Thread: recursion to find a solution for K queen problem (variation)

I'm trying to write a function that can take a partial solution to a problem and give a solution. I am given a board of size MxN , where I should attempt to place K queens given a partial solution of K'

for example the following board:

Q X * *
* * Q *
* * X *
* * X *

the function should take this board, and the indexes 1 for row and 2 for column, the current number of queens on it - 2 and the final number of queens to be on it - 4. I'm only allowed to add queens on the rowth row from the colth column to the end of that row, and any following row for any column in this example - (1,3), (2,0), (2,1), (2,2), (2,3), (3,0), (3,1), (3,2), (3,3) the function should print true because it is possible .. for example the following:

Q X * *
* * Q *
* * X *
* Q X Q

however my function doesn't quite do that and I can't see why..

private static boolean kQueens(int[][] board, int k, int row, int col, int numOfQueens) {
 
        if (numOfQueens == k) return true;
 
        //trying to place a queen in the row at the col+1 ~ row length position.
        // if possible place it and try to place a new one in the next position.
        for (int i = 0; i < board.length; ++i) {
            for (int j = 0; j < board[i].length; j++) {
                //save the value of that position in case we'll need to change it back
                int tmp = board[i][j];
                if (addQueen(board, i, j)) {
                    board[i][j] = QUEEN;
                }
                if (kQueens(board, k, i, j, numOfQueens + 1)) {
 
                    return true;
                }
               //remove the queen and backtrack
                board[i][j]=tmp;
            }
        }
        return false;
    }

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