Reverse Left-Shifted Number Pattern in C#

Beginner
⏱️ 5 min read
📚 Updated: Aug 2025
🎯 2 Code Examples
Nested Loops
Reverse left-shifted number pattern output (54321, 5432, 543, 54, 5) in C# programming

What You’ll Learn

How to print a reverse left-shifted descending number pattern in C#. Each row starts from the maximum number (rows) and counts down to a changing stop value, producing 54321, 5432, 543, 54, and 5.

This pattern is similar to the left-shifted triangle, but the inner loop counts downward and always begins from rows.

⭐ Pattern Output

For rows = 5, the pattern looks like this:

Output
54321
5432
543
54
5
1

Complete C# Program

Outer loop increases the stop value; inner loop prints from rows down to i.

C#
using System;

namespace MyApp
{
    class Program
    {
        static void Main(string[] args)
        {
            int rows = 5;
            int i, j;

            for (i = 1; i <= rows; i++)
            {
                for (j = rows; j >= i; j--)
                {
                    Console.Write(j);
                }
                Console.WriteLine();
            }
        }
    }
}

🧠 How It Works

1

Set the height

int rows = 5; sets both the maximum printed digit and the number of rows.

Setup
2

Outer loop (row index)

for (i = 1; i <= rows; i++) increases the stopping point for each row.

Row control
3

Inner loop (print rows..i)

for (j = rows; j >= i; j--) prints from 5 down to i. Since i grows each row, fewer digits get printed.

Reverse print
4

New line

Console.WriteLine() moves to the next row.

Line break
=

Reverse left-shifted pattern

Total digits printed are 5+4+3+2+1 = n(n+1)/2 for n rows, so time complexity is O(n²).

2

Variation — User Input Version

Read the row count at runtime:

C#
using System;

namespace MyApp
{
    class Program
    {
        static void Main(string[] args)
        {
            Console.Write("Enter the number of rows: ");
            int rows = Convert.ToInt32(Console.ReadLine());

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

💡 Tips for Enhancement

Try These

  • Validate input with int.TryParse to avoid crashes
  • Add spaces between digits for readability
  • Swap bounds to print from rows down to 1 on every row (fixed width)
  • Right-align by printing leading spaces before each row

Avoid

  • Forgetting Console.WriteLine() after each row
  • Using Convert.ToInt32 without handling invalid input
  • Incrementing j (this pattern needs a countdown)

Key Takeaways

1

Each row begins at rows and stops at i.

2

The outer loop increases i, so each row prints fewer digits.

3

Total digits printed are triangular: n(n+1)/2 → O(n²).

4

This is a useful pattern for practicing variable inner-loop stop conditions.

❓ Frequently Asked Questions

On the last row, i = rows, so the inner loop prints just rows once.
Program 3 prints from i down to 1. This program prints from rows down to i, so each row always starts with the maximum digit.
O(n²) for n rows since you print n(n+1)/2 digits.

Explore More C# Number Patterns!

Keep practicing nested loops with inversions, alignments, and alternating number patterns.

All Number Patterns →
Did you know?

When the inner loop prints from rows down to i, the number of printed digits is still triangular: 5+4+3+2+1. Only the starting value changes.

About the author

Mari Selvan M P
Mari Selvan M P 🔗

Developer, cloud engineer, and technical writer

  • Experience 12 years building web and cloud systems
  • Focus Full Stack Development, AWS, and Developer Education

I write practical tutorials so students and working developers can learn by doing—from databases and APIs to deployment on AWS.

AWS Certified Developer – Associate (Credly)

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