Program to print Inverse Diamond pattern

Given an integer n, the task is to print the Inverse Diamond Pattern in 2n-1 rows.
Example:

Input: n = 3
Output:

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

Input: n = 7
Output:

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

Approach:

1. The full Inverse Diamond is of 2n-1 rows for an input of n.
2. The program is divided to print this pattern in two parts: 
   • The first part is the top half of diamond of n rows.
   • This part includes 3 parts- the left triangle of *, the middle triangle of space and the right triangle of *.
   • The second part is the below half of diamond of n-1 rows.
   • This part also includes 3 parts- the left triangle of *, the middle triangle of space and the right triangle of *.
3. Printing each part, the required Inverse Diamond Pattern is obtained.

Below is the implementation of the above approach: 

C++

// C++ Code to print
// the inverse diamond shape

#include<bits/stdc++.h>
 using namespace std;

    // Function to Print Inverse Diamond pattern
    // with 2n-1 rows
void printDiamond(int n)
    {
        cout<<endl;

        int i, j = 0;

        // for top half
        for (i = 0; i < n; i++) {

            // for left *
            for (j = i; j < n; j++)
                cout<<"*";

            // for middle " "
            for (j = 0; j < 2 * i + 1; j++)
                cout<<" ";

            // for right *
            for (j = i; j < n; j++)
                cout<<"*";

            cout<<endl;
        }

        // for below half
        for (i = 0; i < n - 1; i++) {

            // for left *
            for (j = 0; j < i + 2; j++)
                cout<<"*";

            // for middle " "
            for (j = 0; j < 2 * (n - 1 - i) - 1; j++)
                cout<<" ";

            // for right *
            for (j = 0; j < i + 2; j++)
                cout<<"*";

            cout<<endl;
        }

        cout<<endl;
    }

    // Driver Code
    int main()
    {

        // Define n
        int n = 3;
        cout<<"Inverse Diamond Pattern for n = "<<n;
        printDiamond(n);

        n = 7;
        cout<<"\nInverse Diamond Pattern for n = "<<n;
        printDiamond(n);
    }

Java

// Java Code to print
// the inverse diamond shape

import java.util.*;

class GFG {

    // Function to Print Inverse Diamond pattern
    // with 2n-1 rows
    static void printDiamond(int n)
    {
        System.out.println();

        int i, j = 0;

        // for top half
        for (i = 0; i < n; i++) {

            // for left *
            for (j = i; j < n; j++)
                System.out.print("*");

            // for middle " "
            for (j = 0; j < 2 * i + 1; j++)
                System.out.print(" ");

            // for right *
            for (j = i; j < n; j++)
                System.out.print("*");

            System.out.println();
        }

        // for below half
        for (i = 0; i < n - 1; i++) {

            // for left *
            for (j = 0; j < i + 2; j++)
                System.out.print("*");

            // for middle " "
            for (j = 0; j < 2 * (n - 1 - i) - 1; j++)
                System.out.print(" ");

            // for right *
            for (j = 0; j < i + 2; j++)
                System.out.print("*");

            System.out.println();
        }

        System.out.println();
    }

    // Driver Code
    public static void main(String[] args)
    {

        // Define n
        int n = 3;
        System.out.println("Inverse Diamond Pattern for n = " + n);
        printDiamond(n);

        n = 7;
        System.out.println("\nInverse Diamond Pattern for n = " + n);
        printDiamond(n);
    }
}

Python3

#Python3 program to print
# the inverse diamond shape


# Function to Print Inverse
# Diamond pattern
# with 2n-1 rows
def printDiamond(n) :
    print("")
    j = 0

    # for top half
    for i in range(0,n):

        # for left *
        for j in range(i,n):
            print("*",end="")

        # for middle " "
        for j in range(0,2 * i + 1):
            print(" ",end="")

        # for right *
        for j in range(i,n):
            print("*",end="")

        print("")

    # for below half
    for i in range(0,n-1):

        # for left *
        for j in range(0, i+2):
            print("*",end="")

        # for middle " "
        for j in range(0,2 * (n - 1 - i) - 1):
            print(" ",end="")

        # for right *
        for j in range(0, i+2):
            print("*",end="")

        print("")

    print("")


# Driver Code

if __name__=='__main__':

# Define n
    n = 3
    print("Inverse Diamond Pattern for n = ",n)
    printDiamond(n)

    n = 7
    print("\nInverse Diamond Pattern for n = ",n )
    printDiamond(n)

