Program to print reverse character bridge pattern - GeeksforGeeks

28 Mar, 2022
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++
Java
Python3
C#
PHP
Javascript
// CPP program to print reverse character bridge pattern#include <iostream>using namespace std; // Function to print patternvoid 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 Codeint main(){    int n = 6;    ReverseCharBridge(n);    return 0;}
// Java program to print reverse// character bridge patternimport 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 code to print reverse# character bridge pattern # Function to print patterndef 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 Coden = 6ReverseCharBridge(n) # This code is contributed by "Sharad_Bhardwaj".
// C# program to print reverse// character bridge patternusing 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 program to print reverse// character bridge pattern // Function to print patternfunction 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?>
<script>// Javascript program to print reverse character bridge pattern // Function to print patternfunction 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 Codelet n = 6;ReverseCharBridge(n); // This code is contributed by Samim Hossain Mondal.</script>
ABCDEFEDCBA
ABCDE EDCBA
ABCD   DCBA
ABC     CBA
AB       BA
A         A
Time Complexity: O(n2)
Auxiliary Space: O(1)
Mithun Kumar
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simmytarika5
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