Program to Print Diamond Shape Star Pattern
In this article we are going to see how to print the triangle star program.
Example-1 When row value=4 * *** ***** ******* ***** *** *
Example-2: When row value=5 * *** ***** ******* ********* ******* ***** *** *
Now, let’s see program to how print it.
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Approach:
- Enter total row and store it in an integer variable row.
- Take first outer for loop to print all rows.
- Take second/first inner for loop to print column values i.e., first inner for loop will print all the spaces in the column.
- Take third/2nd inner for loop to print column values i.e., second inner for loop will print all the stars in the column.
- Take second outer for loop to print all rows.
- Take second/first inner for loop to print column values i.e., first inner for loop will print all the spaces in the column.
- Take third/2nd inner for loop to print column values i.e., second inner for loop will print all the stars in the column.
JAVA Code:
Method-1 : Static Star Character
import java.util.*;
public class Main
{
public static void main(String args[])
{
// taking variable for loop iteration and row .
int row,r,c,d;
//creating object
Scanner s = new Scanner(System.in);
// entering the number of row
System.out.print("Enter rows : ");
row = s.nextInt();
//counting space
d = row - 1;
//first ireration for printing uper tringle
for (r = 1; r <= row; r++)
{
//iteration for printing space
for (c = 1; c <= d; c++)
System.out.print(" ");
//decreasing the number of space
d--;
//printing number of stars
for (c = 1; c <= 2 * r - 1; c++)
System.out.print("*");
// taking to the new line
System.out.println("");
}
//declaring space as one .
d = 1;
//2nd ireration for printing lower inverted tringle
for (r = 1; r <= row - 1; r++)
{
//iteration for printing space
for (c = 1; c <= d; c++)
System.out.print(" ");
//increasing the number of space
d++;
//printing number of stars
for (c = 1; c <= 2 * (row - r) - 1; c++)
System.out.print("*");
// taking to the new line
System.out.println("");
}
}
}
Output: Enter rows : 5 * *** ***** ******* ********* ******* ***** *** *
Method-2 : User Input Character
import java.util.*;
public class Main
{
public static void main(String args[])
{ // taking variable for loop iteration and row .
int row,r,c,d;
//creating object
Scanner s = new Scanner(System.in);
// entering the number of row
System.out.print("Enter rows : ");
row = s.nextInt();
// enter any character
System.out.print("Enter symbol : ");
char symbol = s.next().charAt(0);
//counting space
d = row - 1;
//first ireration for printing uper tringle
for (r = 1; r <= row; r++)
{
//iteration for printing space
for (c = 1; c <= d; c++)
System.out.print(" ");
//decreasing the number of space
d--;
//printing number of stars
for (c = 1; c <= 2 * r - 1; c++)
System.out.print(symbol);
// taking to the new line
System.out.println("");
}
//declaring space as one .
d = 1;
//2nd ireration for printing lower inverted tringle
for (r = 1; r <= row - 1; r++)
{
//iteration for printing space
for (c = 1; c <= d; c++)
System.out.print(" ");
//increasing the number of space
d++;
//printing number of stars
for (c = 1; c <= 2 * (row - r) - 1; c++)
System.out.print(symbol);
// taking to the new line
System.out.println("");
}
}
}
Output: Enter rows : 5 Enter symbol : @ @ @@@ @@@@@ @@@@@@@ @@@@@@@@@ @@@@@@@ @@@@@ @@@ @
Explanation :
Take a variable to counting space i.e., d .
Assign d= row-1 .
Iteration-1
r=1 (passes the first for loop condition) as it will iterate up to r<=row times.
First inner for loop will print the space d time that means 4 times.
Then space will decrease 1 time .(d=3)
then 2nd inner for loop will print the “* ” 2* r -1 time , that means 1 time .
So, star will be printed one time.
*
Iteration-2
r=2(passes the first for loop condition) as it will iterate up to r<=row times.
First inner for loop will print the space d time that means 3 times.
Then space will decrease 1 time .(d=2)
then 2nd inner for loop will print the “* ” 2* r -1 time , that means 3 time .
Star will be printed 3 times.
