Dado un número N, imprima un cuadrado hueco de lado con N asteriscos (‘*’), y dentro de él imprima un cuadrado hueco de lado N-4 asteriscos (‘*’).
Ejemplos:
Input : 6 Output : ****** * * * ** * * ** * * * ****** Input :9 Output : ********* * * * ***** * * * * * * * * * * * * * * ***** * * * *********
La idea es ejecutar dos bucles anidados de 1 a n para filas y columnas y ahora verificar las condiciones de cuándo imprimir un asterisco (‘*’) y cuándo imprimir un espacio (‘ ‘). La condición para la impresión de asteriscos será:
// This will print asterisks on the boundary (i == 1 || i == n || j == 1 || j == n) // This will print the asterisks on the boundary // of inner square (i >= 3 && i = 3 && j <= n - 2) && (i == 3 || i == n - 2 || j == 3 || j == n - 2))
Todos los índices que no cumplan la condición anterior contendrán espacios.
A continuación se muestra la implementación de la idea anterior:
C++
// C++ program to print square inside
// a square pattern
#include <iostream>
using namespace std;
// Function to print the pattern square
// inside a square
void printPattern(int n)
{
int i, j;
// To access rows of the square
for (i = 1; i <= n; i++)
{
// To access columns of the square
for (j = 1; j <= n; j++)
{
// For printing the required square pattern
if ((i == 1 || i == n || j == 1 || j == n) ||
(i >= 3 && i <= n - 2 && j >= 3 && j <= n - 2) &&
(i == 3 || i == n - 2 || j == 3 || j == n - 2))
{
// Prints star(*)
cout<<"*";
}
else
{
// Prints space(" ")
cout<<" ";
}
}
cout<<endl;
}
}
// Driver Code
int main()
{
int n = 7;
printPattern(n);
return 0;
}
// This code is contributed by Shivam.Pradhan.
C
// C program to print square inside
// a square pattern
#include <stdio.h>
// Function to print the pattern square
// inside a square
void printPattern(int n)
{
int i, j;
// To access rows of the square
for (i = 1; i <= n; i++)
{
// To access columns of the square
for (j = 1; j <= n; j++)
{
// For printing the required square pattern
if ((i == 1 || i == n || j == 1 || j == n) ||
(i >= 3 && i <= n - 2 && j >= 3 && j <= n - 2) &&
(i == 3 || i == n - 2 || j == 3 || j == n - 2))
{
// Prints star(*)
printf("*");
}
else
{
// Prints space(" ")
printf(" ");
}
}
printf("\n");
}
}
// Driver Code
int main()
{
int n = 7;
printPattern(n);
return 0;
}
Java
// Java program to print square inside
// a square pattern
import java.lang.*;
class GFG {
// Function to print the pattern square
// inside a square
static void printPattern(int n) {
// To access rows of the square
for (int i = 1; i <= n; i++) {
// To access columns of the square
for (int j = 1; j <= n; j++) {
/*** For printing the required square pattern ***/
if ((i == 1 || i == n || j == 1 || j == n) ||
(i >= 3 && i <= n - 2 && j >= 3 && j <= n - 2) &&
(i == 3 || i == n - 2 || j == 3 || j == n - 2))
{
// Prints star(*)
System.out.print("*");
}
else
{
// Prints space(" ")
System.out.print(" ");
}
}
System.out.print("\n");
}
}
// Driver code
public static void main(String[] args) {
int n = 7;
printPattern(n);
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python program to
# print square inside
# a square pattern
# Function to print
# the pattern square
# inside a square
def printPattern(n):
# To access rows of the square
for i in range(1,n+1):
# To access columns of the square
for j in range(1,n+1):
# For printing the required square pattern
if ((i == 1 or i == n or j == 1 or j == n) or
(i >= 3 and i <= n - 2 and j >= 3 and j <= n - 2) and
(i == 3 or i == n - 2 or j == 3 or j == n - 2)):
print("*",end="") # Prints star(*)
else:
print(" ",end="") # Prints space(" ")
print()
# Driver code
n = 7
printPattern(n)
# This code is contributed
# by Anant Agarwal.
C#
// C# program to print square inside
// a square pattern
using System;
class GFG {
// Function to print the pattern square
// inside a square
static void printPattern(int n) {
// To access rows of the square
for (int i = 1; i <= n; i++) {
// To access columns of the square
for (int j = 1; j <= n; j++) {
/*** For printing the required
square pattern ***/
if ((i == 1 || i == n || j == 1
|| j == n) || (i >= 3 &&
i <= n - 2 && j >= 3 &&
j <= n - 2) && (i == 3 ||
i == n - 2 || j == 3 ||
j == n - 2))
{
// Prints star(*)
Console.Write("*");
}
else
{
// Prints space(" ")
Console.Write(" ");
}
}
Console.WriteLine();
}
}
// Driver code
public static void Main() {
int n = 7;
printPattern(n);
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program to print square inside
// a square pattern
// Function to print the pattern square
// inside a square
function printPattern($n)
{
// To access rows of the square
for ($i = 1; $i <= $n; $i++)
{
// To access columns of the square
for ($j = 1; $j <= $n; $j++)
{
// For printing the required
// square pattern
if (($i == 1 || $i == $n ||
$j == 1 || $j == $n) ||
($i >= 3 && $i <= $n - 2 &&
$j >= 3 && $j <= $n - 2) &&
($i == 3 || $i == $n - 2 ||
$j == 3 || $j == $n - 2))
{
// Prints star(*)
echo "*";
}
else
{
// Prints space " "
echo " ";
}
}
echo"\n";
}
}
// Driver Code
$n = 7;
printPattern($n);
// This code is contributed by Mithun Kumar
?>
Producción :
******* * * * *** * * * * * * *** * * * *******
Tiempo Complejidad: O(n 2 )
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por priya_1998 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA