Dado un número entero N , la tarea es imprimir el patrón del alfabeto Z como se indica a continuación:
1 2 3 * * * N
*
*
*
3
2
1 2 3 * * * N
Ejemplos:
Input: N = 5
Output:
1 2 3 4 5
4
3
2
1 2 3 4 5
Input: N = 4
Output:
1 2 3 4
3
2
1 2 3 4
Acercarse:
- Imprime la primera fila con números del 1 al N.
- Luego, desde la fila 2 a la (N-1) , imprima 2 * (N – índice – 1) veces los espacios en blanco seguidos del elemento final que es índice – 1 .
- Imprime la última fila con números del 1 al N.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to print the desired
// Alphabet Z Pattern
void alphabetPattern(int N)
{
int index, side_index, size;
// Declaring the values of Right,
// Left and Diagonal values
int Top = 1, Bottom = 1, Diagonal = N - 1;
// Loop for printing the first row
for (index = 0; index < N; index++)
cout << Top++ << " ";
cout << endl;
// Main Loop for the rows from (2 to n-1)
for (index = 1; index < N - 1; index++) {
// Spaces for the diagonals
for (side_index = 0; side_index < 2 * (N - index - 1);
side_index++)
cout << " ";
// Printing the diagonal values
cout << Diagonal--;
cout << endl;
}
// Loop for printing the last row
for (index = 0; index < N; index++)
cout << Bottom++ << " ";
}
// Driver Code
int main()
{
// Number of rows
int N = 5;
alphabetPattern(N);
return 0;
}
C
// C implementation of the approach
#include <stdio.h>
// Function to print the desired
// Alphabet Z Pattern
void alphabet_Z_Pattern(int N)
{
int index, side_index, size;
// Declaring the values of Right,
// Left and Diagonal values
int Top = 1, Bottom = 1, Diagonal = N - 1;
// Loop for printing the first row
for (index = 0; index < N; index++)
printf("%d ", Top++);
printf("\n");
// Main Loop for the rows from (2 to n-1)
for (index = 1; index < N - 1; index++) {
// Spaces for the diagonals
for (side_index = 0; side_index < 2 * (N - index - 1);
side_index++)
printf(" ");
// Printing the diagonal values
printf("%d", Diagonal--);
printf("\n");
}
// Loop for printing the last row
for (index = 0; index < N; index++)
printf("%d ", Bottom++);
}
// Driver Code
int main()
{
// Size of the Pattern
int N = 5;
alphabet_Z_Pattern(N);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to print the desired
// Alphabet Z Pattern
static void alphabetPattern(int N)
{
int index, side_index;
// Declaring the values of Right,
// Left and Diagonal values
int Top = 1, Bottom = 1, Diagonal = N - 1;
// Loop for printing the first row
for (index = 0; index < N; index++)
System.out.print(Top++ + " ");
System.out.println();
// Main Loop for the rows from (2 to n-1)
for (index = 1; index < N - 1; index++)
{
// Spaces for the diagonals
for (side_index = 0;
side_index < 2 * (N - index - 1);
side_index++)
System.out.print(" ");
// Printing the diagonal values
System.out.print(Diagonal--);
System.out.println();
}
// Loop for printing the last row
for (index = 0; index < N; index++)
System.out.print(Bottom++ + " ");
}
// Driver Code
public static void main(String args[])
{
// Number of rows
int N = 5;
alphabetPattern(N);
}
}
// This code is contributed
// by Akanksha Rai
Python3
# Python 3 implementation of the approach
# Function to print the desired
# Alphabet Z Pattern
def alphabetPattern(N):
# Declaring the values of Right,
# Left and Diagonal values
Top, Bottom, Diagonal = 1, 1, N - 1
# Loop for printing the first row
for index in range(N):
print(Top, end = ' ')
Top += 1
print()
# Main Loop for the rows from (2 to n-1)
for index in range(1, N - 1):
# Spaces for the diagonals
for side_index in range(2 * (N - index - 1)):
print(' ', end = '')
# Printing the diagonal values
print(Diagonal, end = '')
Diagonal -= 1
print()
# Loop for printing the last row
for index in range(N):
print(Bottom, end = ' ')
Bottom += 1
# Driver Code
# Number of rows
N = 5
alphabetPattern(N)
# This code is contributed
# by SamyuktaSHegde
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to print the desired
// Alphabet Z Pattern
static void alphabetPattern(int N)
{
int index, side_index;
// Declaring the values of Right,
// Left and Diagonal values
int Top = 1, Bottom = 1, Diagonal = N - 1;
// Loop for printing the first row
for (index = 0; index < N; index++)
Console.Write(Top++ + " ");
Console.WriteLine();
// Main Loop for the rows from (2 to n-1)
for (index = 1; index < N - 1; index++)
{
// Spaces for the diagonals
for (side_index = 0; side_index < 2 * (N - index - 1);
side_index++)
Console.Write(" ");
// Printing the diagonal values
Console.Write(Diagonal--);
Console.WriteLine();
}
// Loop for printing the last row
for (index = 0; index < N; index++)
Console.Write(Bottom++ + " ");
}
// Driver Code
public static void Main()
{
// Number of rows
int N = 5;
alphabetPattern(N);
}
}
// This code is contributed
// by Akanksha Rai
PHP
<?php
// PHP implementation of the approach
// Function to print the desired
// Alphabet Z Pattern
function alphabetPattern($N)
{
$index; $side_index; $size;
// Declaring the values of Right,
// Left and Diagonal values
$Top = 1; $Bottom = 1; $Diagonal = $N - 1;
// Loop for printing the first row
for ($index = 0; $index < $N; $index++)
echo $Top++ . " ";
echo "\n";
// Main Loop for the rows from (2 to n-1)
for ($index = 1; $index < $N - 1; $index++)
{
// Spaces for the diagonals
for ($side_index = 0;
$side_index < 2 * ($N - $index - 1);
$side_index++)
echo " ";
// Printing the diagonal values
echo $Diagonal--;
echo "\n";
}
// Loop for printing the last row
for ($index = 0; $index < $N; $index++)
echo $Bottom++ . " ";
}
// Driver Code
// Number of rows
$N = 5;
alphabetPattern($N);
// This code is contributed by Akanksha Rai
Javascript
<script>
// JavaScript implementation of the approach
// Function to print the desired
// Alphabet Z Pattern
function alphabetPattern(N)
{
var index, side_index, size;
// Declaring the values of Right,
// Left and Diagonal values
var Top = 1,
Bottom = 1,
Diagonal = N - 1;
// Loop for printing the first row
for (index = 0; index < N; index++) {
document.write(Top + " ");
Top++;
}
document.write("<br>");
// Main Loop for the rows from (2 to n-1)
for (index = 1; index < N - 1; index++) {
// Spaces for the diagonals
for (side_index = 0; side_index <
2 * (N - index - 1); side_index++)
document.write(" ");
// Printing the diagonal values
document.write(Diagonal);
Diagonal--;
document.write("<br>");
}
// Loop for printing the last row
for (index = 0; index < N; index++) {
document.write(Bottom + " ");
Bottom++;
}
}
// Driver Code
// Number of rows
var N = 5;
alphabetPattern(N);
</script>
Producción:
1 2 3 4 5
4
3
2
1 2 3 4 5
Complejidad de Tiempo: O(N 2 )
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por VISHAL_PERIYASAMY_R y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA