Program to print double sided Stair-Case Pattern - Coding

Creating a double-sided staircase pattern is a fun and educational exercise that helps practice programming skills, especially loops and conditional statements. In this article, we will walk through the process of writing a program to print a double-sided stair-case pattern

Examples:

Input: 10
Output:

Output for Input: 10

Approach: To solve the problem follow the below idea:

It uses two nested loops: the outer loop iterates from 1 to n, and the inner loops handle space and star printing based on the row number and the condition of odd or even. If a row number is odd, it increments the value of k by 1; otherwise, it keeps it the same. The space and star printing is adjusted accordingly. The output is a staircase-like pattern of stars and spaces.

Below is the implementation of the above approach:

C++

// C++ code for the above approach:
#include <iostream>
 
int main()
{
    int n = 10;
    // for loop for rows
    for (int i = 1; i <= n; ++i) {
        int k = (i % 2 != 0) ? i + 1 : i;
 
        for (int g = 0; g < (n - k); ++g) {
            std::cout << " ";
        }
        // for loop for printing spaces
        for (int j = 0; j < k; ++j) {
            if (j == k - 1) {
                std::cout << "* " << std::endl;
            }
            else {
                std::cout << "* ";
            }
        }
    }
 
    return 0;
}

Java

// Java code for the above approach
public class GFG {
    public static void main(String[] args) {
        int n = 10;
        // for loop for rows
        for (int i = 1; i <= n; ++i) {
            int k = (i % 2 != 0) ? i + 1 : i;
 
            for (int g = 0; g < (n - k); ++g) {
                System.out.print(" ");
            }
            // for loop for printing spaces
            for (int j = 0; j < k; ++j) {
                if (j == k - 1) {
                    System.out.print("*\n");
                } else {
                    System.out.print("* ");
                }
            }
        }
    }
}

Python3

n = 10
 
# for loop for rows
for i in range(1, n + 1):
    k = i + 1 if i % 2 != 0 else i
 
    for g in range(n - k):
        print(" ", end="")
     
    # for loop for printing asterisks
    for j in range(k):
        if j == k - 1:
            print("* ")
        else:
            print("* ", end="")

C#

using System;
 
class GFG {
    public static void Main (string[] args) {
        int n = 10;
        // for loop for rows
        for (int i = 1; i <= n; ++i) {
            int k = (i % 2 != 0) ? i + 1 : i;
            for (int g = 0; g < (n - k); ++g) {
                Console.Write (" ");
            }
            // for loop for printing spaces
            for (int j = 0; j < k; ++j) {
                if (j == k - 1) {
                    Console.WriteLine ("* ");
                }
                else {
                    Console.Write ("* ");
                }
            }
        }
    }
}

Javascript

// JavaScript code for the above approach
function printPattern(n) {
    // for loop for rows
    for (let i = 1; i <= n; ++i) {
        let k = (i % 2 !== 0) ? i + 1 : i;
 
        // for loop for printing spaces
        for (let g = 0; g < (n - k); ++g) {
            process.stdout.write(" ");
        }
 
        for (let j = 0; j < k; ++j) {
            if (j === k - 1) {
                console.log("*");
            } else {
                process.stdout.write("* ");
            }
        }
    }
}
 
// Example: Call the function with n = 10
let n = 10;
printPattern(n);

Output

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

Time Complexity: O(N²)
Auxiliary Space: O(1)

Reffered: https://www.geeksforgeeks.org

DSA

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Category:	Coding
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