Adobe Interview Question Software Engineer in Tests


Country: India
Interview Type: In-Person


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1
of 3 vote

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

- Anonymous on February 14, 2012 | Flag Reply
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1
of 1 vote

#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;
}

- Prathamesh on February 15, 2012 | Flag Reply
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0
of 0 vote

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

- Anonymous on February 14, 2012 | Flag Reply
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0
of 0 votes

To avoid loops? So perhaps he/she was looking for a recursive solution (which is a hidden loop in anyway!) On the other hand to write recursion for this is kind of problems is overkill.

- Selmeczy, P├ęter on February 15, 2012 | Flag
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0
of 0 vote

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;
}

- comonad on February 15, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

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);
}

- Anonymous on February 15, 2012 | Flag Reply
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0
of 0 vote

/*
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]);
}
}

- aragon on February 15, 2012 | Flag Reply
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0
of 0 vote

/*
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]); 
	}

}

- aragon on February 15, 2012 | Flag Reply
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0
of 0 vote

#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;
}

- Random on February 15, 2012 | Flag Reply
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0
of 0 vote

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

- Ashot Madatyan on February 19, 2012 | Flag Reply
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0
of 0 vote

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]);
}

- ashot madatyan on February 19, 2012 | Flag Reply
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0
of 0 vote

//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;
}

- narayan kunal on March 05, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

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]);
}
}

- Rudra on March 13, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

This would do.

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

Complexity:
O(n)

- Pavan Dittakavi on May 03, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I want output diagonal wise
0
13
245
8

- Anonymous on September 06, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I want output diagonal wise
0
13
245
8

- Anonymous on September 06, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

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++;
}
}

- Durgesh on February 14, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
-2
of 2 vote

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

- Ali_BABA on February 14, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

this is wrong... your program prints other diagonal, not the one given in question

- Anonymous on February 14, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

try this..

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

- Random on February 15, 2012 | Flag


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