The idea of pattern based programs is to understand the concept of nesting of for loops and how and where to place the alphabets / numbers / stars to make the desired pattern.
Write to program to print the pattern of numbers in the following manner using for loop
1
232
34543
4567654
567898765
In almost all types of pattern programs, two things that you must take care:
- No. of lines
- If the pattern is increasing or decreasing per line?
Implementation
C++
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n = 5, i, j, num = 1, gap;
gap = n - 1;
for ( j = 1 ; j <= n ; j++ )
{
num = j;
for ( i = 1 ; i <= gap ; i++ )
cout << " ";
gap --;
for ( i = 1 ; i <= j ; i++ )
{
cout << num;
num++;
}
num--;
num--;
for ( i = 1 ; i < j ; i++)
{
cout << num;
num--;
}
cout << "\n";
}
return 0;
}
|
C
#include<stdio.h>
int main()
{
int n = 5, i, j, num = 1, gap;
gap = n - 1;
for ( j = 1 ; j <= n ; j++ )
{
num = j;
for ( i = 1 ; i <= gap ; i++ )
printf(" ");
gap --;
for ( i = 1 ; i <= j ; i++ )
{
printf("%d", num);
num++;
}
num--;
num--;
for ( i = 1 ; i < j ; i++)
{
printf("%d", num);
num--;
}
printf("\n");
}
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main(String args[])
{
int n = 5, i, j, num = 1, gap;
gap = n - 1;
for ( j = 1 ; j <= n ; j++ )
{
num = j;
for ( i = 1 ; i <= gap ; i++ )
System.out.print(" ");
gap --;
for ( i = 1 ; i <= j ; i++ )
{
System.out.print(num);
num++;
}
num--;
num--;
for ( i = 1 ; i < j ; i++)
{
System.out.print(num);
num--;
}
System.out.println();
}
}
}
|
Python3
n = 5
num = 1
gap = n - 1
for j in range(1, n + 1) :
num = j
for i in range(1, gap + 1) :
print(" ", end="")
gap = gap - 1
for i in range(1, j + 1) :
print(num, end="")
num = num + 1
num = num - 2
for i in range(1, j) :
print(num, end="")
num = num - 1
print()
|
C#
using System;
class GFG {
public static void Main()
{
int n = 5, i, j, num = 1, gap;
gap = n - 1;
for (j = 1; j <= n; j++) {
num = j;
for (i = 1; i <= gap; i++)
Console.Write(" ");
gap--;
for (i = 1; i <= j; i++) {
Console.Write(num);
num++;
}
num--;
num--;
for (i = 1; i < j; i++) {
Console.Write(num);
num--;
}
Console.WriteLine();
}
}
}
|
PHP
<?php
$n = 5;
$num = 1;
$gap = $n - 1;
for ($j = 1; $j <= $n; $j++)
{
$num = $j;
for ($i = 1; $i <= $gap; $i++)
printf(" ");
$gap --;
for ($i = 1; $i <= $j; $i++)
{
printf($num);
$num++;
}
$num--;
$num--;
for ($i = 1; $i < $j; $i++)
{
printf($num);
$num--;
}
printf("\n");
}
?>
|
Javascript
<script>
var n = 5,
i,
j,
num = 1,
gap;
gap = n - 1;
for (j = 1; j <= n; j++) {
num = j;
for (i = 1; i <= gap; i++) document.write(" ");
gap--;
for (i = 1; i <= j; i++) {
document.write(num);
num++;
}
num--;
num--;
for (i = 1; i < j; i++) {
document.write(num);
num--;
}
document.write("<br>");
}
</script>
|
Output:
1
232
34543
4567654
567898765
Time Complexity: O(n2), where n represents the given input.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Program for Pyramid Pattern
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Write a program to print the triangle pyramid with numbers
Example
input n=10
Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4 5 6
1 2 3 4 5 6 7
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 10
Approach: using nested loops
This program prints a pyramid pattern of numbers where each row contains numbers from 1 to the row number.
Algorithm
1. Take the input for the number of rows.
2. Using two loops, the first loop is used to iterate over the rows, and the second loop is used to print the numbers in each row.
3. The first inner loop prints the spaces for each row, where the number of spaces is equal to the number of rows minus the current row number.
4. The second inner loop prints the numbers for each row, where the number of numbers is equal to the current row number.
5. The outer loop is used to move to the next row.
Java
public class Main {
public static void main(String[] args) {
int n = 10;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n - i; j++) {
System.out.print(" ");
}
for (int j = 1; j <= i; j++) {
System.out.print(j + " ");
}
System.out.println();
}
}
}
|
Python3
n=10
for i in range(1,n+1):
print(" "*(n-i),end="")
for j in range(1,i+1):
print(j,end=" ")
print()
|
Javascript
const n = 10;
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= n - i; j++) {
process.stdout.write(' ');
}
for (let j = 1; j <= i; j++) {
process.stdout.write(j + ' ');
}
process.stdout.write("\n");
}
|
Output 1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4 5 6
1 2 3 4 5 6 7
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 10 Time complexity: O(n^2), where n is the number of rows. This is because we have two nested loops, and both loops iterate n times.
Auxiliary Space: O(1), as we are not using any additional data structures and are only printing the pattern on the console.