Para un valor N dado, que denota el número de Charters a partir de A, imprima el patrón de puente de caracteres inverso.
Ejemplos:
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
- Para un valor N dado, refleja el número de caracteres que participan en el patrón, comenzando desde A. Para N = 5, el carácter participante sería ABCD E.
- Al usar un bucle for anidado, calcularíamos la lógica. Donde el bucle exterior de ‘i’ oscilaría entre 0 y N y el bucle interior de ‘j’ oscilaría entre 65 (Inicio) y 64 + 2*N.
- Bajo el cual verificaríamos la condición requerida para el diseño del patrón. Para todos los valores de j que son menores que ((64+n)+ i) imprimiría el (char)((64 + n)-( j % (64+n))) y para todos los valores de j <= ((64+n) -i) imprimiría (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>
Producción
ABCDEFEDCBA ABCDE EDCBA ABCD DCBA ABC CBA AB BA A A
Complejidad temporal: O(n 2 )
Espacio Auxiliar: O(1)