Creates A Number Pattern Using Array - c++

Question

So the numbers pattern going like this :
Input : 2
Output :
0 0 0 1 0 0 0
0 0 2 0 12 0 0
0 3 0 0 0 11 0
4 0 0 0 0 0 10
0 5 0 0 0 9 0
0 0 6 0 8 0 0
0 0 0 7 0 0 0
I have solved it using usual method, but when I tried with array, the output is really messed up. Any suggestion how to make this number pattern with array?
This is my code that creates this number pattern :
int input, n, mid, i, j;
cin >> input;
n = (2*input)+3;
mid = (n+1)/2;
for(i = 1; i <= n; i++) {
for (j = 1; j <= mid; j++) {
if (i <= mid && j == mid-i+1) cout << i << " ";
else if (i > mid && j == mid-n+i) cout << i << " ";
else cout << "0 ";
}
for (j = mid+1; j <= n; j++) {
if (i >= mid && j == n+mid-i) cout << (2*n-i) << " ";
else if (i < mid && j == mid+i-1) cout << (2*n-i) << " ";
else cout << "0 ";
}
cout << endl;
}
Thanks in advance.

Answer 1

I think this code(working) might help you.
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main() {
int input, n, mid, i, j;
cin >> input;
n = (2*input)+3;
mid = (n+1)/2;
int arr[n][n];
memset(arr, 0, sizeof(arr));
for(i = 1; i <= n; i++) {
for (j = 1; j <= mid; j++) {
if (i <= mid && j == mid-i+1) arr[i - 1][j - 1] = i;
else if (i > mid && j == mid-n+i) arr[i - 1][j - 1] = i;
}
for (j = mid+1; j <= n; j++) {
if (i >= mid && j == n+mid-i) arr[i - 1][j - 1] = 2 * n - i;
else if (i < mid && j == mid+i-1) arr[i - 1][j - 1] = 2 * n - i;
}
}
}

Answer 2

Supplementing baymaxx, if you desire dynamic memory allocation:
#include <iostream>
#include <stdio.h>
#include <string.h>
int input, n, mid, i, j;
cin >> input;
n = (2*input)+3;
mid = (n+1)/2;
// create array of the specified size.
int** arr = (int**) malloc (n * sizeof(int**));
for (int p = 0; p < n; ++p) {
arr[p] = (int*) malloc(n * sizeof(int));
for (int q = 0; q < n; ++q) {
arr[p][q] = 0;
}
}
// fill array (algorithm attr. baymaxx)
for(i = 1; i <= n; i++) {
for (j = 1; j <= mid; j++) {
if (i <= mid && j == mid-i+1) {
arr[i - 1][j - 1] = i;
}
else if (i > mid && j == mid-n+i) {
arr[i - 1][j - 1] = i;
}
}
for (j = mid+1; j <= n; j++) {
if (i >= mid && j == n+mid-i) {
arr[i - 1][j - 1] = 2 * n - i;
}
else if (i < mid && j == mid+i-1) {
arr[i - 1][j - 1] = 2 * n - i;
}
}
}
// print array
for (int p = 0; p < n; ++p) {
for (int q = 0; q < n; ++q) {
int x = arr[p][q];
std::cout << x << " ";
}
std::cout << endl;
}
// delete array
for (int i = 1; i < n; ++i) {
free(arr[i]);
}
free(arr);