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Programs to print Interesting Patterns

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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 :
\******/
*\****/*
**\**/**
***\/***
***/\***
**/**\**
*/****\*
/******\
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
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
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 is contributed by AJAX
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
C#



// C# program to print the given pattern
using System;
using System.Collections.Generic;
 
class GFG
{
 
  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)) {
            Console.Write("/");
          }
          else {
            Console.Write("\\");
          }
        }
        else
          Console.Write("*");
      }
      Console.Write("\n");
    }
  }
 
  // Driver Code
  static void Main(string[] args)
  {
    pattern(9);
  }
}
Javascript



<script>
 
// JavaScript program to print the given pattern
 
// Function to print the given pattern
function pattern(n)
{
    // for traversing of rows
    for (let i = 1; i <= n; i++)
    {
        // for traversing of columns
        for (let j = 1; j <= n; j++)
          {
            // conditions for left-diagonal and
            // right-diagonal
            if (i == j || i + j == (n + 1)) {
                  if (i + j == (n + 1)) {
                    document.write("/");
                  }
                  else {
                    document.write("\\");
                  }
            }
            else
                  document.write("*");
          }
          document.write("<br>");
    }
}
   
pattern(9);
 
// This code is contributed by lokesh.
 
</script>
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  
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
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)
C#



// C# program to print the given pattern
using System;
using System.Collections.Generic;
 
class GFG
{
    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) {
                    Console.Write(k + " ");
                    k--;
                }
                else {
                    Console.Write("  ");
                }
            }
            Console.WriteLine();
        }
    }
 
 
// Driver Code
    static void Main(string[] args)
    {
        pattern(8);
    }
}
Javascript



// JavaScript program to print the given pattern
function pattern(n) {
 
// for traversing of rows
for (let i = 1; i <= n; i++) {
let k = n - i;
 
// for traversing of columns
for (let j = 1; j <= n; j++) {
if (j <= n + 1 - i) {
console.log(k + " ");
k--;
}
else {
console.log(" ");
}
}
console.log("<br>");
}
}
 
// Driver Code
pattern(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 
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 
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
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
Java



public class Pattern {
 
    public static 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++) {
                System.out.print(p + " ");
                p = p - (n - i + j);
            }
            System.out.println();
            k = k + 1 + (n - i);
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        pattern(7);
    }
}
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)
C#



using System;
 
namespace Pattern
{
    class Program
    {
        static 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++)
                {
                    Console.Write(p + " ");
                    p = p - (n - i + j);
                }
                Console.WriteLine();
                k = k + 1 + (n - i);
            }
        }
 
        // Driver Code
        static void Main(string[] args)
        {
            Pattern(7);
        }
    }
}
Javascript



<script>
 
// JavaScript program to print the given pattern
 
function pattern(n) {
    let k = 1;
         
    // for traversing of rows
    for (let i = 1; i <= n; i++) {
        let p = k;
         
        // for traversing of columns
        for (let j = 1; j <= i; j++) {
            document.write(p + " ");
            p = p - (n - i + j);
        }
        document.write("<br>");
        k = k + 1 + (n - i);
    }
}
 
pattern(7);
 
// This code is contributed by lokesh.
 
</script>
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)
This article is contributed by Nitin Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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Last Updated :
18 Apr, 2023
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