# this code is contributed by Smitha Dinesh Semwal

C#

// C# Code to print
// the inverse diamond shape
using System;

class GFG
{

// Function to Print Inverse
// Diamond pattern with 2n-1 rows
static void printDiamond(int n)
{
    Console.WriteLine();

    int i, j = 0;

    // for top half
    for (i = 0; i < n; i++)
    {

        // for left *
        for (j = i; j < n; j++)
            Console.Write("*");

        // for middle " "
        for (j = 0; j < 2 * i + 1; j++)
            Console.Write(" ");

        // for right *
        for (j = i; j < n; j++)
            Console.Write("*");

        Console.WriteLine();
    }

    // for below half
    for (i = 0; i < n - 1; i++)
    {

        // for left *
        for (j = 0; j < i + 2; j++)
            Console.Write("*");

        // for middle " "
        for (j = 0; j < 2 * (n - 1 - i) - 1; j++)
            Console.Write(" ");

        // for right *
        for (j = 0; j < i + 2; j++)
            Console.Write("*");

        Console.WriteLine();
    }

    Console.WriteLine();
}

// Driver Code
public static void Main()
{

    // Define n
    int n = 3;
    Console.WriteLine("Inverse Diamond " +
                      "Pattern for n = " + n);
    printDiamond(n);

    n = 7;
    Console.WriteLine("\nInverse Diamond " +
                        "Pattern for n = " + n);
    printDiamond(n);
}
}

// This code is contributed
// by inder_verma.

PHP

<?php
// PHP Code to print
// the inverse diamond shape

// Function to Print Inverse
// Diamond pattern with 2n-1 rows
function printDiamond($n)
{
    echo"\n";

    $i; $j = 0;

    // for top half
    for ($i = 0; $i < $n; $i++)
    {

        // for left *
        for ($j = $i; $j < $n; $j++)
            echo("*");

        // for middle " "
        for ($j = 0;
             $j < 2 * $i + 1; $j++)
            echo(" ");

        // for right *
        for ($j = $i; $j < $n; $j++)
            echo("*");

        echo("\n");
    }

    // for below half
    for ($i = 0; $i < $n - 1; $i++)
    {

        // for left *
        for ($j = 0; $j < $i + 2; $j++)
            echo("*");

        // for middle " "
        for ($j = 0;
             $j < 2 * ($n - 1 - $i) - 1; $j++)
            echo(" ");

        // for right *
        for ($j = 0; $j < $i + 2; $j++)
            echo("*");

        echo("\n");
    }

    echo("\n");
}

// Driver Code

// Define n
$n = 3;
echo("Inverse Diamond Pattern for n = " );
echo($n);
printDiamond($n);

$n = 7;
echo("\nInverse Diamond Pattern for n = " );
echo($n);
printDiamond($n);

// This code is contributed
// by inder_verma.
?>

Javascript

<script>

      // JavaScript Code to print
      // the inverse diamond shape

      // Function to Print Inverse Diamond pattern
      // with 2n-1 rows
      function printDiamond(n) {
        document.write("<br><br>");

        var i,
          j = 0;

        // for top half
        for (i = 0; i < n; i++) {
          // for left *
          for (j = i; j < n; j++)
          document.write("*");

          // for middle " "
          for (j = 0; j < 2 * i + 1; j++)
          document.write("  ");

          // for right *
          for (j = i; j < n; j++)
          document.write("*");

          document.write("<br>");
        }

        // for below half
        for (i = 0; i < n - 1; i++) {
          // for left *
          for (j = 0; j < i + 2; j++)
          document.write("*");

          // for middle " "
          for (j = 0; j < 2 * (n - 1 - i) - 1; j++)
            document.write("  ");

          // for right *
          for (j = 0; j < i + 2; j++)
          document.write("*");

          document.write("<br>");
        }

        document.write("<br>");
      }

      // Driver Code
      // Define n
      var n = 3;
      document.write("Inverse Diamond Pattern for n = " + n);
      printDiamond(n);

      n = 7;
      document.write("\nInverse Diamond Pattern for n = " + n);
      printDiamond(n);

</script>

Output:

Inverse Diamond Pattern for n = 3

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


Inverse Diamond Pattern for n = 7

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

Time Complexity : O(n²)
Auxiliary Space : O(1)