Program to print Inverse Diamond pattern
Last Updated :
13 Mar, 2023
Given an integer n, the task is to print the Inverse Diamond Pattern in 2n-1 rows.
Example:
Input: n = 3
Output:
*** ***
** **
* *
** **
*** ***
Input: n = 7
Output:
******* *******
****** ******
***** *****
**** ****
*** ***
** **
* *
** **
*** ***
**** ****
***** *****
****** ******
******* *******
Approach:
- The full Inverse Diamond is of 2n-1 rows for an input of n.
- The program is divided to print this pattern in two parts:
- The first part is the top half of diamond of n rows.
- This part includes 3 parts- the left triangle of *, the middle triangle of space and the right triangle of *.
- The second part is the below half of diamond of n-1 rows.
- This part also includes 3 parts- the left triangle of *, the middle triangle of space and the right triangle of *.
- Printing each part, the required Inverse Diamond Pattern is obtained.
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>
using namespace std;
void printDiamond(int n)
{
cout<<endl;
int i, j = 0;
for (i = 0; i < n; i++) {
for (j = i; j < n; j++)
cout<<"*";
for (j = 0; j < 2 * i + 1; j++)
cout<<" ";
for (j = i; j < n; j++)
cout<<"*";
cout<<endl;
}
for (i = 0; i < n - 1; i++) {
for (j = 0; j < i + 2; j++)
cout<<"*";
for (j = 0; j < 2 * (n - 1 - i) - 1; j++)
cout<<" ";
for (j = 0; j < i + 2; j++)
cout<<"*";
cout<<endl;
}
cout<<endl;
}
int main()
{
int n = 3;
cout<<"Inverse Diamond Pattern for n = "<<n;
printDiamond(n);
n = 7;
cout<<"\nInverse Diamond Pattern for n = "<<n;
printDiamond(n);
}
|
Java
import java.util.*;
class GFG {
static void printDiamond(int n)
{
System.out.println();
int i, j = 0;
for (i = 0; i < n; i++) {
for (j = i; j < n; j++)
System.out.print("*");
for (j = 0; j < 2 * i + 1; j++)
System.out.print(" ");
for (j = i; j < n; j++)
System.out.print("*");
System.out.println();
}
for (i = 0; i < n - 1; i++) {
for (j = 0; j < i + 2; j++)
System.out.print("*");
for (j = 0; j < 2 * (n - 1 - i) - 1; j++)
System.out.print(" ");
for (j = 0; j < i + 2; j++)
System.out.print("*");
System.out.println();
}
System.out.println();
}
public static void main(String[] args)
{
int n = 3;
System.out.println("Inverse Diamond Pattern for n = " + n);
printDiamond(n);
n = 7;
System.out.println("\nInverse Diamond Pattern for n = " + n);
printDiamond(n);
}
}
|
Python3
def printDiamond(n) :
print("")
j = 0
for i in range(0,n):
for j in range(i,n):
print("*",end="")
for j in range(0,2 * i + 1):
print(" ",end="")
for j in range(i,n):
print("*",end="")
print("")
for i in range(0,n-1):
for j in range(0, i+2):
print("*",end="")
for j in range(0,2 * (n - 1 - i) - 1):
print(" ",end="")
for j in range(0, i+2):
print("*",end="")
print("")
print("")
if __name__=='__main__':
n = 3
print("Inverse Diamond Pattern for n = ",n)
printDiamond(n)
n = 7
print("\nInverse Diamond Pattern for n = ",n )
printDiamond(n)
|
C#
using System;
class GFG
{
static void printDiamond(int n)
{
Console.WriteLine();
int i, j = 0;
for (i = 0; i < n; i++)
{
for (j = i; j < n; j++)
Console.Write("*");
for (j = 0; j < 2 * i + 1; j++)
Console.Write(" ");
for (j = i; j < n; j++)
Console.Write("*");
Console.WriteLine();
}
for (i = 0; i < n - 1; i++)
{
for (j = 0; j < i + 2; j++)
Console.Write("*");
for (j = 0; j < 2 * (n - 1 - i) - 1; j++)
Console.Write(" ");
for (j = 0; j < i + 2; j++)
Console.Write("*");
Console.WriteLine();
}
Console.WriteLine();
}
public static void Main()
{
int n = 3;
Console.WriteLine("Inverse Diamond " +
"Pattern for n = " + n);
printDiamond(n);
n = 7;
Console.WriteLine("\nInverse Diamond " +
"Pattern for n = " + n);
printDiamond(n);
}
}
|
PHP
<?php
function printDiamond($n)
{
echo"\n";
$i; $j = 0;
for ($i = 0; $i < $n; $i++)
{
for ($j = $i; $j < $n; $j++)
echo("*");
for ($j = 0;
$j < 2 * $i + 1; $j++)
echo(" ");
for ($j = $i; $j < $n; $j++)
echo("*");
echo("\n");
}
for ($i = 0; $i < $n - 1; $i++)
{
for ($j = 0; $j < $i + 2; $j++)
echo("*");
for ($j = 0;
$j < 2 * ($n - 1 - $i) - 1; $j++)
echo(" ");
for ($j = 0; $j < $i + 2; $j++)
echo("*");
echo("\n");
}
echo("\n");
}
$n = 3;
echo("Inverse Diamond Pattern for n = " );
echo($n);
printDiamond($n);
$n = 7;
echo("\nInverse Diamond Pattern for n = " );
echo($n);
printDiamond($n);
?>
|
Javascript
<script>
function printDiamond(n) {
document.write("<br><br>");
var i,
j = 0;
for (i = 0; i < n; i++) {
for (j = i; j < n; j++)
document.write("*");
for (j = 0; j < 2 * i + 1; j++)
document.write(" ");
for (j = i; j < n; j++)
document.write("*");
document.write("<br>");
}
for (i = 0; i < n - 1; i++) {
for (j = 0; j < i + 2; j++)
document.write("*");
for (j = 0; j < 2 * (n - 1 - i) - 1; j++)
document.write(" ");
for (j = 0; j < i + 2; j++)
document.write("*");
document.write("<br>");
}
document.write("<br>");
}
var n = 3;
document.write("Inverse Diamond Pattern for n = " + n);
printDiamond(n);
n = 7;
document.write("\nInverse Diamond Pattern for n = " + n);
printDiamond(n);
</script>
|
Output:
Inverse Diamond Pattern for n = 3
*** ***
** **
* *
** **
*** ***
Inverse Diamond Pattern for n = 7
******* *******
****** ******
***** *****
**** ****
*** ***
** **
* *
** **
*** ***
**** ****
***** *****
****** ******
******* *******
Time Complexity : O(n2)
Auxiliary Space : O(1)
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