Thread: Hi, I need help with this code.

So besides just dumping your code here and asking us to fix it for free, would you like to pretend to be courteous and try to tell us exactly what you mean when your brain spits out the phrase "It only succeeds sometimes"?

How about error messages.... output..... something to give us an idea where you messed up?

I suggest you to do the following:
- First, initialize the multidimensional array, and insert its values into a new array:

Code:

int newarr[NUM * 2];

- Sort the array (use bubble-sort or merge sort or w/e).
- Go over the array and make a loop that has a condition in it which says: (NOTE that you gotta check the condition of your loop so it doesn't reach newarr[NUM * 2]: for(...;i < (num * 2) - 1;...)

Code:

if(newarr[i] == newarr[i + 1]) return -1;

I'm returning -1 because a number is repeated twice (might be more but we don't care).
If it doesn't return -1 and returns something else it means that the multidimensional array doesn't have "repeating-numbers" (numbers that appear more than one time).

I think the problem is that you are setting the flag on every iteration. You should initialise the flag to 0, then only when you find a match, set it to 1 and break from the loop.

Thank you all...
It was all very helpful

Originally Posted by Livnat

Thank you all...
It was all very helpful
This is what I did eventually...

Code:

#include <stdio.h>
#define NUM 3
int main()
{
int i,j,flag=0,l ;
int mat[NUM][NUM] ;


printf("please enter &#37;d*%d matrix: \n" ,NUM ,NUM);
for (i=0 ; i<NUM ; i++)
    for(j=0 ; j<NUM ; j++)
            scanf("%d" ,&(mat[i][j]));

for (j=0 ; j<NUM && flag==0; j++)
     for(i=0; i<NUM && flag==0; i++)
              for(l=(i+1);l<NUM && flag==0 ;l++)
                                if (mat[i][j]==mat[l][j])
                                   flag=1;
                              

for (i=0 ; i<NUM && flag==0 ; i++)
       for(j=0 ; j<NUM && flag==0; j++)
               for(l=(j+1) ; l<NUM && flag==0; l++)
                           if(mat[i][j]==mat[i][l])
                               flag=1;
                         
if (flag==0)
   printf("YES\n");
else
if (flag==1)
     printf("NO\n");
system("pause");
return 0;
}

That's really not efficient, that works on O(n^3), you could have done it way better (look at my post)
BTW, why did you post these 6 loops, only 3 is needed (not including the initialization)

Code:

for(int i = 0; i < NUM && flag == 0; i++)
 for(int j = 0; j < NUM && flag == 0; j++)
   for(int k = 0; k < NUM && flag == 0; k++)
          if(mat[i][j] == mat[i][k])
                 flag = 1;

Are you trying to check for duplicates across all rows and columns, or just individually? For example:

Code:

please enter 3*3 matrix:
0
1
2
0
3
4
5
6
7
NO
Press any key to continue . . .

Code:

please enter 3*3 matrix:
0
1
2
3
0
4
5
6
7
YES
Press any key to continue . . .

Is this intended behavior? Or should both examples show duplicates?

Originally Posted by eXeCuTeR

That's really not efficient, that works on O(n^3), you could have done it way better (look at my post)

Indeed, though bubble sort was the wrong thing to suggest for improving the complexity.

Originally Posted by MacGyver

Are you trying to check for duplicates across all rows and columns, or just individually?

It seems quite clear that the check is for duplicates across each row and each column, as that is how I interpret "if in every row and column every number appears only once".

Yep, bubble sort is n^2 which is still faster than his code

It is not faster than his code in terms of complexity since the code needs to check for duplicates across each row and column, so a sort merely speeds up the check across a particular row or column. Of course, if the duplicates are to be checked across the entire grid then it would indeed be faster, but a naive check for uniqueness that checks each element against every other element would also take quadratic time (and this is currently the case).