copy constructor [array] C++ - c++

Question

So I'm doing a puzzle game and I came to a problem. My Board will get scrambled later on in the program. I want to make a copy of it before it gets scrambled, to use it in the win-condition for my game. My Idea is to compare the copied board to the scrambled board every time the user moves a tile to see if they succeded (won) or not. But I'm a bit unsure how to do the copy constructor for the board. Here is what I have done(doesn't work as it's supposed to do).
Board::Board(int userInput)
{
this->gameSize = userInput;
int zeroPos[2] = { 0, 0 };
SetTileNumbers();
}
void Board::SetTileNumbers()
{
const int sizeOfGame = gameSize; //Size given from user when the board was created.
int tileNumber = 0; //The value given to each Tile.Number.
int row, column;
boardOfTiles = new Tile* [sizeOfGame];
for (int i = 0; i < sizeOfGame; i++)
{
boardOfTiles[i] = new Tile [sizeOfGame]; //The doublepointer is given an additional dimension
}
for (row = 0; row < sizeOfGame; row++)
{
for (column = 0; column < sizeOfGame; column++)
{
boardOfTiles[row][column].number = tileNumber; //Loops that goes through the entirety to instantiate the board of tiles.
tileNumber++;
}
}
}
void Board::SetTileNumbers(const Board& copy)
{
const int sizeOfGame = copy.gameSize;
int row , column;
boardOfTiles = new Tile*[sizeOfGame];
for (int i = 0; i < sizeOfGame; i++)
{
boardOfTiles[i] = new Tile[sizeOfGame];
}
for (row = 0; row < sizeOfGame; row++)
{
for (column = 0; column < sizeOfGame; column++)
{
boardOfTiles[row][column].number = copy.boardOfTiles[row][column].number;
}
}
}

Answer 1

I hope this does not sound off-topic, but copy-constructor aside, I think you might have a different problem: the way you decide on a solution is very inefficient. You waste cpu time and you waste memory.
For every step to compare two NxN boards is about as ineffciciet as it gets. You can try a different approach: assign a number to each position on the board and you know that the solution is to have the numbers sorted in a particular order.
For a 3x3 board, 'THE' solution would look like:
1 2 3
4 5 6
7 8 9
Furthermore, a 2d array is stored in memory as a continuous array, you you can treat the solution as an array:
1 2 3 4 5 6 7 8 9
Any other arrangement will not represent an solution.
Checking weather an array is sorted is a much easier problem to solve. Worst case scenario you are still looking at O(n^2) to check if the array is sorted but without the memory overhead. Plus I'm sure you can figure out a more efficient algorithm to check if the array is sorted.
I hope this helps...