chars you need inside the loop to create the pyramid
For example:
"Stream".charAt(0);
Will print the char S
"Stream".charAt(3);
Will print the char e.
More info here: String class reference
Related
Help with this question:
Take a positive integer n and form n triangles from stars with their base down of size n each.
For example, for input 3, the following output will be obtained:
* * *
** ** **
*** *** ***
Here's what I've tried.
import java.util.Scanner;
public class ex3 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("enter n");
int n = input.nextInt();
while (n <= 0) {
System.out.println("error");
n = input.nextInt();
}
for (int r = 1; r <= n; r++) {
for (int c = 1; c <= n - r; c++)
System.out.print(" ");
for (int c = 1; c <= r; c++)
System.out.print("*");
System.out.println();
}
}
}
Here's my solution, with explanations in comments
import java.util.Scanner;
class Main {
public static void main(String[] args) {
try(Scanner input = new Scanner(System.in)) {
System.out.println("enter n");
int n = input.nextInt();
while (n <= 0) {
System.out.println("error");
n = input.nextInt();
}
for (int r = 1; r <= n; r++) { // <-- we will have to print n rows
printLine(n, r);
}
}
}
static void printLine(int n, int lineNumber) {
StringBuffer line = new StringBuffer();
for(int i = 0; i < n; i ++) { // <-- each line will have '*'s for n triangles
for(int j = lineNumber; j > 0; j--) { // <-- each line has as many '*'s as its line number, so print those first
line.append("*");
}
for(int j = 0; j < n - lineNumber + 1; j ++) { // <-- we then have to add enough spaces to leave room for the next triangle's '*'s
line.append(" ");
}
}
System.out.println(line.toString()); // <-- print the line we've built so far
}
}
EDIT:
Here's a replit that avoids one of the loops by using a modulo to print an entire line at once, and also uses recursion, for no real reason, in place of the outer-most loop: https://replit.com/#anqit/MicroExtrovertedTrace#Main.java
First, the blanks follow the stars, then, you have to repeat n times the two loops:
for (int r = 1; r <= n; r++) {
for (int t = 1; t <= n; t++) {
for (int c = 1; c <= r; c++)
System.out.print("*");
for (int c = 1; c <= n - r; c++)
System.out.print(" ");
}
System.out.println();
}
You have one for loop that prints out a single triangle of stars using nested for loops. The outer loop is responsible for the number of rows and the inner loops are responsible for printing the spaces and stars.
In my updated code, I added an outer loop that runs n times, and each time it runs, it prints out a triangle of stars. The inner loops are responsible for printing the stars of the triangle and spaces between the triangles.
The main difference is that in your first code, only one triangle is printed, while in my updated code, n triangles are printed. In addition, the indentation of the inner loops has been adjusted to align the triangles correctly on the same line.
import java.util.Scanner;
public class ex3 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("enter n:");
int n = input.nextInt();
while (n <= 0) {
System.out.println("error: please enter a positive number:");
n = input.nextInt();
}
for (int r = 1; r <= n; r++) {
for (int t = 1; t <= n; t++) {
for (int c = 1; c <= r; c++) {
System.out.print("*");
}
for (int i = 0; i < n - r + 1; i++) {
System.out.print(" ");
}
}
System.out.println();
}
}
}
Here is a 3 forloop variant
int n = 4;
for (int i = 1; i <= n; i++) {
StringBuilder b = new StringBuilder("");
for (int s = 0; s < n; s++) {
b.append(s < i ? "*" : " ");
}
String line = b.append(" ").toString();//space between
for (int j = 1; j < n; j++) {
System.out.print(line);
}
System.out.println();
}
produces
> * * *
> ** ** **
> *** *** ***
> **** **** ****
Code below is for pattern a, feel like once I get a I could get the others.
I like the array[int].length syntax in Java and was helpful to get the pattern to print as shown in the picture. But I do not think such a thing exists in C#.
class Main {
public static void main(String[] args)
{
char[][] arr = new char[10][10];
int starCount = 10;
for(int i = 0; i < arr.length; i++)
{
for(int j = 0; j < starCount; j++)
{
arr[i][j] = '*';
}
for(int k = starCount; k < arr[i].length; k++)
{
arr[i][k] = '-';
}
starCount--;
}
for(int a = 0; a < arr.length; a++)
{
for(int b = 0; b < arr[a].length; b++)
{
System.out.print(arr[a][b]);
}
System.out.println();
}
}
}
This code prints the * in a decreasing fashion but I am struggling with how to replace the empty elements of the array with the - character as shown in the image.
