# 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

## Recommended: Please solve it on PRACTICE first, before moving on to the solution.

• 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 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

 = (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  ?>

Output :

ABCDEFEDCBA
ABCDE EDCBA
ABCD   DCBA
ABC     CBA
AB       BA
A         A

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