Program to print reverse character bridge pattern
For a given value N, denoting the number of Charters starting from the A, print reverse character bridge pattern.
Examples :
Input : n = 5 Output : ABCDEDCBA ABCD DCBA ABC CBA AB BA A A Input : n = 8 Output : ABCDEFGHGFEDCBA ABCDEFG GFEDCBA ABCDEF FEDCBA ABCDE EDCBA ABCD DCBA ABC CBA AB BA A A
- For a given value N, reflect the number of characters taking part in the pattern, starting from A. For N = 5, Participating character would be A B C D E.
- By using a nested for loop we would compute the logic. Where the outer loop of ‘i’ would range from 0 to N and the inner loop of ‘j’ would range from 65(Start) to 64 + 2*N.
- Under which we would check the required condition for the pattern design. For all the values of j which are less than ((64+n)+ i) it would print the (char)((64 + n)-( j % (64+n))) and for all the values of j <= ((64+n) -i) it would print (char)j.
C++
// CPP program to print reverse character bridge pattern #include <iostream> using namespace std; // Function to print pattern void ReverseCharBridge( int n) { for ( int i = 0; i < n; i++) { for ( int j = 'A' ; j < 'A' + (2 * n) - 1; j++) { if (j >= ( 'A' + n - 1) + i) cout << ( char )(( 'A' + n - 1) - (j % ( 'A' + n - 1))); else if (j <= ( 'A' + n - 1) - i) cout << ( char )j; else cout << " " ; } cout << endl; } } // Driver Code int main() { int n = 6; ReverseCharBridge(n); return 0; } |
Java
// Java program to print reverse // character bridge pattern import java.io.*; class GFG { // Function to print pattern static void ReverseCharBridge( int n) { for ( int i = 0 ; i < n; i++) { for ( int j = 'A' ; j < 'A' + ( 2 * n) - 1 ; j++) { if (j >= ( 'A' + n - 1 ) + i) System.out.print(( char )(( 'A' + n - 1 ) - (j % ( 'A' + n - 1 )))); else if (j <= ( 'A' + n - 1 ) - i) System.out.print(( char )j); else System.out.print( " " ); } System.out.println(); } } // Driver Code public static void main(String args[]) { int n = 6 ; ReverseCharBridge(n); } } /*This code is contributed by Nikita Tiwari.*/ |
Python3
# Python3 code to print reverse # character bridge pattern # Function to print pattern def ReverseCharBridge( n ): for i in range ( n ): for j in range ( ord ( 'A' ), ord ( 'A' ) + ( 2 * n) - 1 ): if j > = ( ord ( 'A' ) + n - 1 ) + i: print ( chr (( ord ( 'A' ) + n - 1 ) - (j % ( ord ( 'A' ) + n - 1 ))), end = '') elif j < = ( ord ( 'A' ) + n - 1 ) - i: print ( chr (j), end = '') else : print (end = " " ) print ( "\n" , end = '') # Driver Code n = 6 ReverseCharBridge(n) # This code is contributed by "Sharad_Bhardwaj". |
C#
// C# program to print reverse // character bridge pattern using System; class GFG { // Function to print pattern static void ReverseCharBridge( int n) { for ( int i = 0; i < n; i++) { for ( int j = 'A' ; j < 'A' + (2 * n) - 1; j++) { if (j >= ( 'A' + n - 1) + i) Console.Write(( char )(( 'A' + n - 1) - (j % ( 'A' + n - 1)))); else if (j <= ( 'A' + n - 1) - i) Console.Write(( char )j); else Console.Write( " " ); } Console.WriteLine(); } } // Driver Code public static void Main() { int n = 6; ReverseCharBridge(n); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to print reverse // character bridge pattern // Function to print pattern function ReverseCharBridge( $n ) { //Ascii of A is 65 for ( $i = 0; $i < $n ; $i ++) { for ( $j = 65; $j < 65 + (2 * $n ) - 1; $j ++) { if ( $j >= (65 + $n - 1) + $i ) echo chr ((65 + $n - 1) - ( $j % (65 + $n - 1))); else if ( $j <= (65 + $n - 1) - $i ) echo chr ( $j ); else echo " " ; } echo "\n" ; } } // Driver Code $n = 6; ReverseCharBridge( $n ); // This code is contributed by mits ?> |
Javascript
<script> // Javascript program to print reverse character bridge pattern // Function to print pattern function ReverseCharBridge(n) { for (let i = 0; i < n; i++) { for (let j = 65; j < 65 + (2 * n) - 1; j++) { if (j >= (65 + n - 1) + i) document.write(String.fromCharCode((65 + n - 1) - (j % (65 + n - 1)))); else if (j <= (65 + n - 1) - i) document.write(String.fromCharCode(j)); else document.write( " " ); } document.write( "\n" ); } } // Driver Code let n = 6; ReverseCharBridge(n); // This code is contributed by Samim Hossain Mondal. </script> |
Output
ABCDEFEDCBA ABCDE EDCBA ABCD DCBA ABC CBA AB BA A A
Time Complexity: O(n2)
Auxiliary Space: O(1)
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