Question

in java

Given n >= 0, create an array with length n*n with following the pattern, shown for n=3 { 0, 1, 2, 1, 0, 3, 2, 3, 0} (Spaces have been added to show the grouping of elements of the array more "obvious.") Note: The values in the array are the sum of the "row" and "column" values, except along the diagonal, where it is zero.

diagonal(3) ? [0, 1, 2, 1, 0, 3, 2, 3, 0]
diagonal(2) ? [0, 1, 1, 0]
diagonal(4) ? [0, 1, 2, 3, 1, 0, 3, 4, 2, 3, 0, 5, 3, 4, 5, 0]

Answer 1

import java.util.Scanner;

public class Sample{
  
public static void main(String args[]){
  
int n, i, j;
Scanner obj = new Scanner(System.in);
n = obj.nextInt();
int a[][] = new int[n][n];
for (i = 0; i < n; i++){
for(j = 0; j < n; j++){
if(i == j){
a[i][j] = 0;
}else{
a[i][j] = i + j;
}
}
}
int b[] = new int[n*n];

int k=0;
for (i = 0; i < n; i++){
for(j = 0; j < n; j++){
b[k++] = a[i][j];

}
}
}
}