Print reverse character bridge pattern
In this article, we will explore the problem of printing a reverse character bridge pattern. The pattern consists of characters arranged in a bridge-like structure, with each row displaying characters in a reverse order. We will provide an explanation of the problem, a suitable example, an algorithm, pseudocode, and an explanation of the resultant output with the time complexity of the code.
Introduction
The problem is to print a bridge pattern formed by characters in reverse order. The pattern starts with 'A' and gradually builds up to the given input 'n', displaying the characters in reverse order on each subsequent row. The pattern is symmetric, with a bridge-like structure formed by the characters.
Problem Statement
Given a positive integer 'n', we need to print a reverse character bridge pattern using the first 'n' characters of the English alphabet.
The pattern consists of 'n' rows, with each row displaying a series of characters. The first row starts with 'A' and ends with 'A'. The second row starts with 'A' and ends with 'B'. This pattern continues until the last row, which starts and ends with the same character, 'A'. The characters between the start and end characters form an intermediate pattern.
The task is to write a program that takes an input 'n' and prints the reverse character bridge pattern accordingly.
Example
Let's take an example where 'n' is 7:
Given n: 7 ABCDEFGGFEDCBA ABCDEF FEDCBA ABCDE EDCBA ABCD DCBA ABC CBA AB BA A A
In this example, the pattern consists of 7 rows. Each row is displayed with the characters in reverse order, forming a bridge-like structure. The intermediate spaces are included to maintain the symmetry of the pattern.
Algorithm
The algorithm for printing the reverse character bridge pattern can be summarized as follows:
- Check if the input 'n' is within the valid range (greater than 0 and less than or equal to 26).
- Iterate through 'i' from 0 to 'n-1', representing the current row.
- Print the left side pattern by iterating through 'j' from 0 to 'n-i-1' and printing the character ('A' + j).
- Print the intermediate spaces by iterating through 'j' from 0 to 'i+i' and printing a space character.
- Print the right side pattern by iterating through 'j' from 'A' + (n - i - 1) down to 'A' and printing the character.
- Print a new line to move to the next row.
- Repeat steps 3-6 for each row.
Pseudocode
function printBridgePattern(n):
if n <= 0 or n > 26:
return
for i in range(0, n):
for j in range(0, n - i):
print character ('A' + j)
for j in range(0, i + i):
print a space character
for j in range('A' + (n - i - 1), 'A' - 1, -1):
print character j
print a new line
Code Solution
Here given code implementation process.
// C Program
// Print reverse character bridge pattern
#include <stdio.h>
void printBridgePattern(int n)
{
if (n <= 0 || n > 26)
{
return;
}
// Loop controlling variables
int i = 0;
int j = 0;
// Display given n
printf("\nGiven n : %d \n\n", n);
// iterating the loop through by size n
for (i = 0; i < n; ++i)
{
// Print left side pattern
for (j = 0; j < n - i; ++j)
{
printf("%c", ('A' + j));
}
for (j = 0; j < i + i; ++j)
{
// Include intermediate space
printf(" ");
}
// Print right side pattern
for (j = 'A' + (n - i - 1); j >= 'A'; --j)
{
printf("%c", j);
}
// Include new line
printf("\n");
}
}
int main(int argc, char const *argv[])
{
// Test cases
printBridgePattern(7);
printBridgePattern(10);
return 0;
}input
Given n : 7
ABCDEFGGFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A
Given n : 10
ABCDEFGHIJJIHGFEDCBA
ABCDEFGHI IHGFEDCBA
ABCDEFGH HGFEDCBA
ABCDEFG GFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A// Java program for
// Print reverse character bridge pattern
public class BridgePattern
{
public void printBridgePattern(int n)
{
if (n <= 0 || n > 26)
{
return;
}
// Loop controlling variables
int i = 0;
int j = 0;
// Display given n
System.out.print("\nGiven n : " + n + " \n\n");
// iterating the loop through by size n
for (i = 0; i < n; ++i)
{
// Print left side pattern
for (j = 0; j < n - i; ++j)
{
System.out.print( (char)('A' + j));
}
for (j = 0; j < i + i; ++j)
{
// Include intermediate space
System.out.print(" ");
}
// Print right side pattern
for (j = 'A' + (n - i - 1); j >= 'A'; --j)
{
System.out.print((char) j);
}
// Include new line
System.out.print("\n");
}
}
public static void main(String[] args)
{
BridgePattern task = new BridgePattern();
// Test cases
task.printBridgePattern(7);
task.printBridgePattern(10);
}
}input
Given n : 7
ABCDEFGGFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A
Given n : 10
ABCDEFGHIJJIHGFEDCBA
ABCDEFGHI IHGFEDCBA
ABCDEFGH HGFEDCBA
ABCDEFG GFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A// Include header file
#include <iostream>
using namespace std;
// C++ program for
// Print reverse character bridge pattern
class BridgePattern
{
public: void printBridgePattern(int n)
{
if (n <= 0 || n > 26)
{
return;
}
// Loop controlling variables
int i = 0;
int j = 0;
// Display given n
cout << "\nGiven n : " << n << " \n\n";
// iterating the loop through by size n
for (i = 0; i < n; ++i)
{
// Print left side pattern
for (j = 0; j < n - i; ++j)
{
cout << (char)('A' + j);
}
for (j = 0; j < i + i; ++j)
{
// Include intermediate space
cout << " ";
}
// Print right side pattern
for (j = 'A' + (n - i - 1); j >= 'A'; --j)
{
cout << (char) j;
}
// Include new line
cout << "\n";
}
}
};
int main()
{
BridgePattern *task = new BridgePattern();
// Test cases
task->printBridgePattern(7);
task->printBridgePattern(10);
return 0;
}input
Given n : 7
ABCDEFGGFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A
Given n : 10
ABCDEFGHIJJIHGFEDCBA
ABCDEFGHI IHGFEDCBA
ABCDEFGH HGFEDCBA
ABCDEFG GFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A// Include namespace system
using System;
// Csharp program for
// Print reverse character bridge pattern
public class BridgePattern
{
public void printBridgePattern(int n)
{
if (n <= 0 || n > 26)
{
return;
}
// Loop controlling variables
int i = 0;
int j = 0;
// Display given n
Console.Write("\nGiven n : " + n + " \n\n");
// iterating the loop through by size n
for (i = 0; i < n; ++i)
{
// Print left side pattern
for (j = 0; j < n - i; ++j)
{
Console.Write((char)('A' + j));
}
for (j = 0; j < i + i; ++j)
{
// Include intermediate space
Console.Write(" ");
}
// Print right side pattern
for (j = 'A' + (n - i - 1); j >= 'A'; --j)
{
Console.Write((char) j);
}
// Include new line
Console.Write("\n");
}
}
public static void Main(String[] args)
{
BridgePattern task = new BridgePattern();
// Test cases
task.printBridgePattern(7);
task.printBridgePattern(10);
}
}input
Given n : 7
ABCDEFGGFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A
Given n : 10
ABCDEFGHIJJIHGFEDCBA
ABCDEFGHI IHGFEDCBA
ABCDEFGH HGFEDCBA
ABCDEFG GFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A<?php
// Php program for
// Print reverse character bridge pattern
class BridgePattern
{
public function printBridgePattern($n)
{
if ($n <= 0 || $n > 26)
{
return;
}
// Loop controlling variables
$i = 0;
$j = 0;
// Display given n
echo("\nGiven n : ".$n.
" \n\n");
// iterating the loop through by size n
for ($i = 0; $i < $n; ++$i)
{
// Print left side pattern
for ($j = 0; $j < $n - $i; ++$j)
{
echo(chr((ord('A') + $j)));
}
for ($j = 0; $j < $i + $i; ++$j)
{
// Include intermediate space
echo(" ");
}
// Print right side pattern
for ($j = ord('A') + ($n - $i - 1); $j >= ord('A'); --$j)
{
echo(chr($j));
}
// Include new line
echo("\n");
}
}
}
function main()
{
$task = new BridgePattern();
// Test cases
$task->printBridgePattern(7);
$task->printBridgePattern(10);
}
main();input
Given n : 7
ABCDEFGGFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A
Given n : 10
ABCDEFGHIJJIHGFEDCBA
ABCDEFGHI IHGFEDCBA
ABCDEFGH HGFEDCBA
ABCDEFG GFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A// Node JS program for
// Print reverse character bridge pattern
class BridgePattern
{
printBridgePattern(n)
{
if (n <= 0 || n > 26)
{
return;
}
// Loop controlling variables
var i = 0;
var j = 0;
// Display given n
process.stdout.write("\nGiven n : " + n + " \n\n");
// iterating the loop through by size n
for (i = 0; i < n; ++i)
{
// Print left side pattern
for (j = 0; j < n - i; ++j)
{
process.stdout.write(
String.fromCharCode(('A'.charCodeAt(0) + j)));
}
for (j = 0; j < i + i; ++j)
{
// Include intermediate space
process.stdout.write(" ");
}
// Print right side pattern
for (j = 'A'.charCodeAt(0) + (n - i - 1);
j >= 'A'.charCodeAt(0);
--j)
{
process.stdout.write(String.fromCharCode(j));
}
// Include new line
process.stdout.write("\n");
}
}
}
function main()
{
var task = new BridgePattern();
// Test cases
task.printBridgePattern(7);
task.printBridgePattern(10);
}
main();input
Given n : 7
ABCDEFGGFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A
Given n : 10
ABCDEFGHIJJIHGFEDCBA
ABCDEFGHI IHGFEDCBA
ABCDEFGH HGFEDCBA
ABCDEFG GFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A# Python 3 program for
# Print reverse character bridge pattern
class BridgePattern :
def printBridgePattern(self, n) :
if (n <= 0 or n > 26) :
return
# Loop controlling variables
i = 0
j = 0
# Display given n
print("\nGiven n : ", n ," \n")
# iterating the loop through by size n
while (i < n) :
# Print left side pattern
j = 0
while (j < n - i) :
print(chr((ord('A') + j)), end = "")
j += 1
j = 0
while (j < i + i) :
# Include intermediate space
print(" ", end = "")
j += 1
# Print right side pattern
j = ord('A') + (n - i - 1)
while (j >= ord('A')) :
print(chr(j), end = "")
j -= 1
# Include new line
print(end = "\n")
i += 1
def main() :
task = BridgePattern()
# Test cases
task.printBridgePattern(7)
task.printBridgePattern(10)
if __name__ == "__main__": main()input
Given n : 7
ABCDEFGGFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A
Given n : 10
ABCDEFGHIJJIHGFEDCBA
ABCDEFGHI IHGFEDCBA
ABCDEFGH HGFEDCBA
ABCDEFG GFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A# Ruby program for
# Print reverse character bridge pattern
class BridgePattern
def printBridgePattern(n)
if (n <= 0 || n > 26)
return
end
# Loop controlling variables
i = 0
j = 0
# Display given n
print("\nGiven n : ", n ," \n\n")
# iterating the loop through by size n
while (i < n)
# Print left side pattern
j = 0
while (j < n - i)
print((('A'.ord + j)).chr)
j += 1
end
j = 0
while (j < i + i)
# Include intermediate space
print(" ")
j += 1
end
# Print right side pattern
j = 'A'.ord + (n - i - 1)
while (j >= 'A'.ord)
print((j).chr)
j -= 1
end
# Include new line
print("\n")
i += 1
end
end
end
def main()
task = BridgePattern.new()
# Test cases
task.printBridgePattern(7)
task.printBridgePattern(10)
end
main()input
Given n : 7
ABCDEFGGFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A
Given n : 10
ABCDEFGHIJJIHGFEDCBA
ABCDEFGHI IHGFEDCBA
ABCDEFGH HGFEDCBA
ABCDEFG GFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A
// Scala program for
// Print reverse character bridge pattern
class BridgePattern()
{
def printBridgePattern(n: Int): Unit = {
if (n <= 0 || n > 26)
{
return;
}
// Loop controlling variables
var i: Int = 0;
var j: Int = 0;
// Display given n
print("\nGiven n : " + n + " \n\n");
// iterating the loop through by size n
while (i < n)
{
// Print left side pattern
j = 0;
while (j < n - i)
{
print(('A'.toInt + j).toChar);
j += 1;
}
j = 0;
while (j < i + i)
{
// Include intermediate space
print(" ");
j += 1;
}
// Print right side pattern
j = 'A'.toInt + (n - i - 1);
while (j >= 'A'.toInt)
{
print(j.toChar);
j -= 1;
}
// Include new line
print("\n");
i += 1;
}
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: BridgePattern = new BridgePattern();
// Test cases
task.printBridgePattern(7);
task.printBridgePattern(10);
}
}input
Given n : 7
ABCDEFGGFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A
Given n : 10
ABCDEFGHIJJIHGFEDCBA
ABCDEFGHI IHGFEDCBA
ABCDEFGH HGFEDCBA
ABCDEFG GFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A// Swift 4 program for
// Print reverse character bridge pattern
class BridgePattern
{
func printBridgePattern(_ n: Int)
{
if (n <= 0 || n > 26)
{
return;
}
// Loop controlling variables
var i = 0;
var j = 0;
// Display given n
print("\nGiven n : ", n ," \n");
// iterating the loop through by size n
while (i < n)
{
// Print left side pattern
j = 0;
while (j < n - i)
{
print(Character(UnicodeScalar(65 + j)!), terminator: "");
j += 1;
}
j = 0;
while (j < i + i)
{
// Include intermediate space
print(" ", terminator: "");
j += 1;
}
// Print right side pattern
j = 65 + (n - i - 1);
while (j >= 65)
{
print(Character(UnicodeScalar(j)!), terminator: "");
j -= 1;
}
// Include new line
print(terminator: "\n");
i += 1;
}
}
}
func main()
{
let task = BridgePattern();
// Test cases
task.printBridgePattern(7);
task.printBridgePattern(10);
}
main();input
Given n : 7
ABCDEFGGFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A
Given n : 10
ABCDEFGHIJJIHGFEDCBA
ABCDEFGHI IHGFEDCBA
ABCDEFGH HGFEDCBA
ABCDEFG GFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A// Kotlin program for
// Print reverse character bridge pattern
class BridgePattern
{
fun printBridgePattern(n: Int): Unit
{
if (n <= 0 || n > 26)
{
return;
}
// Loop controlling variables
var i: Int = 0;
var j: Int = 0;
// Display given n
print("\nGiven n : " + n + " \n\n");
// iterating the loop through by size n
while (i < n)
{
// Print left side pattern
while (j < n - i)
{
print(('A'.toInt() + j).toChar());
j += 1;
}
j = 0;
while (j < i + i)
{
// Include intermediate space
print(" ");
j += 1;
}
// Print right side pattern
j = 'A'.toInt() + (n - i - 1);
while (j >= 'A'.toInt())
{
print(j.toChar());
j -= 1;
}
// Include new line
print("\n");
i += 1;
j = 0;
}
}
}
fun main(args: Array < String > ): Unit
{
val task: BridgePattern = BridgePattern();
// Test cases
task.printBridgePattern(7);
task.printBridgePattern(10);
}input
Given n : 7
ABCDEFGGFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A A
Given n : 10
ABCDEFGHIJJIHGFEDCBA
ABCDEFGHI IHGFEDCBA
ABCDEFGH HGFEDCBA
ABCDEFG GFEDCBA
ABCDEF FEDCBA
ABCDE EDCBA
ABCD DCBA
ABC CBA
AB BA
A AThe output consists of the reverse character bridge pattern as described earlier. Each row is printed, and the intermediate spaces are included to maintain symmetry.
Time Complexity
The time complexity of the code is O(n^2), where 'n' is the given input. The nested loops iterate through 'n' rows and 'n' columns, resulting in a quadratic time complexity.
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