Given an integer N, the task is to print the given pattern.
Examples:
Input: 3
Output:
1 2 4
3 5 7
6 8 9
Input: 4
Output:
1 2 4 7
3 5 8 11
6 9 12 14
10 13 15 16
Approach:
- Create a matrix of size N X N which will store the pattern before printing.
- Store the elements in the upper triangle of the pattern. As observed the row index increases by 1 and column index decreases by 1 as you move down the diagonal.
- Once the upper triangle is completed then store the elements of the lower triangle in similar way as the upper triangle i.e. row index increases by 1 and column index decreases by 1 as you move down the diagonal.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void printPattern( int n)
{
int arr[n][n], k, i, j, p = 1, f;
for (k = 0; k < n; k++) {
j = k;
i = 0;
while (j >= 0) {
arr[i][j] = p;
p++;
i = i + 1;
j = j - 1;
}
}
for (k = 1; k < n; k++) {
i = k;
j = n - 1;
f = k;
while (j >= f) {
arr[i][j] = p;
p++;
i = i + 1;
j = j - 1;
}
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
cout << arr[i][j] << " " ;
}
cout << endl;
}
}
int main()
{
int n = 3;
printPattern(n);
return 0;
}
|
Java
public class GFG{
static void printPattern( int n)
{
int arr[][] = new int [n][n] ;
int k, i, j, p = 1 , f ;
for (k = 0 ; k < n; k++) {
j = k;
i = 0 ;
while (j >= 0 ) {
arr[i][j] = p;
p++;
i = i + 1 ;
j = j - 1 ;
}
}
for (k = 1 ; k < n; k++) {
i = k;
j = n - 1 ;
f = k;
while (j >= f) {
arr[i][j] = p;
p++;
i = i + 1 ;
j = j - 1 ;
}
}
for (i = 0 ; i < n; i++) {
for (j = 0 ; j < n; j++) {
System.out.print(arr[i][j] + " " ) ;
}
System.out.println() ;
}
}
public static void main(String []args)
{
int n = 3 ;
printPattern(n);
}
}
|
Python3
def printPattern(n):
arr = [[ 0 for i in range (n)]
for j in range (n)]
p = 1
for k in range (n):
j = k
i = 0
while (j > = 0 ):
arr[i][j] = p
p + = 1
i = i + 1
j = j - 1
for k in range ( 1 , n, 1 ):
i = k
j = n - 1
f = k
while (j > = f):
arr[i][j] = p
p + = 1
i = i + 1
j = j - 1
for i in range ( 0 , n, 1 ):
for j in range ( 0 , n, 1 ):
print (arr[i][j], end = " " )
print ( "\n" , end = "")
if __name__ = = '__main__' :
n = 3
printPattern(n)
|
C#
using System;
public class GFG{
static void printPattern( int n)
{
int [,]arr = new int [n,n] ;
int k, i, j, p = 1, f ;
for (k = 0; k < n; k++) {
j = k;
i = 0;
while (j >= 0) {
arr[i,j] = p;
p++;
i = i + 1;
j = j - 1;
}
}
for (k = 1; k < n; k++) {
i = k;
j = n - 1;
f = k;
while (j >= f) {
arr[i,j] = p;
p++;
i = i + 1;
j = j - 1;
}
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
Console.Write(arr[i,j] + " " ) ;
}
Console.WriteLine() ;
}
}
public static void Main()
{
int n = 3;
printPattern(n);
}
}
|
PHP
<?php
function printPattern( $n )
{
$arr [][] = array ( $n , $n );
$k ; $i ; $j ; $p = 1; $f ;
for ( $k = 0; $k < $n ; $k ++)
{
$j = $k ;
$i = 0;
while ( $j >= 0)
{
$arr [ $i ][ $j ] = $p ;
$p ++;
$i = $i + 1;
$j = $j - 1;
}
}
for ( $k = 1; $k < $n ; $k ++)
{
$i = $k ;
$j = $n - 1;
$f = $k ;
while ( $j >= $f )
{
$arr [ $i ][ $j ] = $p ;
$p ++;
$i = $i + 1;
$j = $j - 1;
}
}
for ( $i = 0; $i < $n ; $i ++)
{
for ( $j = 0; $j < $n ; $j ++)
{
echo ( $arr [ $i ][ $j ] . " " );
}
echo ( "\n" );
}
}
$n = 3;
printPattern( $n );
?>
|
Javascript
<script>
function printPattern(n)
{
let arr = new Array(n);
for (let i = 0; i < n; i++)
{
arr[i] = new Array(n);
for (let j = 0; j < n; j++)
{
arr[i][j] = 0;
}
}
let k, i, j, p = 1, f ;
for (k = 0; k < n; k++) {
j = k;
i = 0;
while (j >= 0) {
arr[i][j] = p;
p++;
i = i + 1;
j = j - 1;
}
}
for (k = 1; k < n; k++) {
i = k;
j = n - 1;
f = k;
while (j >= f) {
arr[i][j] = p;
p++;
i = i + 1;
j = j - 1;
}
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
document.write(arr[i][j] + " " ) ;
}
document.write( "<br>" ) ;
}
}
let n = 3;
printPattern(n);
</script>
|
Output:
1 2 4
3 5 7
6 8 9
Time complexity: O(n^2) for given n*n matrix
Auxiliary space: O(n^2) because using space for array arr
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Last Updated :
13 Sep, 2022
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