Interview Question: Print both diagonals of a n*n matrix

Given a matrix of order N*N, write code to print both the diagonals of that matrix. For example: the matrix and its 2 diagonals are given below:

In the above diagram, I have colored the elements in first diagonal as red and elements in 2nd diagonal as green.

All the elements of first diagonals are at (i,i) positions, do it is easy to print the first diagonal

for (int i=0; i<n; i++)
   cout << arr[i][i];

Note that some of my students have written the code like this:

for(int i=0; i<n; i++)
    for(int j=0; j<n; j++)
        if(i==j)
            cout << arr[i][j];

But, this is unnecessarily traversing all the elements of the matrix (even elements which are not diagonal). Hence the complexity of this code is O(n²). Where as the code above (with one loop) takes O(n) time.

The elements in the second diagonal are at following positions

(3,0)  (2,1)  (1,0)   (0,3)

or, if I generalize:

(n-1, 0) (n-2, 1) ..... (0, n-1)

Hence, the code will be:

for(int i=0; i<n; i++)
    cout << arr[n-1-i][i];

#define N 4
void printdiagonal(int arr[N][N])
{
   cout<< "First Diagonal :";
   for(int i=0;i<N;i++)
      cout<< " "<<arr[i][i];
   
   cout<<"\n Second Diagonal :";
   for(int i=0;i<N;i++)
      cout<< " "<<arr[N-i-1][i];
}

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