class MainClass {
public static void Main (string[] args)
{
char[ , ] arr = new char[10,10];
int starCount = 10;
for(int i = 0; i < arr.Length; i++)
{
for (int j = 0; j < starCount; j++)
{
arr[i , j] = '*';
}
for (int k = 0; ) //IDK WHAT TO DO TO ASSIGN ARR[I , K] = '-';
starCount--;
}
for (int i = 0; i < 10; i++)
{
for (int j = 0; j < 10; j++)
{
Console.Write(arr[i , j]);
}
Console.WriteLine();
}
}
}
You can use two nested for loops and one if else statement to print all of these patterns. This is Java code, but I think it can be easily converted to C#:
int n = 5;
for (int i = -n; i <= n; i++) {
for (int j = -n; j <= n; j++) {
// a) if (i + j <= 0)
// b) if (i + j >= 0)
// c) if (i <= j)
// d) if (Math.abs(i) + Math.abs(j) <= n)
if (i + j <= 0)
System.out.print("*");
else
System.out.print("-");
}
System.out.println();
}
Output (combined):
a) b) c) d)
*********** ----------* *********** -----*-----
**********- ---------** -********** ----***----
*********-- --------*** --********* ---*****---
********--- -------**** ---******** --*******--
*******---- ------***** ----******* -*********-
******----- -----****** -----****** ***********
*****------ ----******* ------***** -*********-
****------- ---******** -------**** --*******--
***-------- --********* --------*** ---*****---
**--------- -********** ---------** ----***----
*---------- *********** ----------* -----*-----
See also: Output an ASCII diamond shape using loops
Here's a different approach to the original problem, which might be easier for you to convert.
Build a string of 10 stars and 9 dashes, e.g. hard-coded that would be:
String line = "**********---------";
Now print a 10 rows with substrings of that string:
for (int i = 0; i < 10; i++)
System.out.println(line.substring(i, i + 10));
Output
**********
*********-
********--
*******---
******----
*****-----
****------
***-------
**--------
*---------
If the size is dynamic, based on an int value in variable starCount, then in Java 11+ you can use the repeat() method:
String line = "*".repeat(starCount) + "-".repeat(starCount - 1);
for (int i = 0; i < starCount; i++)
System.out.println(line.substring(i, i + starCount));
That one should be easy to do in C#. See: Best way to repeat a character in C#.
In versions of Java below 11, you can build a char[]:
char[] line = new char[2 * starCount - 1];
Arrays.fill(line, 0, starCount, '*');
Arrays.fill(line, starCount, line.length, '-');
for (int i = 0; i < starCount; i++)
System.out.println(new String(line, i, starCount));
The easiest way would be to look at the C# documentation for the Array class. There you would find that the Array class has a GetLength() method, that returns what the length property of a Java array returns.
Using that method you can change your code to
class MainClass {
public static void Main (string[] args)
{
char[ , ] arr = new char[10,10];
int starCount = 10;
for(int i = 0; i < arr.GetLength(0); i++)
{
for (int j = 0; j < starCount; j++)
{
arr[i , j] = '*';
}
for (int k = starCount; k < arr.GetLength(1); k++)
{
arr[i , k] = '*';
}
starCount--;
}
for (int i = 0; i < arr.GetLength(0); i++)
{
for (int j = 0; j < arr.GetLength(1); j++)
{
Console.Write(arr[i , j]);
}
Console.WriteLine();
}
}
}
For some reason when you switched from java to C# you went from using jagged arrays:
//java
char[][] arr = new char[10][10];
To rectangular arrays:
//c#
char[ , ] arr = new char[10,10];
This undoubtedly makes your life more hard work because it means a lot more has to change. C# supports jagged arrays in exactly the same way Java does, and in fact if I hadn't written "//java" above you wouldn't have been able to tell whether it was C# or java because they're the same
I like the array[int].length syntax in Java ... But I do not think such a thing exists in C#.
It absolutely does, and you need to change just one single character to get it: properties in C# are named in Pascal case, so you want Length, not length
In fact, the logic of that entire block of java you have will work just fine in C# - just paste it in and change the following minor changes:
length -> Length
System.out.print -> Console.Write
System.out.println -> Console.WriteLine
I want to basically hide characters following three constant dots (...), the pattern goes like this:
Inputs a phrase from the user and outputs the phrase followed by three dots (...), then the phrase minus one character followed by three dots (...), then the phrase minus two characters followed by the dots, and so on until only one dot is left.
Note: This has to be done using nested for loops only
Sample input
1
disappear
Expected output:
disappear...
disappea...
disappe...
disapp...
disap...
disa...
dis...
di...
d...
...
..
.
This is my attempt:
Problem: I am unable to make it so the phrase decreases each time (minus 1 each time)
I tried using the charAt(); method, but it wouldn't work, I am sure that you would need a for loop separate for each of the dots or a whole set of dots, in this case.
import java.util.Scanner;
public class Dissappear{
public static void main(String[]args){
Scanner keyboard = new Scanner(System.in);
int option = keyboard.nextInt();
String phrase = keyboard.next();
if (option == 1){
for (int x = 0; x <= phrase.length(); x++){
System.out.print(phrase + "...");
for (int y = 0; y <= phrase.length(); y++){
char n = phrase.charAt(y);
System.out.print(n+"...");
}
}
}
}
}
This is how I got it to work:
public class Disappear {
public static void main(String... args) {
String word = "disappear";
int originalLength = word.length();
for(int i = 0; i < originalLength; i++) {
System.out.println(word.substring(0, originalLength - i) + "...");
}
for(int i = 0; i < 3; i++) {
for(int j = 0; j < 3 - i; j++) {
System.out.print(".");
}
System.out.println();
}
}
}
Without substring:
public class Disappear {
public static void main(String... args) {
String word = "disappear";
int originalLength = word.length();
for(int i = 0; i < originalLength; i++) {
for(int j = 0; j < originalLength - i; j++) {
System.out.print(word.charAt(j));
}
System.out.println("...");
}
for(int i = 0; i < 3; i++) {
for(int j = 0; j < 3 - i; j++) {
System.out.print(".");
}
System.out.println();
}
}
}
You can do it with StringBuilder:
StringBuilder stringBuilder = new StringBuilder(str);
System.out.println(str + "...");
for (int i = 0; i < length; i++) {
stringBuilder.deleteCharAt(stringBuilder.length() - 1);
System.out.println(stringBuilder.toString() + "...");
if (i == length - 1) {
for (int j = 0; j < 2; j++) {
for (int k = j; k < 2; k++) {
System.out.print(".");
}
System.out.println();
}
}
Ok! Nested for loops. But the outer one is only included to meet the requirement. Probably not in the spirit of the assignment though. Just keep decrementing k until it is zero and then latch it there until the StringBuilder length is 0 and the inner loop terminates.
StringBuilder sb = new StringBuilder("disappear...");
for (;;) {
for (int k = sb.length() - 4; sb.length() > 0;) {
System.out.println(sb);
sb.delete(k, k + 1);
k = k > 0 ? --k : 0;
}
break;
}
I am a newbie to java programming and I am working on this excercise from my textbook. The goal is to print a V shape pattern of numbers. From the picture below, you can see what the output should look like. I am having trouble creating the other half of numbers. I have pasted my code down below for reference.
for (int i = 7; i >= 1; i--) {
for (int j = 1; j <= i; j++) {
System.out.print(" ");
}
System.out.print(i);
for (int k = 1; k >= i*2; k++) {
System.out.print(" ");
}
System.out.println(i);
Use the following code (just made a few modifications to your code, did not check its efficiency):
public static void main(String[] args) {
for (int i = 7; i >= 1; i--) {
for (int k = 7; k >= i; k--) {
System.out.print(" "); // Print 7-i number of spaces before start of each line
}
System.out.print(i); // Print i
for (int j = 1; j <= i*2; j++) {
System.out.print(" "); // Print i*2 number of spaces after printing i
}
System.out.println(i); // Print i
}
}
Rather then nesting loops (and iterating backwards), I would decompose the generating of white-space with a method to repeat a given String a given number of times. Like,
private static String repeat(String s, int n) {
return Stream.generate(() -> s).limit(n).collect(Collectors.joining());
}
Then I would prefer a StringBuilder and a single call to println like
public static void main(String[] args) {
int start = 6;
for (int i = 0; i < start; i++) {
int v = start - i;
StringBuilder sb = new StringBuilder();
sb.append(repeat(" ", i)).append(v);
sb.append(repeat(" ", 2 * v)).append(v);
System.out.println(sb);
}
}
I want to get a number from user and calculate how many different triangles can be formed with the given length for example :
5 (2-2-1)
Answer: 1
12 (5,5,2)(3,4,5)(4,4,4)
Answer: 3
I've wrote some codes but I want a faster way to do that.
Here is my codes:
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int t = input.nextInt();
int value = 0;
for (int i = 1; i < t; i++) {
for (int j = i; j < t; j++) {
for (int h = j; h < t; h++) {
if (i+h+j == t & i+j > h & i+h > j & h+j > i) value++;
}
}
}
System.out.println(value);
}
You can do it in O(1):
int n = input.nextInt();
long value = Math.round(((long)n*n)/12d) - ((long)n/4)*(((long)n + 2)/4);
using Alcuin's Sequence.
You can make it O(n^2) easily.
for (int i = 1; i < t; i++) {
for (int j = i; j < t; j++) {
int h=t-i-j;
//check in O(1)
}
}