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)