Inverted Right Triangle Star Pattern
Problem
Read n and print an upside-down right triangle of stars. Row 1 has n stars and every next row has one star less.
Constraints
- 1 <= n <= 50
Examples
Input
3
Output
*** ** *
Solutions
For row r the number of stars is n - r + 1, so the first row is full and the last row has one star.
Java 8 has no String.repeat, so we use a small helper method. In Java 11 and newer you can write "*".repeat(count).
/*
* Problem : Inverted right triangle of stars.
* Input : One integer n (number of rows).
* Output : n lines of stars.
* Approach: Count the stars down from n to 1.
* Time : O(n^2)
* Space : O(1)
* Java : 8 or newer.
*/
import java.util.Scanner;
public class Main {
// This method returns the text repeated "count" times. Java 11 has String.repeat for this.
static String repeat(String text, int count) {
StringBuilder result = new StringBuilder();
for (int i = 0; i < count; i++) {
result.append(text);
}
return result.toString();
}
public static void main(String[] args) {
int rows = new Scanner(System.in).nextInt();
// Step 1: Repeat once for each row. Row numbers go from 1 to rows.
for (int row = 1; row <= rows; row++) {
// Step 2: Row 1 has n stars. Each next row has one star less. Example: n = 4 -> 4, 3, 2, 1.
int stars = rows - row + 1;
// Step 3: Print the stars on one line.
System.out.println(repeat("*", stars));
}
}
}
Hints (2)
- Use one outer loop for the rows. What changes from one row to the next?
- Write the pattern for n = 3 on paper and count the stars in each row.