Y shaped pattern

Try it on GfG Practice

Print a ‘Y’ shaped pattern from asterisks in N number of lines.

Examples:

Input: N = 12
Output:

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

Input: 8
Output: 

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

Approach:

Follow the steps to solve this problem:

• Initialize two variable s = N / 2 and t = N / 2.
• Traverse a loop on i from 0 till N-1
• Check if i > s
  • Traverse a loop on j from 0 till s-1 and print " " i.e., space
• Else,  
  • Iterate on j from 0 till i-1 and print " " i.e., space
  • Then print "*" i.e., asterisk
  • Iterate on k from 0 till 2*t-1 and print " " i.e., space
  • Decrement t by 1.
• Print " *" at the end of each iteration.

Below is the implementation of the above approach:

// C++ code for the above approach

#include <iostream>
using namespace std;

int main()
{

    // Given integer N
    int N = 12;

    // Initialize s and t
    int s = N / 2;
    int t = s;

    // Traverse from 0 till N
    for (int i = 0; i < N; i++) {

        if (i > s) {

            for (int j = 0; j < s; j++)
                cout << " ";
        }
        else {

            for (int j = 0; j < i; j++)
                cout << " ";

            cout << "*";

            for (int k = 0; k < 2 * t; k++)
                cout << " ";

            // Decrement t
            t--;
        }

        cout << " *" << endl;
    }

    return 0;
}

// Java code for the above approach
import java.io.*;

class GFG {
    public static void main(String[] args)
    {
        // Given integer N
        int N = 12;

        // Initialize s and t
        int s = N / 2;
        int t = s;

        // Traverse from 0 till N
        for (int i = 0; i < N; i++) {

            if (i > s) {

                for (int j = 0; j < s; j++)
                    System.out.print(" ");
            }
            else {

                for (int j = 0; j < i; j++)
                    System.out.print(" ");

                System.out.print("*");

                for (int k = 0; k < 2 * t; k++)
                    System.out.print(" ");

                // Decrement t
                t--;
            }

            System.out.println(" *");
        }
    }
}

// This code is contributed by Rohit Pradhan

# Given integer N
N = 12

# Initialize s and t
s = N / 2
t = s

# Traverse from 0 till N
for i in range(0, N):

  if (i > s):
    for j in range(0, int(s)):
        print(" ", end = "")
  else:
    for j in range(0, i):
        print(" ", end = "")

    print("*", end = "")
    
    for k in range(0, (2*int(t))):
        print(" ", end = "")

    # Decrement t
    t = t - 1

  print(" *")
  
 # This code is contributed by akashish__

// C# code to implement the approach
using System;
using System.Collections.Generic;

class GFG
{

  public static void Main()
  {
    
    // Given integer N
    int N = 12;

    // Initialize s and t
    int s = N / 2;
    int t = s;

    // Traverse from 0 till N
    for (int i = 0; i < N; i++) {

      if (i > s) {

        for (int j = 0; j < s; j++)
          Console.Write(" ");
      }
      else {

        for (int j = 0; j < i; j++)
          Console.Write(" ");

        Console.Write("*");

        for (int k = 0; k < 2 * t; k++)
          Console.Write(" ");

        // Decrement t
        t--;
      }

      Console.WriteLine(" *");
    }
  }
}

// This code is contributed by code_hunt.

<script>

// JavaScript implementation of the approach

// Given integer N
        let N = 12;

        // Initialize s and t
        let s = Math.floor(N / 2);
        let t = s;

        // Traverse from 0 till N
        for (let i = 0; i < N; i++) {

            if (i > s) {

                for (let j = 0; j < s; j++)
                    document.write(" ");
            }
            else {

                for (let j = 0; j < i; j++)
                    document.write(" ");

                document.write("*");

                for (let k = 0; k < 2 * t; k++)
                    document.write(" ");

                // Decrement t
                t--;
            }

            document.write(" *" + "<br/>");
        }

// This code is contributed by sanjoy_62.
</script>

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

Time Complexity: O(N²), for using nested loops.
Auxiliary Space: O(1), as constant extra space is required.