Hollow diamond pattern

#1

import java.util.;
public class Main {
public static void main(String args[]) {
// Your Code Here
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int presentrow,nst=(n/2),cst,nsp=1,csp,i;
for(presentrow=1;presentrow<=n;presentrow++)
{
if(presentrow==1||presentrow==n){
for(i=1;i<=n;i++)
System.out.println();
}
else{
for(cst=1;cst<=nst;cst++){
System.out.print("");
}
for(csp=1;csp<=nsp;csp++){
System.out.print(" “);
}
for(cst=1;cst<=nst;cst++){
System.out.print(”*");
}
if(presentrow<=(n/2)){
nst–;
nsp=nsp+2;
}
else{
nst++;
nsp=nsp-2;
System.out.println();
}
}
}
}
}

where i went wrong plz tell me

#2

@karthik1989photos,
nst = (n/2 + 1), cst = 1, rows = 1, nsp = -1,total rows = n;

  • The given pattern is seen as first stars then spaces and then stars .So for each row do the work of stars first , then that of spaces and then that of stars. And then change the variables accordingly for next iterations.
  • But this work will not help u to pass the test cases as there will be 1 extra star in the first line and last line. For e.g.,
       * * * * * *  //Extra star
       * *   * *
       *       *
       * *   * *
       * * * * * * // Extra Star
  • To handle this condition all you need to do is run your second look 1 time less only when control is in first and last line.
    if(rows == 1 || rows == n) cst = 2;

https://ide.codingblocks.com/s/262426 you can refer to this as well

#3

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