Programs to print Interesting Patterns (original) (raw)

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 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

Python

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

JavaScript

PHP

i;i; i;j; // This is upper half of pattern for ($i = 1; i<=i <= i<=n; $i++) { for ($j = 1; j<=(2∗j <= (2 * j<=(2∗n); $j++) { // Left part of pattern if ($i > ($n - $j + 1)) echo " "; else echo "*"; // Right part of pattern if (($i + n)>n) > n)>j) echo " "; else echo "*"; } printf("\n"); } // This is lower half of pattern for ($i = 1; i<=i <= i<=n; $i++) { for ($j = 1; j<=(2∗j <= (2 * j<=(2∗n); $j++) { // Right Part of pattern if ($i < $j) echo " "; else echo "*"; // Left Part of pattern if ($i <= ((2 * n)−n) - n)−j)) echo " "; else echo "*"; } echo "\n"; } } // Driver Code pattern(7); // This code is contributed by m_kit ?>

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

Python

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

JavaScript

PHP

i;i; i;j; // This is upper half // of pattern for ($i = 1; i<=i <= i<=n; $i++) { for ($j = 1; j<=(2∗j <= (2 * j<=(2∗n); $j++) { // Left part of pattern if ($i < $j) echo " "; else echo "*"; // Right part of pattern if ($i <= ((2 * n)−n) - n)−j)) echo " "; else echo "*"; } echo "\n"; } // This is lower half of pattern for ($i = 1; i<=i <= i<=n; $i++) { for ($j = 1;$j <= (2 * n);n); n);j++) { // Left part of pattern if ($i > ($n - $j + 1)) echo " "; else echo "*"; // Right part of pattern if (($i + n)>n) > n)>j) echo " "; else echo "*"; } echo "\n"; } } // Driver Code pattern(7); // This code is contributed by aj_36 ?>

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

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

Python

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

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

Python

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("
"); } }

// 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

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)F { 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);
}

}

Python

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

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)