* * *
Iteration-3
r=3 (passes the first for loop condition) as it will iterate up to r<=row times.
First inner for loop will print the space d time that means 2 times.
Then space will decrease 1 time .(d=1)
then 2nd inner for loop will print the “* ” 2* r -1 time , that means 5 time .
Star will be printed 5 times.
* * * * *
Iteration-4
r=4 (passes the first for loop condition) as it will iterate up to r<=row times.
First inner for loop will print the space d time that means 1 time.
Then space will decrease 1 time .(d=0)
then 2nd inner for loop will print the “* ” 2* r -1 time , that means 7 time .
Star will be printed 7 times.
* * * * * * *
Iteration-5
r=5(passes the first for loop condition) as it will iterate up to r<=row times.
First inner for loop will print the space d time that means 0 time because it fails the condition c<=d .
Then space will decrease 1 time .(d=-1)
then 2nd inner for loop will print the “* ” 2* r -1 time , that means 9 time .
Star will be printed 9 times.
* * * * * * * * *
Now when r=6, first for loop condition will fail so no other loops will be executed.
For second outer for loop .
Assign d= 1 .
Iteration-1
r=1 (passes the first for loop condition) as it will iterate up to r<=row-1 times.
First inner for loop will print the space 1 time because loop will iterate up to c<=d time .
Then space will increase 1 time .(d=2)
then 2nd inner for loop will print the “* ” 2* (row -r)-1 time , that means 7 time .
So, star will be printed 7 time.
* * * * * * *
Iteration-2
r=2 (passes the first for loop condition) as it will iterate up to r<=row-1 times.
First inner for loop will print the space 2 time because loop will iterate up to c<=d time .
Then space will increase 1 time .(d=3)
then 2nd inner for loop will print the “* ” 2* (row -r)-1 time , that means 5 time .
So, star will be printed 5 time.
* * * * *
Iteration-3
r=3 (passes the first for loop condition) as it will iterate up to r<=row-1 times.
First inner for loop will print the space 3 time because loop will iterate up to c<=d time .
Then space will increase 1 time .(d=4)
then 2nd inner for loop will print the “* ” 2* (row -r)-1 time , that means 3 time .
So, star will be printed 3 time.
* * *
Iteration-4
r=4 (passes the first for loop condition) as it will iterate up to r<=row-1 times.
First inner for loop will print the space 4 time because loop will iterate up to c<=d time .
Then space will increase 1 time .(d=5)
then 2nd inner for loop will print the “* ” 2* (row -r)-1 time , that means 1 time .
So, star will be printed 1 time.
*
Now when r=5 , second outer for loop condition will fail so no other loops will be executed.
At last, we will see a pattern like this as output.
* *** ***** ******* ********* ******* ***** *** *
C Code:
#include <stdio.h>
int main() {
int r, row, c ,d;
printf("Enter rows: ");
scanf("%d", &row);
d = row - 1;
for (r = 1; r <= row; r++)
{
for (c = 1; c <= d; c++)
printf(" ");
d--;
for (c = 1; c <= 2 * r - 1; c++)
printf("*");
printf("\n");
}
d = 1;
for (r = 1; r <= row - 1; r++)
{
for (c = 1; c <= d; c++)
printf(" ");
d++;
for (c = 1; c <= 2 * (row - r) - 1; c++)
printf("*");
printf("\n");
}
return 0;
}
Output: Enter rows : 5 * *** ***** ******* ********* ******* ***** *** *
C++ Code:
#include <iostream>
using namespace std;
int main()
{
int row, r , c ,d ;
cout << "Enter rows: ";
cin >> row;
d = row - 1;
for (r = 1; r <= row; r++)
{
for (c = 1; c <= d; c++)
cout << " ";
d--;
for (c = 1; c <= 2 * r - 1; c++)
cout << "*";
cout << "\n";
}
d = 1;
for (r = 1; r <= row - 1; r++)
{
for (c = 1; c <= d; c++)
cout << " ";
d++;
for (c = 1; c <= 2 * (row - r) - 1; c++)
cout << "*";
cout << "\n";
}
return 0;
}
Output: Enter rows : 5 * *** ***** ******* ********* ******* ***** *** *
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