Why does the output look right-aligned?

Because we print spaces while j > i. That adds leading whitespace on smaller rows, pushing the numbers to the right.

How are the numbers printed in descending order?

The inner loop runs j from 5 down to 1. When j <= i, we print j, which naturally prints i..1 on each row.

What is the time complexity?

O(n^2) for n rows because each row performs a fixed-width inner loop.

Right-Aligned Descending Number Triangle in C#

Q: What is the time complexity?

O(n^2) for n rows because each row performs a fixed-width inner loop.

Beginner

⏱️ 6 min read

📚 Updated: Aug 2025

🎯 2 Code Examples

Condition + Loops

Right-aligned descending number triangle output (1, 21, 321, 4321, 54321) in C# programming

What You’ll Learn

How to print a right-aligned triangle where each row prints numbers in descending order: 1, then 21, then 321, and so on.

⭐ Pattern Output

For rows = 5, the pattern looks like this:

Output

Complete C# Program

This matches the reference logic: loop j from 5 down to 1, printing spaces while j > i, otherwise print j.

using System;

namespace MyApp
{
    class Program
    {
        static void Main(string[] args)
        {
            int i, j;

            for (i = 1; i <= 5; i++)
            {
                for (j = 5; j >= 1; j--)
                {
                    if (j > i)
                        Console.Write(" ");
                    else
                        Console.Write(j);
                }
                Console.WriteLine();
            }
        }
    }
}

Variation — User Input rows

Make the number of rows configurable:

using System;

namespace MyApp
{
    class Program
    {
        static void Main(string[] args)
        {
            Console.Write("Enter rows: ");
            int rows = Convert.ToInt32(Console.ReadLine());
            if (rows < 1) return;

            for (int i = 1; i <= rows; i++)
            {
                for (int j = rows; j >= 1; j--)
                    Console.Write((j > i) ? " " : j.ToString());
                Console.WriteLine();
            }
        }
    }
}