Dado un número N que representa el último término del Patrón Triángulo. La tarea es imprimir el patrón de triángulos del 1 al N, de modo que cada fila esté completa.
El patrón de triángulo se da como:
1 2 3 4 5 6 7 8 9 10 . .
Ejemplos:
Input: N = 3 Output: 1 2 3 Input: N = 7 Output: 1 2 3 4 5 6 7 will not be printed as it would result in an incomplete row
Acercarse:
- Encuentre el número de filas completas del último término dado N.
- Como:
Para Max Height = 1, el último término sería 1
Para Max Height = 2, el último término sería 3
Para Max Height = 3, el último término sería 6 - Así que el último término forma un patrón: 1, 3, 6, 10, 15,…
- Por tanto, el n-ésimo término de la serie 1, 3, 6, 10, 15,…
A(n) = 1 + 2 + 3 + 4… + (n – 1) + n
= n(n + 1) / 2
es decir, A(n) es la suma de los primeros n números naturales. - así que en
- Como:
A(n) = n(n + 1) / 2
A(n) represents the last term (as per our problem),
and n represents the max height of the Triangle
- Por lo tanto, esto puede verse como:
Last term = height (height + 1) / 2
- Por lo tanto,
height = (-1 + sqrt(1 + 8*lastTerm)) / 2
- Después de encontrar la altura máxima, el patrón de triángulo se puede imprimir fácilmente.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ code for printing the
// Triangle Pattern using last term N
#include <bits/stdc++.h>
using namespace std;
// Function to demonstrate printing pattern
void triangle(int n)
{
// number of spaces
int k = 2 * n - 2;
// character to be printed
int ch = 1;
// outer loop to handle number of rows
// n in this case
for (int i = 0; i < n; i++) {
// inner loop to handle number spaces
// values changing acc. to requirement
for (int j = 0; j < k; j++)
cout << " ";
// decrementing k after each loop
k = k - 1;
// inner loop to handle number of columns
// values changing acc. to outer loop
for (int j = 0; j <= i; j++) {
// printing stars
cout << ch++ << " ";
}
// ending line after each row
cout << endl;
}
}
// Function to find the max height
// or the number of lines
// in the triangle pattern
int maxHeight(int n)
{
return (((int)sqrt(1 + 8.0 * n)) - 1) / 2;
}
// Driver Function
int main()
{
int N = 9;
triangle(maxHeight(N));
return 0;
}
Java
// Java code for printing the
// Triangle Pattern using last term N
import java.util.*;
class GFG
{
// Function to demonstrate printing pattern
static void triangle(int n)
{
// number of spaces
int k = 2 * n - 2;
// character to be printed
int ch = 1;
// outer loop to handle number of rows
// n in this case
for (int i = 0; i < n; i++)
{
// inner loop to handle number spaces
// values changing acc. to requirement
for (int j = 0; j < k; j++)
System.out.print(" ");
// decrementing k after each loop
k = k - 1;
// inner loop to handle number of columns
// values changing acc. to outer loop
for (int j = 0; j <= i; j++)
{
// printing stars
System.out.print(ch++ + " ");
}
// ending line after each row
System.out.println();
}
}
// Function to find the max height
// or the number of lines
// in the triangle pattern
static int maxHeight(int n)
{
return (((int)Math.sqrt(1 + 8.0 * n)) - 1) / 2;
}
// Driver Code
public static void main(String[] args)
{
int N = 9;
triangle(maxHeight(N));
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 code for printing the
# Triangle Pattern using last term N
from math import sqrt
# Function to demonstrate printing pattern
def triangle(n) :
# number of spaces
k = 2 * n - 2;
# character to be printed
ch = 1;
# outer loop to handle number of rows
# n in this case
for i in range(n) :
# inner loop to handle number spaces
# values changing acc. to requirement
for j in range(k) :
print(" ", end = "");
# decrementing k after each loop
k = k - 1;
# inner loop to handle number of columns
# values changing acc. to outer loop
for j in range(i + 1) :
# printing stars
print(ch, end = " ");
ch += 1;
# ending line after each row
print()
# Function to find the max height
# or the number of lines
# in the triangle pattern
def maxHeight(n) :
ans = (sqrt(1 + 8.0 * n) - 1) // 2;
return int(ans);
# Driver Code
if __name__ == "__main__" :
N = 9;
triangle(maxHeight(N));
# This code is contributed by AnkitRai01
C#
// C# code for printing the
// Triangle Pattern using last term N
using System;
class GFG
{
// Function to demonstrate printing pattern
static void triangle(int n)
{
// number of spaces
int k = 2 * n - 2;
// character to be printed
int ch = 1;
// outer loop to handle number of rows
// n in this case
for (int i = 0; i < n; i++)
{
// inner loop to handle number spaces
// values changing acc. to requirement
for (int j = 0; j < k; j++)
Console.Write(" ");
// decrementing k after each loop
k = k - 1;
// inner loop to handle number of columns
// values changing acc. to outer loop
for (int j = 0; j <= i; j++)
{
// printing stars
Console.Write(ch++ + " ");
}
// ending line after each row
Console.WriteLine();
}
}
// Function to find the max height
// or the number of lines
// in the triangle pattern
static int maxHeight(int n)
{
return (((int)Math.Sqrt(1 + 8.0 * n)) - 1) / 2;
}
// Driver Code
public static void Main(String[] args)
{
int N = 9;
triangle(maxHeight(N));
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// Javascript code for printing the
// Triangle Pattern using last term N
// Function to demonstrate printing pattern
function triangle(n)
{
// number of spaces
var k = 2 * n - 2;
// character to be printed
var ch = 1;
// outer loop to handle number of rows
// n in this case
for (var i = 0; i < n; i++) {
// inner loop to handle number spaces
// values changing acc. to requirement
for (var j = 0; j < k; j++)
document.write(" ");
// decrementing k after each loop
k = k - 1;
// inner loop to handle number of columns
// values changing acc. to outer loop
for (var j = 0; j <= i; j++) {
// printing stars
document.write(ch++ + " ");
}
// ending line after each row
document.write("<br>");
}
}
// Function to find the max height
// or the number of lines
// in the triangle pattern
function maxHeight(n)
{
return parseInt(((parseInt(Math.sqrt(1 + 8.0 * n))) - 1) / 2);
}
// Driver Function
var N = 9;
triangle(maxHeight(N));
// This code is contributed by noob2000.
</script>
Producción:
1 2 3 4 5 6
Complejidad temporal: O(n 2 )
Espacio Auxiliar: O(1)