Programs to print Interesting Patterns

Program to print the following pattern: 

Examples :

Input : 5
Output: 
* * * * *  * * * * *
* * * *      * * * *
* * *          * * *
* *              * *
*                  *
*                  *
* *              * *
* * *          * * *
* * * *      * * * *
* * * * *  * * * * *

This program is divided into four parts.

C++

// C++ program to print
// the given pattern
#include<iostream>
using namespace std;

void pattern(int n)
{
    int i, j;

    // This is upper half of pattern
    for(i = 1; i <= n; i++)
    {
        for(j = 1; j <= (2 * n); j++)
        {

            // Left part of pattern
            if (i > (n - j + 1))
                cout << " ";
            else
                cout << "*";

            // Right part of pattern
            if ((i + n) > j)
                cout << " ";
            else
                cout << "*";
        }
        cout << endl ;
    }

    // This is lower half of pattern
    for(i = 1; i <= n; i++)
    {
        for(j = 1; j <= (2 * n); j++)
        {

            // Right Part of pattern
            if (i < j)
                cout << " ";
            else
                cout << "*";

            // Left Part of pattern
            if (i <= ((2 * n) - j))
                cout << " ";
            else
                cout << "*";
        }
        cout << endl;
    }
}

// Driver Code
int main()
{
    pattern(7);

    return 0;
}

// This code is contributed by bunnyram19

C

// C program to print
// the given pattern

#include<stdio.h>
void pattern(int n)
{
    int i,j;

    // This is upper half of pattern
    for (i=1; i<=n; i++)
    {
        for (j=1; j<=(2*n); j++)
        {
            // Left part of pattern
            if (i>(n-j+1))
                printf(" ");
            else
                printf("*");

            // Right part of pattern
            if ((i+n)>j)
                printf(" ");
            else
                printf("*");
        }
        printf("\n");
    }

    // This is lower half of pattern
    for (i=1; i<=n; i++)
    {
        for (j=1; j<=(2*n); j++)
        {
            // Right Part of pattern
            if (i<j)
                printf(" ");
            else
                printf("*");

            // Left Part of pattern
            if (i<=((2*n)-j))
                printf(" ");
            else
                printf("*");
        }
        printf("\n");
    }
}

// Driver Code
int main()
{
    pattern(7);
    return 0;
}

Java

// Java program to print
// the given pattern
import java.io.*;

class GFG {

    static void pattern(int n)
    {
        int i, j;

        // This is upper half of pattern
        for (i = 1; i <= n; i++) {
            for (j = 1; j <= (2 * n); j++) {

                // Left part of pattern
                if (i > (n - j + 1))
                    System.out.print(" ");
                else
                    System.out.print("*");

                // Right part of pattern
                if ((i + n) > j)
                    System.out.print(" ");
                else
                    System.out.print("*");
            }

            System.out.println("");
        }

        // This is lower half of pattern
        for (i = 1; i <= n; i++) {
            for (j = 1; j <= (2 * n); j++) {

                // Right Part of pattern
                if (i < j)
                    System.out.print(" ");
                else
                    System.out.print("*");

                // Left Part of pattern
                if (i <= ((2 * n) - j))
                    System.out.print(" ");
                else
                    System.out.print("*");
            }

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

    // Driver Code
    public static void main(String[] args)
    {
        pattern(7);
    }
}

// This code is contributed by vt_m

Python3

# Python3 program to print
# the given pattern

def pattern(n):

    # This is upper half of pattern
    for i in range (1, n + 1):
        for j in range (1, 2 * n):

            # Left part of pattern
            if i > (n - j + 1):
                print("", end = ' ');
            else:
                print("*", end = '');

            # Right part of pattern
            if i + n - 1 > j:
                print("", end = ' ');
            else:
                print("*", end = '');
        print("");

    # This is lower half of pattern
    for i in range (1, n + 1):
        for j in range (1, 2 * n):
            #Left part of pattern
            if i < j:
                print("", end = ' ');
            else:
                print("*", end = '');

            # Right part of pattern
            if i < 2 * n - j:
                print("", end = ' ');
            else:
                print("*", end = '');
        print("");

# Driver Code
pattern(7);

# This code is contributed by mits

C#

// C# program to print
// the given pattern
using System;

class GFG
{
    static void pattern(int n)
    {
        int i, j;

        // This is upper
        // half of pattern
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= (2 * n); j++)
            {

                // Left part of pattern
                if (i > (n - j + 1))
                    Console.Write(" ");
                else
                    Console.Write("*");

                // Right part of pattern
                if ((i + n) > j)
                    Console.Write(" ");
                else
                    Console.Write("*");
            }

            Console.WriteLine("");
        }

        // This is lower
        // half of pattern
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= (2 * n); j++)
            {

                // Right Part of pattern
                if (i < j)
                    Console.Write(" ");
                else
                    Console.Write("*");

                // Left Part of pattern
                if (i <= ((2 * n) - j))
                    Console.Write(" ");
                else
                    Console.Write("*");
            }

            Console.WriteLine("");
        }
    }

    // Driver Code
    static public void Main ()
    {
        pattern(7);
    }
}

// This code is contributed by ajit

PHP

<?php
// PHP program to print
// the given pattern

function pattern($n)
{
    $i; $j;

    // This is upper half of pattern
    for ($i = 1; $i <= $n; $i++)
    {
        for ($j = 1; $j <= (2 * $n); $j++)
        {
            // Left part of pattern
            if ($i > ($n - $j + 1))
                echo " ";
            else
                echo "*";

            // Right part of pattern
            if (($i + $n) > $j)
                echo " ";
            else
                echo "*";
        }
        printf("\n");
    }

    // This is lower half of pattern
    for ($i = 1; $i <= $n; $i++)
    {
        for ($j = 1; $j <= (2 * $n); $j++)
        {
            // Right Part of pattern
            if ($i < $j)
                echo " ";
            else
                echo "*";

            // Left Part of pattern
            if ($i <= ((2 * $n) - $j))
                echo " ";
            else
                echo "*";
        }
        echo "\n";
    }
}

// Driver Code
pattern(7);

// This code is contributed by m_kit
?>

Javascript

<script>

      // JavaScript program to print
      // the given pattern

      function pattern(n) {
        var i, j;

        // This is upper half of pattern
        for (i = 1; i <= n; i++) {
          for (j = 1; j <= 2 * n; j++) {
            // Left part of pattern
            if (i > n - j + 1)
            document.write("  ");
            else
            document.write("*");

            // Right part of pattern
            if (i + n > j)
            document.write("  ");
            else
            document.write("*");
          }
          document.write("<br>");
        }

        // This is lower half of pattern
        for (i = 1; i <= n; i++) {
          for (j = 1; j <= 2 * n; j++) {
            // Right Part of pattern
            if (i < j)
            document.write("  ");
            else
            document.write("*");

            // Left Part of pattern
            if (i <= 2 * n - j)
            document.write("  ");
            else
            document.write("*");
          }
          document.write("<br>");
        }
      }

      // Driver Code
      pattern(7);

    </script>

Output

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

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

Program to print following pattern: 

Examples :

Input : 5
Output: 
*                  *
* *              * *
* * *          * * *
* * * *      * * * *
* * * * *  * * * * *
* * * * *  * * * * *
* * * *      * * * *
* * *          * * *
* *              * *
*                  *

This program is divided into four parts. 

C++

// C++ program to print the
// given pattern
#include <bits/stdc++.h>
using namespace std;

void pattern(int n)
{
    int i, j;

    // This is upper half of pattern
    for (i = 1; i <= n; i++)
    {
        for (j = 1; j <= (2 * n); j++)
        {
            // Left part of pattern
            if (i < j)
                cout << " ";
            else
                cout << "*";

            // Right part of pattern
            if (i <= ((2 * n) - j))
                cout << " ";
            else
                cout << "*";
        }
        cout << "\n";
    }

    // This is lower half of pattern
    for (i = 1; i <= n; i++)
    {
        for (j = 1; j <= (2 * n); j++)
        {
            // Left part of pattern
            if (i > (n - j + 1))
                cout <<" ";
            else
                cout <<"*";

            // Right part of pattern
            if ((i + n) > j)
                cout << " ";
            else
                cout << "*";
        }
        cout << "\n";
    }
}

// Driver Code
int main()
{
    pattern(7);
    return 0;
}

// This code is contributed by shivanisinghss2110

C

// C program to print the
// given pattern

#include<stdio.h>
void pattern(int n)
{
    int i,j;

    // This is upper half of pattern
    for (i=1; i<=n; i++)
    {
        for (j=1; j<=(2*n); j++)
        {
            // Left part of pattern
            if (i<j)
                printf(" ");
            else
                printf("*");

            // Right part of pattern
            if (i<=((2*n)-j))
                printf(" ");
            else
                printf("*");
        }
        printf("\n");
    }

    // This is lower half of pattern
    for (i=1; i<=n; i++)
    {
        for (j=1;j<=(2*n);j++)
        {
            // Left part of pattern
            if (i>(n-j+1))
                printf(" ");
            else
                printf("*");

            // Right part of pattern
            if ((i+n)>j)
                printf(" ");
            else
                printf("*");
        }
        printf("\n");
    }
}

// Driver Code
int main()
{
    pattern(7);
    return 0;
}

Java

// Java program to print the
// given pattern

import java.io.*;

class GFG {

    static void pattern(int n)
    {
        int i, j;

        // This is upper half of pattern
        for (i = 1; i <= n; i++) {
            for (j = 1; j <= (2 * n); j++) {

                // Left part of pattern
                if (i < j)
                    System.out.print(" ");
                else
                    System.out.print("*");

                // Right part of pattern
                if (i <= ((2 * n) - j))
                    System.out.print(" ");
                else
                    System.out.print("*");
            }

            System.out.println("");
        }

        // This is lower half of pattern
        for (i = 1; i <= n; i++) {
            for (j = 1; j <= (2 * n); j++) {

                // Left part of pattern
                if (i > (n - j + 1))
                    System.out.print(" ");
                else
                    System.out.print("*");

                // Right part of pattern
                if ((i + n) > j)
                    System.out.print(" ");
                else
                    System.out.print("*");
            }

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

    // Driver Code
    public static void main(String[] args)
    {
        pattern(7);
    }
}

// This code is contributed by vt_m

Python3

# Python3 program to
# print the given pattern

def pattern(n):

    # This is upper
    # half of pattern
    for i in range(1, n + 1):
        for j in range(1, 2 * n + 1):

            # Left part of pattern
            if (i < j):
                print("", end = " ");
            else:
                print("*", end = "");

            # Right part of pattern
            if (i <= ((2 * n) - j)):
                print("", end = " ");
            else:
                print("*", end = "");
        print("");

    # This is lower
    # half of pattern
    for i in range(1, n + 1):
        for j in range(1, 2 * n + 1):

            # Left part of pattern
            if (i > (n - j + 1)):
                print("", end = " ");
            else:
                print("*", end = "");

            # Right part of pattern
            if ((i + n) > j):
                print("", end = " ");
            else:
                print("*", end = "");
        print("");

# Driver Code
pattern(7);

# This code is contributed
# by mits

C#

// C# program to print
// the given pattern
using System;

class GFG
{
    static void pattern(int n)
    {
        int i, j;

        // This is upper
        // half of pattern
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= (2 * n); j++)
            {

                // Left part of pattern
                if (i < j)
                    Console.Write(" ");
                else
                    Console.Write("*");

                // Right part of pattern
                if (i <= ((2 * n) - j))
                    Console.Write(" ");
                else
                    Console.Write("*");
            }

            Console.WriteLine("");
        }

        // This is lower
        // half of pattern
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= (2 * n); j++)
            {

                // Left part of pattern
                if (i > (n - j + 1))
                    Console.Write(" ");
                else
                    Console.Write("*");

                // Right part of pattern
                if ((i + n) > j)
                    Console.Write(" ");
                else
                    Console.Write("*");
            }

            Console.WriteLine("");
        }
    }

    // Driver Code
    static public void Main ()
    {
        pattern(7);
    }
}

// This code is contributed by ajit

PHP

<?php
// PHP program to print
// the given pattern
function pattern($n)
{
    $i; $j;

    // This is upper half
    // of pattern
    for ($i = 1; $i <= $n; $i++)
    {
        for ($j = 1; $j <= (2 * $n); $j++)
        {
            // Left part of pattern
            if ($i < $j)
                echo " ";
            else
                echo "*";

            // Right part of pattern
            if ($i <= ((2 * $n) - $j))
                echo " ";
            else
                echo "*";
        }
        echo "\n";
    }

    // This is lower half of pattern
    for ($i = 1; $i <= $n; $i++)
    {
        for ($j = 1;$j <= (2 * $n); $j++)
        {
            // Left part of pattern
            if ($i > ($n - $j + 1))
                echo " ";
            else
                echo "*";

            // Right part of pattern
            if (($i + $n) > $j)
                echo " ";
            else
                echo "*";
        }
    echo "\n";
    }
}

// Driver Code
pattern(7);

// This code is contributed by aj_36
?>

Javascript

<script>

// Javascript program to print the
// given pattern
function pattern(n)
{
    var i, j;

    // This is upper half of pattern
    for(i = 1; i <= n; i++)
    {
        for(j = 1; j <= (2 * n); j++)
        {

            // Left part of pattern
            if (i < j)
                document.write("  ");
            else
                document.write("*");

            // Right part of pattern
            if (i <= ((2 * n) - j))
                document.write("  ");
            else
                document.write("*");
        }
        document.write('<br>');
    }

    // This is lower half of pattern
    for(i = 1; i <= n; i++)
    {
        for(j = 1; j <= (2 * n); j++)
        {

            // Left part of pattern
            if (i > (n - j + 1))
                document.write("  ");
            else
                document.write("*");

            // Right part of pattern
            if ((i + n) > j)
                document.write("  ");
            else
                document.write("*");
        }
        document.write('<br>');
    }
}

// Driver Code
pattern(7);

// This code is contributed by Princi Singh

</script>

Output

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

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

Program to print the following pattern:

Examples:

Input : 9 [For Odd number]
Output: 
\*******/
*\*****/*
**\***/**
***\*/***
****/****
***/*\***
**/***\**
*/*****\*
/*******\

Input : 8 [For Even number]
Output :
\******/
*\****/*
**\**/**
***\/***
***/\***
**/**\**
*/****\*
/******\

Code implementation to print the given pattern:

C++

// C++ program to print the given pattern
#include <bits/stdc++.h>
using namespace std;

void pattern(int n)
{
    // for traversing of rows
    for (int i = 1; i <= n; i++) {
        // for traversing of columns
        for (int j = 1; j <= n; j++) {
            // conditions for left-diagonal and
            // right-diagonal
            if (i == j || i + j == (n + 1)) {
                if (i + j == (n + 1)) {
                    cout << "/";
                }
                else {
                    cout << "\\";
                }
            }
            else
                cout << "*";
        }
        cout << endl;
    }
}

// Driver Code
int main()
{
    pattern(9);
    return 0;
}
// This code is contributed by Nitin Kumar

Python3

# Python3 program to print the given pattern
def pattern(n):
  # For traversing of rows
  for i in range(1, n+1):
    # For traversing of columns
    for j in range(1, n+1):
      # Conditions for left-diagonal and right-diagonal
      if i == j or i+j == n+1:
        if i+j == (n+1):
          print('/', end = '')
        else:
          print('\\', end = '')
      else:
        print('*', end = '')
    print('')

#Driver Code
if __name__ == '__main__':
  n = 8
  pattern(n)

  # This code is contributed by Mahendra Varma

Java

// Java program to print the given pattern
import java.io.*;

class pattern
{
    // Function to print the given pattern
    static void pattern(int n)
    {
        // for traversing of rows
        for (int i = 1; i <= n; i++) {
            // for traversing of columns
            for (int j = 1; j <= n; j++) {
                // conditions for left-diagonal and
                // right-diagonal
                if (i == j || i + j == (n + 1)) {
                    if (i + j == (n + 1)) {
                        System.out.print("/");
                    }
                    else {
                        System.out.print("\\");
                    }
                }
                else
                    System.out.print("*");
            }
            System.out.println();
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        pattern(9);
    }
}
// This code contributed by AJAX

Output

\*******/
*\*****/*
**\***/**
***\*/***
****/****
***/*\***
**/***\**
*/*****\*
/*******\

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

Program to print the following pattern:

Examples  :

Input : 8
Output : 
7 6 5 4 3 2 1 0 
6 5 4 3 2 1 0   
5 4 3 2 1 0     
4 3 2 1 0       
3 2 1 0         
2 1 0           
1 0             
0  

Code implementation to print the given pattern:

C++

// C++ program to print the given pattern
#include <bits/stdc++.h>
using namespace std;

void pattern(int n)
{
    // for traversing of rows
    for (int i = 1; i <= n; i++) {
        int k = n - i;
        // for traversing of columns
        for (int j = 1; j <= n; j++) {
            if (j <= (n + 1) - i) {
                cout << k << " ";
                k--;
            }
            else {
                cout << "  ";
            }
        }
        cout << endl;
    }
}

// Driver Code
int main()
{
    pattern(8);
    return 0;
}

// This code is contributed by Nitin Kumar

Java

// Java program to print the given pattern
import java.util.*;

public class Main {
    public static void pattern(int n) {
        // for traversing of rows
        for (int i = 1; i <= n; i++) {
            int k = n - i;
            // for traversing of columns
            for (int j = 1; j <= n; j++) {
                if (j <= (n + 1) - i) {
                    System.out.print(k + " ");
                    k--;
                }
                else {
                    System.out.print("  ");
                }
            }
            System.out.println();
        }
    }

    // Driver Code
    public static void main(String args[]) {
        pattern(8);
    }
}
// this code contributed by SRJ2777

Python3

def pattern(n):
  # for traversing of rows
  for i in range(1, n+1):
    # inner loop for decrement in i values
    for j in range(n - i, -1, -1):
        print(j, end=' ')
    print()

#Driver Code
if __name__ == '__main__':
  n = 8
  pattern(n)

Output

7 6 5 4 3 2 1 0 
6 5 4 3 2 1 0 
5 4 3 2 1 0 
4 3 2 1 0 
3 2 1 0 
2 1 0 
1 0 
0 

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

Program to print the following pattern :

Examples:

Input: 7
Output: 
1 
8 2 
14 9 3 
19 15 10 4 
23 20 16 11 5 
26 24 21 17 12 6 
28 27 25 22 18 13 7 

Code implementation to print the given pattern:

C++

// C++ program to print the given pattern
#include <bits/stdc++.h>
using namespace std;

void pattern(int n)
{
    int p, k = 1;
    // for traversing of rows
    for (int i = 1; i <= n; i++) {
        p = k;
        // for traversing of columns
        for (int j = 1; j <= i; j++) {
            cout << p << " ";
            p = p - (n - i + j);
        }
        cout << endl;
        k = k + 1 + (n - i);
    }
}

// Driver Code
int main()
{
    pattern(7);
    return 0;
}

// This code is contributed by Nitin Kumar

Python3

# code
def pattern(n):
  k=1
  # for traversing of rows
  for i in range(1, n+1):
    p=k
    # for traversing of columns
    for j in range(1,i+1):
        print(p, end=' ')
        p=p-(n-i+j)
    print()
    k=k+1+(n-i)

#Driver Code
if __name__ == '__main__':
  n = 7
  pattern(n)

Output

1 
8 2 
14 9 3 
19 15 10 4 
23 20 16 11 5 
26 24 21 17 12 6 
28 27 25 22 18 13 7 

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

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