CareerCup

void printDiagonal(int a[][],int n)
{
for(int i=n-1,j=0;i>=0;i--,j++)
cout<<" "<<a[i][j];
}

#include<stdio.h>
#include<stdlib.h>

int main()
{
    int i, j, row, col, a[10][10];
    printf("\nEnter the number of rows and columns\n");
    scanf("%d %d", &row, &col);
    printf("\nEnter the elements row-wise\n");
    for(i = 0; i < row; i++)
          for(j = 0; j < col; j++)
                scanf("%d", &a[i][j]);
    for(i = 0; i < row; i++)
          {
               for(j = 0; j < col; j++)
                     printf("%d\t", a[i][j]);
               printf("\n");
          }
    printf("\nFollowing are the elements from left-bottom to right-top\n");
    for(i = row - 1, j = 0; i >= 0; i--, j++)
        {        
          printf("%d\t", a[i][j]);
        }
    system("pause");
    return 0;
}

I am very sure the question is not as simple as it seems. The interviewer wanted me to avoid loops.

template<class M>
void printDiagonal(const M& rowmajor,std::size_t n)
{
  for(std::size_t r=n,c=0;r--;++c)
    std::cout << rowmajor[r][c] << std::endl;
}

Below code is without using loops
int n=5;
printDiagonal(a, 0, n-1);

void printDiagonal(int a[][5], int i, int n)
{
    if(i > n)
        return;
    cout << a[n-i][i] << " ";
    i++;
    printDiagonal(a, i, n);
}

/*
matrix with order(nxn)
*/
for(int i=n-1; i>=0; ) {
for(int j=0; j<n; j++, i--) {
System.out.println(matrix[i][j]);
}
}

/*
matrix with order(nxn)
*/
for(int i=n-1; i>=0; ) { 
	for(int j=0; j<n; j++, i--) { 
		System.out.println(matrix[i][j]); 
	}

}

#include<iostream>
using namespace std;


int printDiagonal(int a[100][100], int n, int m)
{
    if(m!=0)
            printDiagonal(a, n, m-1);
    
            cout<<a[n-m][m]<<"\t";
            return 0;    
}

int main()
{
    int a[100][100];
    int n;
    
    cout<<"Enter the value of N in NxN matrix : ";
    cin>>n;
    
    for(int i=0; i<n; i++)
    {
            for(int j=0; j<n; j++)
            {
                    a[i][j] = i*n +j;
            }
    }
        
    for(int i=0; i<n; i++)
    {
            for(int j=0; j<n; j++)
            {
                    cout<<a[i][j]<<"  ";
            }cout<<"\n";
    }cout<<"\n\n";
                    //
    printDiagonal(a, n-1, n-1);
    
    system("pause");
    return 0;
}

void print_bl_tr_diag(int a[][], int n)
{
    for (int j = n - 1; j >= 0; j--)
        printf("%d\n", a[j][n - j - ]);
}

When the interviewer said "wo loops", I presume he meant wo nested loops. Anyway, below is the single loop solution, though we could also implement it with recursion, which as noted earlier, would be an overkill.

void print_bl_tr_diag(int a[][], int n)
{
    for (int j = n - 1; j >= 0; j--)
        printf("%d\n", a[j][n - j - 1]);
}

//Given an n*n matrix. Print the main diagonal elements starting from bottom-left to top-right.
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
void diag(int a[][10],int i,int j,int n)
{
if(i<0 || j>n-1)
{
getch();
exit(0);
}
printf("%d\t",a[i][j]);
diag(a,i-1,j+1,n);
}
int main()
{
int n,i,j,a[10][10];
printf("Enter the value of n(n*n)\n");
scanf("%d",&n);
printf("Enter the matrix\n");
for(i=0;i<n;i++)
for(j=0;j<n;j++)
scanf("%d",&a[i][j]);
diag(a,n-1,0,n);
return 0;
}

As the given matrix is a square matrix , so we can print the diagonal matrix as follows-

for(i=0;i<row;i++)
{
for(j=0;j<col;j++)
{
if(i==j)
printf("%d ",a[i][j]);
}
}

This would do.

Code:
for(int i=0;i<n;i++)
sum = sum + a[i][n-1-i];

Complexity:
O(n)

I want output diagonal wise
0
13
245
8

I want output diagonal wise
0
13
245
8

void printDiagonal(int (*arr)[4], int row){
int j = 0;
int i ;
for(i =( row-1) ;i >=0 ; i--){
printf("%d",arr[i][j]);
j++;
}
}

void printDiagonal(int a[][],int n)
{
for(int i=n-1;i>=0;i--)
cout<<" "<<a[i][i];
}