Dado un número entero N que es el número de filas , la tarea es dibujar el patrón numérico en forma de flecha de dos puntas .
Requisito previo: el patrón es un patrón de tipo de crecimiento y reducción y, por lo tanto, se requieren conocimientos básicos para ejecutar bucles para comprender el tema y el código en cualquier idioma. La forma geométrica se puede visualizar como-
Ejemplos:
Input: R = 7
Output:
1
2 1 1 2
3 2 1 1 2 3
4 3 2 1 1 2 3 4
3 2 1 1 2 3
2 1 1 2
1
Input: R = 9
Output:
1
2 1 1 2
3 2 1 1 2 3
4 3 2 1 1 2 3 4
5 4 3 2 1 1 2 3 4 5
4 3 2 1 1 2 3 4
3 2 1 1 2 3
2 1 1 2
1
Acercarse:
- En el ejemplo dado, N=7 y el número de FILAS es 7.
- VERTICALMENTE, el patrón CRECE hasta FILA=N/2 y luego SE ENCOGE.
- FILA 1 tiene 4 ” “(ESPACIO) caracteres y luego un valor.
- El número de caracteres de ESPACIO disminuye mientras que los NÚMEROS aumentan en el conteo en cada fila sucesiva.
- Además, tenga en cuenta que el primer valor del número colocado en cada fila es el mismo que el número de la fila.
- También HORIZONTALMENTE el patrón tiene NÚMEROS, luego ESPACIOS y luego NÚMEROS.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to print the required pattern
void drawPattern(int N)
{
int n = N;
int row = 1;
// 'nst' is the number of values
int nst = 1;
// 'nsp' is the number of spaces
int nsp1 = n - 1;
int nsp2 = -1;
int val1 = row;
int val2 = 1;
while (row <= n) {
// Here spaces are printed
// 'csp' is the count of spaces
int csp1 = 1;
while (csp1 <= nsp1) {
cout << " "
<< " ";
csp1 = csp1 + 1;
}
// Now, values are printed
// 'cst' is the count of stars
int cst1 = 1;
while (cst1 <= nst) {
cout << val1 << " ";
val1 = val1 - 1;
cst1 = cst1 + 1;
}
// Again spaces have to be printed
int csp2 = 1;
while (csp2 <= nsp2) {
cout << " "
<< " ";
csp2 = csp2 + 1;
}
// Again values have to be printed
if (row != 1 && row != n) {
int cst2 = 1;
while (cst2 <= nst) {
cout << val2 << " ";
val2 = val2 + 1;
cst2 = cst2 + 1;
}
}
cout << endl;
// Move to the next row
if (row <= n / 2) {
nst = nst + 1;
nsp1 = nsp1 - 2;
nsp2 = nsp2 + 2;
val1 = row + 1;
val2 = 1;
}
else {
nst = nst - 1;
nsp1 = nsp1 + 2;
nsp2 = nsp2 - 2;
val1 = n - row;
val2 = 1;
}
row = row + 1;
}
}
// Driver code
int main()
{
// Number of rows
int N = 7;
drawPattern(N);
return 0;
}
Java
// Java implementation of the approach
class GFG {
// Function to print the required pattern
static void drawPattern(int N)
{
int n = N;
int row = 1;
// 'nst' is the number of values
int nst = 1;
// 'nsp' is the number of spaces
int nsp1 = n - 1;
int nsp2 = -1;
int val1 = row;
int val2 = 1;
while (row <= n) {
// Here spaces are printed
// 'csp' is the count of spaces
int csp1 = 1;
while (csp1 <= nsp1) {
System.out.print(" ");
csp1 = csp1 + 1;
}
// Now, values are printed
// 'cst' is the count of stars
int cst1 = 1;
while (cst1 <= nst) {
System.out.print(val1 + " ");
val1 = val1 - 1;
cst1 = cst1 + 1;
}
// Again spaces have to be printed
int csp2 = 1;
while (csp2 <= nsp2) {
System.out.print(" ");
csp2 = csp2 + 1;
}
// Again values have to be printed
if (row != 1 && row != n) {
int cst2 = 1;
while (cst2 <= nst) {
System.out.print(val2 + " ");
val2 = val2 + 1;
cst2 = cst2 + 1;
}
}
System.out.println();
// Move to the next row
if (row <= n / 2) {
nst = nst + 1;
nsp1 = nsp1 - 2;
nsp2 = nsp2 + 2;
val1 = row + 1;
val2 = 1;
}
else {
nst = nst - 1;
nsp1 = nsp1 + 2;
nsp2 = nsp2 - 2;
val1 = n - row;
val2 = 1;
}
row = row + 1;
}
}
// Driver code
public static void main(String args[])
{
// Number of rows
int N = 7;
drawPattern(N);
}
}
Python3
# Python3 implementation of the approach
# Function to print the required pattern
def drawPattern(N) :
n = N;
row = 1;
# 'nst' is the number of values
nst = 1;
