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> |
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
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> |
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
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> |
\*******/ *\*****/* **\***/** ***\*/*** ****/**** ***/*\*** **/***\** */*****\* /*******\
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); |
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> |
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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