Given an array, print the array in 2D form where upper triangle has values 0 and lower triangle has values increasing prefix sizes (First row has prefix of size 1, second row has prefix of size 2, ..)
Examples :
Input : 1 2 3 4 5
Output : 1 0 0 0 0
1 2 0 0 0
1 2 3 0 0
1 2 3 4 0
1 2 3 4 5
Input : 1 2 3
Output : 1 0 0
1 2 0
1 2 3
C++
#include <bits/stdc++.h>
using namespace std;
void printPattern( int arr[], int n)
{
int b[n][n];
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j < n; j++)
if (i >= j)
b[i][j] = arr[j];
else
b[i][j] = 0;
}
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j < n; j++)
cout << b[i][j] << " " ;
cout << endl;
}
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
printPattern(arr, n);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static void printPattern( int arr[], int n)
{
int b[][] = new int [n][n];
for ( int i = 0 ; i < n; i++)
{
for ( int j = 0 ; j < n; j++)
if (i >= j)
b[i][j] = arr[j];
else
b[i][j] = 0 ;
}
for ( int i = 0 ; i < n; i++)
{
for ( int j = 0 ; j < n; j++)
System.out.print(b[i][j] + " " );
System.out.println();
}
}
public static void main (String[] args)
{
int arr[] = { 1 , 2 , 3 , 4 , 5 };
int n = arr.length;
printPattern(arr, n);
}
}
|
Python3
def printPattern(arr, n):
b = [[ 0 for i in range (n)]
for i in range (n)]
for i in range ( 0 , n):
for j in range ( 0 , n):
if (i > = j):
b[i][j] = arr[j];
else :
b[i][j] = 0 ;
for i in range ( 0 , n):
for j in range ( 0 , n):
print (b[i][j], end = " " );
print ();
arr = [ 1 , 2 , 3 , 4 , 5 ];
n = len (arr);
printPattern(arr, n);
|
C#
using System;
namespace Array
{
public class GFG
{
static void printPattern( int []arr, int n)
{
int [,] b = new int [n,n];
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j < n; j++)
if (i >= j)
b[i,j] = arr[j];
else
b[i,j] = 0;
}
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j < n; j++)
Console.Write(b[i,j] + " " );
Console.WriteLine();
}
}
public static void Main()
{
int []arr = { 1, 2, 3, 4, 5 };
int n = arr.Length;
printPattern(arr, n);
}
}
}
|
PHP
<?php
function printPattern( array $arr , $n )
{
$b ;
for ( $i = 0; $i < $n ; $i ++)
{
for ( $j = 0; $j < $n ; $j ++)
if ( $i >= $j )
$b [ $i ][ $j ] = $arr [ $j ];
else
$b [ $i ][ $j ] = 0;
}
for ( $i = 0; $i < $n ; $i ++)
{
for ( $j = 0; $j < $n ; $j ++)
echo $b [ $i ][ $j ]. " " ;
echo "\n" ;
}
}
$arr = array (1, 2, 3, 4, 5);
$n = sizeof( $arr ) / sizeof( $arr [0]);
printPattern( $arr , $n );
?>
|
Javascript
<script>
function printPattern(arr, n)
{
var b = Array.from(Array(n), ()=>Array(n));
for ( var i = 0; i < n; i++)
{
for ( var j = 0; j < n; j++)
if (i >= j)
b[i][j] = arr[j];
else
b[i][j] = 0;
}
for ( var i = 0; i < n; i++)
{
for ( var j = 0; j < n; j++)
document.write( b[i][j] + " " );
document.write( "<br>" );
}
}
var arr = [1, 2, 3, 4, 5];
var n = arr.length;
printPattern(arr, n);
</script>
|
Output:
1 0 0 0 0
1 2 0 0 0
1 2 3 0 0
1 2 3 4 0
1 2 3 4 5
Time complexity: O(n^2) for given n, where n is size of given array
Auxiliary space: O(n^2)
Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!
Last Updated :
20 Feb, 2023
Like Article
Save Article