# 'nsp' is the number of spaces
nsp1 = n - 1;
nsp2 = -1;
val1 = row;
val2 = 1;
while (row <= n) :
# Here spaces are printed
# 'csp' is the count of spaces
csp1 = 1;
while (csp1 <= nsp1) :
print(" ",end= " ");
csp1 = csp1 + 1;
# Now, values are printed
# 'cst' is the count of stars
cst1 = 1;
while (cst1 <= nst) :
print(val1,end = " ");
val1 = val1 - 1;
cst1 = cst1 + 1;
# Again spaces have to be printed
csp2 = 1;
while (csp2 <= nsp2) :
print(" ",end = " ");
csp2 = csp2 + 1;
# Again values have to be printed
if (row != 1 and row != n) :
cst2 = 1;
while (cst2 <= nst) :
print(val2,end = " ");
val2 = val2 + 1;
cst2 = cst2 + 1;
print()
# Move to the next row
if (row <= n // 2) :
nst = nst + 1;
nsp1 = nsp1 - 2;
nsp2 = nsp2 + 2;
val1 = row + 1;
val2 = 1;
else :
nst = nst - 1;
nsp1 = nsp1 + 2;
nsp2 = nsp2 - 2;
val1 = n - row;
val2 = 1;
row = row + 1;
# Driver code
if __name__ == "__main__" :
# Number of rows
N = 7;
drawPattern(N);
# This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to print the required pattern
static void drawPattern(int N)
{
int n = N;
int row = 1;
// 'nst' is the number of values
int nst = 1;
// 'nsp' is the number of spaces
int nsp1 = n - 1;
int nsp2 = -1;
int val1 = row;
int val2 = 1;
while (row <= n)
{
// Here spaces are printed
// 'csp' is the count of spaces
int csp1 = 1;
while (csp1 <= nsp1)
{
Console.Write(" ");
csp1 = csp1 + 1;
}
// Now, values are printed
// 'cst' is the count of stars
int cst1 = 1;
while (cst1 <= nst)
{
Console.Write(val1 + " ");
val1 = val1 - 1;
cst1 = cst1 + 1;
}
// Again spaces have to be printed
int csp2 = 1;
while (csp2 <= nsp2)
{
Console.Write(" ");
csp2 = csp2 + 1;
}
// Again values have to be printed
if (row != 1 && row != n)
{
int cst2 = 1;
while (cst2 <= nst)
{
Console.Write(val2 + " ");
val2 = val2 + 1;
cst2 = cst2 + 1;
}
}
Console.WriteLine();
// Move to the next row
if (row <= n / 2)
{
nst = nst + 1;
nsp1 = nsp1 - 2;
nsp2 = nsp2 + 2;
val1 = row + 1;
val2 = 1;
}
else
{
nst = nst - 1;
nsp1 = nsp1 + 2;
nsp2 = nsp2 - 2;
val1 = n - row;
val2 = 1;
}
row = row + 1;
}
}
// Driver code
public static void Main()
{
// Number of rows
int N = 7;
drawPattern(N);
}
}
// This code is contributed by AnkitRai01
Javascript
<script>
// JavaScript implementation of the approach
// Function to print the required pattern
function drawPattern(N) {
var n = N;
var row = 1;
// 'nst' is the number of values
var nst = 1;
// 'nsp' is the number of spaces
var nsp1 = n - 1;
var nsp2 = -1;
var val1 = row;
var val2 = 1;
while (row <= n) {
// Here spaces are printed
// 'csp' is the count of spaces
var csp1 = 1;
while (csp1 <= nsp1) {
document.write(" " + " ");
csp1 = csp1 + 1;
}
// Now, values are printed
// 'cst' is the count of stars
var cst1 = 1;
while (cst1 <= nst) {
document.write(val1 + " ");
val1 = val1 - 1;
cst1 = cst1 + 1;
}
// Again spaces have to be printed
var csp2 = 1;
while (csp2 <= nsp2) {
document.write(" " + " ");
csp2 = csp2 + 1;
}
// Again values have to be printed
if (row != 1 && row != n) {
var cst2 = 1;
while (cst2 <= nst) {
document.write(val2 + " ");
val2 = val2 + 1;
cst2 = cst2 + 1;
}
}
document.write("<br>");
// Move to the next row
if (row <= n / 2) {
nst = nst + 1;
nsp1 = nsp1 - 2;
nsp2 = nsp2 + 2;
val1 = row + 1;
val2 = 1;
} else {
nst = nst - 1;
nsp1 = nsp1 + 2;
nsp2 = nsp2 - 2;
val1 = n - row;
val2 = 1;
}
row = row + 1;
}
}
// Driver code
// Number of rows
var N = 7;
drawPattern(N);
</script>
Producción:
1
2 1 1 2
3 2 1 1 2 3
4 3 2 1 1 2 3 4
3 2 1 1 2 3
2 1 1 2
1
Complejidad de tiempo: O(N 2 )
Complejidad de espacio: O(1)
Publicación traducida automáticamente
Artículo escrito por medha130101 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA