Dado un número N, la tarea es imprimir el siguiente patrón:-
Ejemplos:
Input : N = 4
Output : 4 4 4 4 4 4 4
4 3 3 3 3 3 4
4 3 2 2 2 3 4
4 3 2 1 2 3 4
4 3 2 2 2 3 4
4 3 3 3 3 3 4
4 4 4 4 4 4 4
Input : N = 2
Output : 2 2 2
2 1 2
2 2 2
Enfoque 1: La observación común es que el cuadrado así formado será de tamaño (2*N-1)x(2*N-1). Llene la primera fila y columna, la última fila y columna con N, y luego disminuya gradualmente N y llene las filas y columnas restantes de manera similar. Disminuya N cada vez que complete 2 filas y 2 columnas.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to print the
// spiral pattern
#include <bits/stdc++.h>
using namespace std;
// Function to print the pattern
void pattern(int value)
{
// Declare a square matrix
int row = 2 * value - 1;
int column = 2 * value - 1;
int arr[row][column];
int i, j, k;
for (k = 0; k < value; k++) {
// store the first row
// from 1st column to last column
j = k;
while (j < column - k) {
arr[k][j] = value - k;
j++;
}
// store the last column
// from top to bottom
i = k + 1;
while (i < row - k) {
arr[i][row - 1 - k] = value - k;
i++;
}
// store the last row
// from last column to 1st column
j = column - k - 2;
while (j >= k) {
arr[column - k - 1][j] = value - k;
j--;
}
// store the first column
// from bottom to top
i = row - k - 2;
while (i > k) {
arr[i][k] = value - k;
i--;
}
}
// print the pattern
for (i = 0; i < row; i++) {
for (j = 0; j < column; j++) {
cout << arr[i][j] << " ";
}
cout << endl;
}
}
// Driver code
int main()
{
int n = 5;
pattern(n);
return 0;
}
Java
// Java program to print
// the spiral pattern
class GFG {
// Function to print the pattern
static void pattern(int value)
{
// Declare a square matrix
int row = 2 * value - 1;
int column = 2 * value - 1;
int[][] arr = new int[row][column];
int i, j, k;
for (k = 0; k < value; k++) {
// store the first row
// from 1st column to last column
j = k;
while (j < column - k) {
arr[k][j] = value - k;
j++;
}
// store the last column
// from top to bottom
i = k + 1;
while (i < row - k) {
arr[i][row - 1 - k] = value - k;
i++;
}
// store the last row
// from last column
// to 1st column
j = column - k - 2;
while (j >= k) {
arr[column - k - 1][j] = value - k;
j--;
}
// store the first column
// from bottom to top
i = row - k - 2;
while (i > k) {
arr[i][k] = value - k;
i--;
}
}
// print the pattern
for (i = 0; i < row; i++) {
for (j = 0; j < column; j++) {
System.out.print(arr[i][j] + " ");
}
System.out.println();
}
}
// Driver code
public static void main(String[] args)
{
int n = 5;
pattern(n);
}
}
// This code is contributed
// by ChitraNayal
Python3
# Python3 program to print # the spiral pattern # Function to print the pattern def pattern(value): # Declare a square matrix row = 2 * value - 1 column = 2 * value - 1 arr = [[0 for i in range(row)] for j in range (column)] for k in range( value): # store the first row # from 1st column to # last column j = k while (j < column - k): arr[k][j] = value - k j += 1 # store the last column # from top to bottom i = k + 1 while (i < row - k): arr[i][row - 1 - k] = value - k i += 1 # store the last row # from last column # to 1st column j = column - k - 2 while j >= k : arr[column - k - 1][j] = value - k j -= 1 # store the first column # from bottom to top i = row - k - 2 while i > k : arr[i][k] = value - k i -= 1 # print the pattern for i in range(row): for j in range(column): print(arr[i][j], end = " ") print() # Driver code if __name__ == "__main__": n = 5 pattern(n) # This code is contributed # by ChitraNayal
C#
// C# program to print
// the spiral pattern
using System;
class GFG {
// Function to print the pattern
static void pattern(int value)
{
// Declare a square matrix
int row = 2 * value - 1;
int column = 2 * value - 1;
int[, ] arr = new int[row, column];
int i, j, k;
for (k = 0; k < value; k++) {
// store the first row
// from 1st column to
// last column
j = k;
while (j < column - k) {
arr[k, j] = value - k;
j++;
}
// store the last column
// from top to bottom
i = k + 1;
while (i < row - k) {
arr[i, row - 1 - k] = value - k;
i++;
}
// store the last row
// from last column
// to 1st column
j = column - k - 2;
while (j >= k) {
arr[column - k - 1, j] = value - k;
j--;
}
// store the first column
// from bottom to top
i = row - k - 2;
while (i > k) {
arr[i, k] = value - k;
i--;
}
}
// print the pattern
for (i = 0; i < row; i++) {
for (j = 0; j < column; j++) {
Console.Write(arr[i, j] + " ");
}
Console.Write("\n");
}
}
// Driver code
public static void Main()
{
int n = 5;
pattern(n);
}
}
// This code is contributed
// by ChitraNayal
PHP
<?php
// PHP program to print
// the spiral pattern
// Function to print the pattern
function pattern($value)
{
// Declare a square matrix
$row = 2 * $value - 1;
$column = 2 * $value - 1;
$arr = array(array());
for ($k = 0; $k < $value; $k++)
{
// store the first row
// from 1st column to
// last column
$j = $k;
while ($j < $column - $k)
{
$arr[$k][$j] = $value - $k;
$j++;
}
// store the last column
// from top to bottom
$i = $k + 1;
while ($i < $row - $k)
{
$arr[$i][$row - 1 - $k] = $value - $k;
$i++;
}
// store the last row
// from last column
// to 1st column
$j = $column - $k - 2;
while ($j >= $k)
{
$arr[$column - $k - 1][$j] = $value - $k;
$j--;
}
// store the first column
// from bottom to top
$i = $row - $k - 2;
while ($i > $k)
{
$arr[$i][$k] = $value - $k;
$i--;
}
}
// print the pattern
for ($i = 0; $i < $row; $i++)
{
for ($j = 0; $j < $column; $j++)
{
echo $arr[$i][$j] . " ";
}
echo "\n";
}
}
// Driver code
$n = 5;
pattern($n);
// This code is contributed
// by ChitraNayal
?>
Javascript
<script>
// Javascript program to print the
// spiral pattern
// Function to print the pattern
function pattern(value)
{
// Declare a square matrix
let row = 2 * value - 1;
let column = 2 * value - 1;
let arr = new Array(row);
for(let i = 0; i < row; i++)
{
arr[i] = new Array(column);
}
let i, j, k;
for(k = 0; k < value; k++)
{
// Store the first row
// from 1st column to last column
j = k;
while (j < column - k)
{
arr[k][j] = value - k;
j++;
}
// Store the last column
// from top to bottom
i = k + 1;
while (i < row - k)
{
arr[i][row - 1 - k] = value - k;
i++;
}
// Store the last row
// from last column to 1st column
j = column - k - 2;
while (j >= k)
{
arr[column - k - 1][j] = value - k;
j--;
}
// Store the first column
// from bottom to top
i = row - k - 2;
while (i > k)
{
arr[i][k] = value - k;
i--;
}
}
// Print the pattern
for(i = 0; i < row; i++)
{
for(j = 0; j < column; j++)
{
document.write(arr[i][j] + " ");
}
document.write("<br>");
}
}
// Driver code
let n = 5;
pattern(n);
// This code is contributed by subham348
</script>
Producción:
5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 5 5 4 3 3 3 3 3 4 5 5 4 3 2 2 2 3 4 5 5 4 3 2 1 2 3 4 5 5 4 3 2 2 2 3 4 5 5 4 3 3 3 3 3 4 5 5 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5
Enfoque 2: Comenzando la indexación desde i = 1 y j = 1 , se puede observar que cada valor de la array requerida será max(abs(i – n), abs(j – n)) + 1 .
A continuación se muestra la implementación del enfoque anterior:
C++14
// C++ implementation of the approach
#include <algorithm>
#include <iostream>
using namespace std;
// Function to print the required pattern
void pattern(int n)
{
// Calculating boundary size
int p = 2 * n - 1;
for (int i = 1; i <= p; i++) {
for (int j = 1; j <= p; j++) {
// Printing the values
cout << max(abs(i - n), abs(j - n)) + 1 << " ";
}
cout << endl;
}
}
// Driver code
int main()
{
int n = 5;
pattern(n);
return 0;
}
// This code is contributed by : Vivek kothari
C
// C implementation of the approach
#include <stdio.h>
#include <stdlib.h>
// Function Declaration
int max(int, int);
// Function to print the required pattern
void pattern(int n)
{
// Calculating boundary size
int size = 2 * n - 1;
for(int i = 1; i <= size; i++)
{
for(int j = 1; j <= size; j++)
{
// Printing the values
printf("%d ", max(abs(i - n),
abs(j - n)) + 1);
}
printf("\n");
}
}
// Function to return maximum value
int max(int val1, int val2)
{
if (val1 > val2)
return val1;
return val2;
}
// Driver code
int main()
{
int n = 5;
pattern(n);
return 0;
}
// This code is contributed by Yuvaraj R
Java
// Java implementation of the approach
import java.io.*;
import java.util.*;
class GFG{
// Function to print the required pattern
public static void pattern(int n)
{
// Calculating boundary size
int size = 2 * n - 1;
for(int i = 1; i <= size; i++)
{
for(int j = 1; j <= size; j++)
{
// Printing the values
System.out.print(Math.max(
Math.abs(i - n),
Math.abs(j - n)) + 1 + " ");
}
System.out.println();
}
}
// Driver code
public static void main(String[] args)
{
int n = 5;
pattern(n);
}
}
// This code is contributed by Yuvaraj R
Python3
# Python3 implementation of the approach # Function to print the required pattern def pattern(n): # Calculating boundary size p = 2 * n - 1 for i in range(1, p + 1): for j in range(1, p + 1): # Printing the values print(max(abs(i - n), abs(j - n)) + 1, " ", end="") print() # Driver code n = 5 pattern(n) # This code is contributed by subhammahato348
C#
// C# implementation of the approach
using System;
class GFG{
// Function to print the required pattern
static void pattern(int n)
{
// Calculating boundary size
int size = 2 * n - 1;
for(int i = 1; i <= size; i++)
{
for(int j = 1; j <= size; j++)
{
// Printing the values
Console.Write(Math.Max(Math.Abs(i - n),
Math.Abs(j - n)) +
1 + " ");
}
Console.WriteLine();
}
}
// Driver code
public static void Main()
{
int n = 5;
pattern(n);
}
}
// This code is contributed by subhammahato348
Javascript
<script>
// Javascript implementation of the approach
// Function to print the required pattern
function pattern(n)
{
// Calculating boundary size
let p = 2 * n - 1;
for (let i = 1; i <= p; i++) {
for (let j = 1; j <= p; j++) {
// Printing the values
document.write((Math.max(Math.abs(i - n),
Math.abs(j - n)) + 1) + " ");
}
document.write("<br>");
}
}
// Driver code
let n = 5;
pattern(n);
</script>
Producción
5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 5 5 4 3 3 3 3 3 4 5 5 4 3 2 2 2 3 4 5 5 4 3 2 1 2 3 4 5 5 4 3 2 2 2 3 4 5 5 4 3 3 3 3 3 4 5 5 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5
Publicación traducida automáticamente
Artículo escrito por Naman_Garg y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA