2021/quals/xs_and_os_sol.md

For any given row or column, if it has no **O**s and has \(k\) empty spaces, then you can win by placing \(k\) **X**s in it (filling in the empty spaces). This covers all possible optimal ways in which you might win, as you should never place additional **X**s which don't directly contribute to winning in a single row or column. The first part of the answer is therefore the minimum value of \(k\) across all such rows and columns, and it's impossible to win if there are none.

If it's possible to win, then the second part of the answer is equal to the number of distinct sets of empty cells (in which **X**s will be placed) in those minimum-\(k\) rows/columns. If the minimum \(k\) is at least 2, this is simply equal to the number of such rows/columns, as no two of them can correspond to the same set of empty cells. On the other hand, if the minimum \(k\) is 1, then the second part of the answer may be smaller than that row/column count, as filling a row and filling a column might both involve placing an **X** in the same cell as one another. Therefore, for this case, we need to compute the set of distinct single empty cells found in all rows/columns which have \(k=1\), with the second part of the answer then being the size of that set.

[See David Harmeyer's solution video here](https://youtu.be/tm77EHakpxE).