How can i get the pattern as like this?????

Question

https://code.sololearn.com/cMrh2IQlsi2D/?ref=app

output--
*111*
**111
111**
*111*
Helga
Arun Tomar
Dan Walker
Dev
Hans Larry
‎ ‏‏‎Sreejith 
HAWKEYE
Fata1 Err0r
Gaurav Agrawal
[A^dr3w]

Answer 1

ok, the problem is this could be a sub-pattern of many patterns.
1. the no. ones and * could be constant
2. no. of 1's increase with varying input
3. no. of * increases with varying input

now i'll be considering no. of 1's as constant.

Harsh Agrawal
here's the code...
Is this what you wanted?
https://code.sololearn.com/cVsY8vtzK5Dz/?ref=app

Answer 2

Probably the most automated way:

public class Program
{
public static void main(String[] args) {
printRow(0,4);
printRow(0,1);
printRow(3,4);
printRow(0,4);
}

static void printRow(int x, int y){
for(int j=0;j<5;j++){
if(j==x || j==y)
System.out.print("*");
else
System.out.print("1");
}
System.out.println();
}
}

Answer 3

Harsh Agrawal, what do you mean? You can't make it with two nested loops as in your example

Answer 4

we can hardcode it in main, but that’s the worst solution. In this case a method for printing a row is the most efficient solution

Answer 5

Harsh Agrawal i need some more examples for this pattern, for different input

Answer 6

harsh do you want a general code for that pattern ?

Answer 7

Harsh Agrawal
first take a single line and divide it into multiple parts,

-> each line has 3 one's and 2 *, only order changes
-> stars on the right decreases whereas the star on left increases, once they reach the limit, pattern restarts
-> 111 is constant throughout the series.
-> input can be no. of lines,
.. no. of 1's = 3 [constant]
.. no. of stars in a line = (input/2)+1 (divided among left and right side)
.. no. of elements in a line = input - 1

Answer 8

Hi

public class Program
{
public static void main(String[] args)
{
char arr[]={'*','1','1','1','*','*','*','1','1','1'};
for(int a = 0; a <=9; a++)
{
System.out.print(arr[a]);
if(a == 4 || a == 9)
{
System.out.println("");
}
}
for(int a = 9; a >=0; a--)
{
System.out.print(arr[a]);
if(a == 5)
{
System.out.println("");
}
}
}
}

Answer 9

or
public class Program
{
public static void main(String[] args)
{
int count=0;
String arr[]={"*111*","**111","111**"};
for(int a = 0; a <=3; a++)
{
System.out.println(arr[count]);
if (count == 2) count = 0;
count++;
}
}
}

Answer 10

thnxzzz All of u

Answer 11

MarJan
can u make it simple??????

Answer 12

MarJan i mean can we cant make this without calling it four time?????

Answer 13

okkk thnxzzz MarJan

Answer 14

‎ ‏‏‎Sreejith  what is dat?????

can u give me???

Answer 15

‎ ‏‏‎Sreejith  ya i want or u wll also give suggestion about this!!!

Answer 16

‎ ‏‏‎Sreejith  thnxzzzz but one thing i want to know that how u solve the pattern???? Is der any sequence to solve pattern easily n quickly????

Answer 17

ohh!!!

Answer 18

harsh agarwal can you solve this pattern in simple way so plzz this pattern program make and make a public

Answer 19

Raj gupta what do u mean?????
can u elaborate what u want????

Answer 20

i means make a code for this program and makes privacy public