Program to Print Inverted Right Half Pyramid Pattern (Star Pattern)

Given an integer N, print N rows of inverted right half pyramid pattern. In inverted right half pattern of N rows, the first row has N number of stars, second row has (N - 1) number of stars and so on till the Nth row which has only 1 star.

Examples:

Input: n = 5
Output:
*****
****
***
**
*

Input: n = 3
Output:
***
**
*

Approach:

The problem can be solved using two nested loops. The outer loop will run for the rows and the inner loop will print the stars.

Step-by-step algorithm:

• Run an outer loop from i = 1 to the number of rows.
  • Run an inner loop from j = 1 to N - i + 1.
    • Print an asterisk in each iteration of the inner loop.
  • Print a newline character ("\n") to move to the next row.
• After N iterations, we will have the inverted right half pyramid.

Below is the implementation of the above approach:

#include <iostream>
using namespace std;

int main()
{
    // Number of rows
    int N = 5;

    // Outer loop runs N times, once for each row
    for (int i = 1; i <= N; i++) {
        // Inner loop prints 'N - i + 1' stars
        for (int j = 1; j <= N - i + 1; j++) {
            cout << "*";
        }
        // Move to the next line
        cout << "\n";
    }

    return 0;
}

import java.util.Scanner;

public class Main {
    public static void main(String[] args)
    {

        // Number of rows
        int N = 5;

        // Outer loop runs N times, once for each row
        for (int i = 1; i <= N; i++) {
            // Inner loop prints 'N - i + 1' stars
            for (int j = 1; j <= N - i + 1; j++) {
                System.out.print("*");
            }
            // Move to the next line
            System.out.println();
        }
    }
}

# Number of rows
N = 5

# Outer loop runs N times, once for each row
for i in range(1, N + 1):
    # Inner loop prints 'N - i + 1' stars
    for j in range(1, N - i + 2):
        print("*", end="")
    # Move to the next line
    print()

using System;

class MainClass
{
    public static void Main(string[] args)
    {
        // Number of rows
        int N = 5;

        // Outer loop runs N times, once for each row
        for (int i = 1; i <= N; i++)
        {
            // Inner loop prints 'N - i + 1' stars
            for (int j = 1; j <= N - i + 1; j++)
            {
                Console.Write("*");
            }

            // Move to the next line
            Console.WriteLine();
        }
    }
}

// Number of rows
const N = 5;

// Outer loop runs N times, once for each row
for (let i = 1; i <= N; i++) {
    // Inner loop prints 'N - i + 1' stars
    for (let j = 1; j <= N - i + 1; j++) {
        process.stdout.write("*");
    }
    // Move to the next line
    process.stdout.write("\n");
}

*****
****
***
**
*

Time Complexity: O(N^2), where N is the number of rows in the pattern.
Auxiliary Space: O(1)

Inverted Triangle in Python

Visit Course