Traversing through a 2D array?

[hopesfall]

I have a 2D array of ints and I want to check that within this array, only one value of each number 1 through N exist (for example, N being 9, that only 1-9 exist once and only once). I was thinking of approaching this by using some sort of loop and testing each element value against the loop counter. Anyone got some suggestions?

[ResidentBiscuit]

Not sure what you're going after here. The whole array can only have once instance of a specific number? I'll recommend using a 1D array, you can use them the same way as a 2D array, and it's much easier to work with

[hopesfall]

well I have something like this:

[1][2][3]
[4][5][6]
[7][8][9]

and I want to ensure that there's only one occurrence of each value 1-9.

[ResidentBiscuit]

I'm assuming you want it sorted it too? I'd do the sorting first, and then replace the duplicates with whatever number it should be.

[nooblet]

simple, yet i am unsure if it is the most efficient. use 2 pointers.

one pointer looks at a certain location

array[0][0]

the second pointer looks at a different location then compares

array[1][0]

you can use a loop as u said to increment either location.

array[loopVar1] [loopVar2]

[Lynx876]

Off the top of my head, I came up with the following code. It needs some tweaking, so that the loop skips a future location, if it has had a match before.

There's probably a better/cleaner way to do it, but here it is:

#include <iostream>		// cout


int main()
{
	int ary[ 3 ][ 3 ] = { { 1, 2, 7 },
			      { 3, 5, 9 },
			      { 4, 6, 5 } };

	// print the array
	for( int i = 0; i < 3; ++i )
	{
		for( int j = 0; j < 3; ++j )
			std::cout << ary[ i ][ j ] << ' ';

		std::cout << '\n';
	}

	// location_i & location_j will stay the same until the iner loop
	// has traversed through the whole array once.
	for( int location_i = 0; location_i < 3; ++location_i )
	{
		for( int location_j = 0; location_j < 3; ++location_j )
		{

			// loop which will check against the location element
			for( int check_i = 0; check_i < 3; ++check_i )
			{
				for( int check_j = 0; check_j < 3; ++check_j )
				{
					// if the location and check are at the same element
					// the number will be the same, so skip it( continue; )
					if( ( location_i == check_i ) && ( location_j == check_j ) )
						continue;

					// if the check matches the location
					if( ary[ location_i ][ location_j ] == ary[ check_i ][ check_j ] )
					{
						std::cout << "Match at: " << location_i << ' ' << location_j;
						std::cout << "\nNumber was: " << ary[ location_i ][ location_j ] << '\n';
					}
				}
			}
		}
	}
    
    return 0;
}